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Physics Work Problems for High Schools

In this tutorial, we want to practice some problems on work in physics. All these questions are easy and helpful for your high school homework. 

Work Problems: Constant Force

Problem (1): A constant force of 1200 N is required to push a car along a straight line. A person displaces the car by 45 m. How much work is done by the person?

Solution : If a constant force $F$ acts on an object over a distance of $d$, and $F$ is parallel to $d$, then the work done by force $F$ is the product of the force times distance. 

Physics work problems

In this case, a force of $1200\,{\rm N}$ displaces the car $45\,{\rm m}$. The pushing force is parallel to the displacement. So, the work done by the person is equal to \[W=Fd=1200\times 45=54000\,{\rm J}\] The SI unit of work is the joule, ${\rm J}$. 

Problem (2): You lift a book of mass $2\,{\rm kg}$ at constant speed straight upward a distance of $2\,{\rm m}$. How much work is done during this lifting by you?

Solution : The force you apply to lift the book must be balanced with the book's weight. So, the exerted force on the book is \[F=mg=2\times 10=20\quad{\rm N}\] The book is lifted 2 meters vertically. The force and displacement are both parallel to each other, so the work done by the person is the product of them. \[W=Fd=20\times 2=40\quad {\rm J}\]

Problem (3): A force of $F=20\,{\rm N}$ at an angle of $37^\circ$ is applied to a 3-kg object initially at rest. The object has displaced a distance of $25\,{\rm m}$ over a frictionless horizontal table. Determine the work done by  (a) The applied force (b) The normal force exerted by the table (c) The force of gravity

Angled force in physics work problem

Solution : In this problem, the force makes an angle with the displacement. In such cases, we should use the work formula $W=Fd\cos\theta$ where $\theta$ is the angle between force $F$ and displacement. To this object, an external force $F$, normal force $F_N$, and a gravity force $w=mg$ are applied.

(a) Using vector decompositions, the component of the force parallel to the displacement is found to be $F_{\parallel}=F\cos \theta$. Thus, the product of this component parallel to the displacement times the magnitude of displacement gives us the work done by external force $F$ as below \begin{align*} W_F&=\underbrace{F\cos\theta}_{F_{\parallel}}d\\\\ &=(20\times \cos 37^\circ)(25)\\\\&=400\quad {\rm J}\end{align*}  (b) Now, we want to find the work done by the normal force. But let's define what the normal force is.

In physics, ''normal'' means perpendicular. When an object is in contact with a surface, a contact force is exerted on the object. The component of the contact force perpendicular to the surface is called the normal force. 

Thus, by definition, the normal force is always perpendicular to the displacement. So, the angle between $F_N$ and displacement $d$ is $90^\circ$. Hence, the work done by the normal force is determined to be \[W_N=F_N d\cos\theta=(30)(25)\cos 90^\circ=0\] (c) The weight of the object is the same as the force of gravity. This force applies to the object vertically downward, and the displacement of the object is horizontal. So, again, the angle between these two vectors is $\theta=90^\circ$. Hence, the work done by the force of gravity is zero. 

Problem (4): A person pulls a crate using a force of $56\,{\rm N}$ which makes an angle of $25^\circ$ with the horizontal. The floor is frictionless. How much work does he do in pulling the crate over a horizontal distance of $200\,{\rm m}$?

Pulling force in physics work problems

Solution : The component of the external force parallel to the displacement does work on an object over a distance of $d$. In all work problems in physics, this force component parallel to the displacement is found by the formula $F_{\parallel}=F\cos \theta$. Thus, the work done by this force is computed as below \begin{align*} W&=F_{\parallel}d\\&=(F\cos\theta)d\\&=(56\cos 25^\circ)(200) \\&=10080\quad {\rm J}\end{align*} We could use the work formula from the beginning $W=Fd\cos\theta$ where $\theta$ is the angle between $F$ and $d$. 

Problem (5): A worker pushes a cart with a force of $45\,{\rm N}$ directed at an angle of $32^\circ$ below the horizontal. The cart moves at a constant speed.  (a) Find the work done by the worker as the cart moves a straight distance of $50\,{\rm m}$.  (b) What is the net work done on the cart?

Solution :(a) All information to find the work done by the worker is given, so we have \begin{align*} W&=Fd\cos\theta\\&=(45)(50) \cos 32^\circ\\&=1912.5\quad {\rm J}\end{align*}  (b) ''net'' means "total". In all work problems in physics, there are two equivalent methods to find the net work. Identify all forces that are applied to the cart, find their resultant force, and then compute the work done by this net force over a specific distance. 

Or compute all works done on the object across a distance individually, then sum them algebraically. 

Usually, the second method is easier. We take this approach here. 

The cart moves in a straight horizontal path. All forces apply on it are, the worker force $F$, the normal force $F_N$, and the force of gravity or its weight $F_g=mg$. The work done by normal and gravity forces in a horizontal displacement is always zero since the angle between these forces and the displacement is $90^\circ$. So, $W_N=W_g=0$. Hence, the net (total) work done on the object is \[W_{total}=W_N+W_g+W_F=1912.5\,{\rm J}\]

Problem (6): A $1200-{\rm kg}$ box is at rest on a rough floor. How much work is required to move it $5\,{\rm m}$ at a constant speed (a) along the floor against a $230\,{\rm N}$ friction force, (b) vertically?

Solution : In this problem, we want to displace a box $5\,{\rm m}$ horizontally and vertically. In the horizontal direction, there is also kinetic friction. 

(a) At constant speed , means there is no acceleration in the course of displacement, so according to Newton's second law $\Sigma F=ma$, the net force on the box must be zero. To meet this condition, the external force $F_p$ applied by a person must cancel out the friction force $f_k$. So, \[F_p=f_k=230\quad {\rm N}\] The force $F_p$ and displacement are both parallel, so their product get the work done by $F_p$ \[W_p=F_p d=230\times 5=1150\quad {\rm J}\] (b) In the vertical path, two forces act on the box. One is the external lifting force, and the other is the force of gravity. Since the box is moving at constant speed vertically, there is no acceleration, and thus this lifting external force $F_p$ must be balanced with the weight of the box. \[F_p=F_g=mg=(1200)(10)=12000\,{\rm J}\] Assume the box is moved vertically upward. In this case, the lifting force and displacement are parallel, so the angle between them is zero $\theta=0$, and the work done by this force is \[W_p=F_p d\cos 0=12000\times 5=60\,{\rm kJ}\] On the other side, the weight force, or force of gravity $F_g=mg$ is always downward, so the angle between the box's weight and upward displacement is $180^\circ$. So, the work done by the weight of the box is \begin{align*}W_g&=F_g d\cos 180^\circ \\\\ &=(1200)(10)(5)(-1) \\\\ &=-60\,{\rm kJ}\end{align*} In such cases where the angle between $F$ and $d$ is $180^\circ$, they are called antiparallel. 

Problem (7): A 40-kg crate is pushed using a force of 150 N at a distance of $6\,{\rm m}$ on a rough surface. The crate moves at a constant speed. Find (a) the work done by the external force on the crate. (b) The coefficient of kinetic friction between the crate and the floor? 

Solution : (a) the crate is moved horizontally through a distance of $6\,{\rm m}$ by a force parallel to its displacement. So, the work done by this external force is \[W_p=F_p d \cos\theta=(150)(6)\cos 0=900\,{\rm J}\] where subscript $p$ denotes the person or any external agent.

(b) According to the definition of the kinetic friction force formula, $f_k=\mu_k F_N$, to find the coefficient of kinetic friction $\mu_k$, we must have both the friction force and normal force $F_N$.

In the question, we are told that the crate moves at a constant speed, so there is no acceleration, and thus, the net force applied to it must be zero. 

When the friction force, which opposes the motion, is equal to the external force $F_p$, then this condition is satisfied. So, \[f_k=F_p=150\,{\rm N}\] On the other side, the crate is not lifted off the floor, so there is no motion vertically. 

Balancing all forces applied vertically, the weight force and the normal force $F_N$, we can find the normal force $F_N$ as below \begin{gather*} F_N-F_g=0\\ F_N=F_g\\ \Rightarrow F_N=mg=40\times 10=400\quad {\rm N}\end{gather*} Therefore, the coefficient of kinetic friction is found to be \[f_k=\frac{f_k}{F_N}=\frac{150}{400}=0.375\] 

Practice these questions to understand friction force Problems on the coefficient of friction

Problem (8): A 18-kg packing box is pulled at constant speed by a rope inclined at $20^\circ$. The box moves a distance of 20 m over a rough horizontal surface. Assume the coefficient of kinetic friction between the box and the surface to be $0.5$.  (a) Find the tension in the rope? (b) How much work is done by the rope on the box?

Solution : The aim of this problem is to find the work done by the tension in the rope. The magnitude of the tension in the rope is not given. So, we must first find it. 

(a) We are told the box moves at a constant speed, so, as previously mentioned, the net force on the box must be zero to produce no acceleration. But what forces are acting horizontally on the box? The horizontal component of tension in the rope, $T_{\parallel}=T\cos\theta$, and the kinetic friction force $f_k$ in the opposite direction of motion are the forces acting on the box horizontally. 

If these two forces are equal in magnitude but opposite in direction, then their resultant (net) becomes zero, and consequently, the box will move at a constant speed. \begin{align*} f_k&=T_{\parallel}\\\mu_k F_N&=T\cos\theta\quad (I) \end{align*} The forces in the vertical direction must also cancel each other since there is no motion vertically. As you can see in the figure, we have \[F_N=T\sin\theta+F_g\] Substituting this into the relation (I), rearranging and solving for $T$, yields \begin{gather*} \mu_k (T\sin\theta+mg)=T\cos\theta \\\\ \Rightarrow T=\frac{\mu_k mg}{\cos\theta-\mu_k\sin\theta}\end{gather*} Substituting the numerical values into the above expression, we find the tension in the rope. \[T=\frac{(0.5)(18)(10)}{\cos 20^\circ-(0.5) \sin20^\circ}=117\quad {\rm N}\]  (b) The only force that causes the box to move some distance is the horizontal component of the tension in the rope, $T_{\parallel}=T\cos\theta$. So, the work done by the tension in the rope is \begin{align*} W&=T_{\parallel}d\\ &=(117)( \cos 20^\circ)(20) \\&=2199\quad {\rm J}\end{align*} 

Problem (9): A table of mass 40 kg is accelerated from rest at a constant rate of $2\,{\rm m/s^2}$ for $4\,{\rm s}$ by a constant force. What is the net work done on the table?

Solution : This is a combination of a  kinematics problem and a physics work problem. Here, first, we must find the distance over which the box is displaced. The given information is: initial speed $v_0=0$, acceleration $2\,{\rm m/s^2}$, time taken $t=4\,{\rm s}$. Using this data, we can find the total displacement by applying the kinematics equation $\Delta x=\frac 12 at^2+v_0t$, \begin{align*} \Delta x&=\frac 12 at^2+v_0t\\\\&=\frac 12 (2)(4)^2+0(4)\\\\&=16\quad {\rm m}\end{align*} So, this constant force causes the table to move a distance of 16 meters across the surface. To find the work done, we need a force, as well. The force is mass times acceleration, $F=ma$, so we have \[F=ma=40\times 2=80\,{\rm N}\] Now that we have both the force and displacement, the net work done on the table is the product of force along the displacement times the magnitude of displacement \[W=80\times 16=128\quad {\rm J}\] 

Work problems in a uniform circular motion 

Problem (10): A 5-kg object is held at the end of a string and undergoes uniform circular motion around a circle of radius $5\,{\rm m}$. If the tangential speed of the object around the circle is $15\,{\rm m/s}$, how much work was done on the object by the centripetal force?

Solution : Here, an object moves around a circle, so we encounter a uniform circular motion problem .

In such motions around a curve or circle, that force in the radial direction exerting on the object is called the centripetal force. 

On the other hand, a movement around a circle is tangent to the path at any instant of time. Thus, we conclude that in any uniform circular motion, a force is applied to the whirling object that is perpendicular to its motion at any moment of time.

So, the angle between the centripetal force and displacement at any instant is always zero, $\theta=0$. Using the work formula $W=Fd\cos\theta$, we find that the work done by the centripetal force is always zero. 

This is another example of zero work in physics.  

Work problems on an incline

Problem (11): A $5-{\rm kg}$ box, initially at rest, slides $2.5\,{\rm m}$ down a ramp of angle $30^\circ$. The coefficient of friction between the box and the incline is $\mu_k=0.435$. Determine (a) the work done by the gravity force, (b) the work done by the frictional force, and (c) the work done by the normal force exerted by the surface.

Solution : This part is related to problems on inclined plane surfaces . The forces acting on a box on an inclined plane are shown in the figure. As you can see, the forces along the direction of motion are the parallel component of the weight $W_{\parallel}$, and the friction force $f_k$. 

(a) In the figure, you realize that the angle between the object's weight (the same as the force of gravity) and downward displacement $d$ is zero, $\theta=30^\circ$.

incline in physics work problems

So, the work done by the force of gravity on the box is found using work formula as below \begin{align*} W&=Fd\cos\theta\\&=(mg)(d) \cos 30^\circ\\&=(5\times 10)(2.5) \cos 30^\circ\\&=108.25\quad {\rm J}\end{align*}  (b) To find the work done by friction, we need to know its magnitude. From the kinetic friction force formula, $f_k=\mu_k F_N$, we must determine, first, the normal force acting on the box. 

There is no motion in the direction perpendicular to the incline, so the resultant of forces acting in this direction must be zero. Equating the same direction forces, we will have \[F_N=mg\sin\alpha=(5)(10) \sin 30^\circ=25\,{\rm N}\] Substituting this into the above equation for kinetic friction, we can find its magnitude as \[f_k=\mu_k F_N=(0.435)(25)=10.875\,{\rm N}\] The friction force and the displacement of the box down the ramp are parallel, i.e., $\theta=0$. The work done by friction is \begin{align*} W_f&=f_kd\cos\theta \\\\ &=(10.875)(2.5) \cos 0\\\\ &=27.1875\,{\rm J}\end{align*}  (c) By definition, the normal force is the same contact force that is applied to the object from the surface perpendicularly. On the other hand, the object moves along the incline, so its displacement is perpendicular to the normal force, $\theta=90^\circ$. Hence, the work done by the normal force is zero. \[W_N=F_N d\cos\theta=F_N d\cos 90^\circ=0\]

Problem (12): We want to push a $950-{\rm kg}$ heavy object 650 m up along a $7^\circ$ incline at a constant speed. How much work do we do over this distance? Ignore friction.

Solution : When it comes to constant speed in all work problems in physics, you must remember that all forces in the same direction must be equal to the opposing forces. This condition ensures that there is no acceleration in the motion. 

In this case, all forces acting on the object are: the pushing force along the incline upward $F_p$, and the parallel component of the force of gravity (weight) along the incline downward, $W_{\parallel}=mg\sin\alpha$. Thus, we can find the pushing force as \begin{align*}F_p&=mg\sin 7^\circ\\\\ &=(950)(10) \sin 7^\circ \\\\& =1159\quad {\rm N}\end{align*} In the question, we are told that the object is moving up the incline, so the angle between its displacement and upward pushing force is zero, $\theta=0$. Hence, the work done by the person to push the object along the incline upward is \begin{align*} W_p&=F_p d\cos\theta\\&=1159\times 650 \cos 0\\&=753350\quad{\rm J}\end{align*} 

Problem (13): Consider an electron moving at a constant speed of $1.1\times 10^6\,\rm m/s$ in a straight line. How much energy is required to stop this electron? (Take the electron's mass, $m_e=9.11\times 10^{-31}\,\rm kg$.

Solution : In this problem on work, we cannot use the work formula directly, since none of the work variables, i.e., $F$, $d$, $\theta$, are given except the velocity. In these cases, we have a problem on the work-energy theorem . 

According to this rule, the net work done over a distance by a constant force on an object of mass $m$ equals the change in its kinetic energy \[W_{net}=\underbrace{\frac 12 mv_f^2-\frac 12 mv_i^2}_{\Delta K}\] Substituting the numerical values given in this problem, we get the required work to stop this fast-moving electron. \begin{align*}W_{net}&=\frac 12 m(v_f^2-v_i^2) \\\\ &=\frac 12 (9.11\times 10^{-31}) \left(0^2-(1.1\times 10^6)^2 \right) \\\\ &=-5.5\times 10^{-19}\,\rm J\end{align*} where we set the final velocity $v_f=0$ since the electron is to stop.

Problem (14): How much power is needed to lift a $25-\rm kg$ weight $1\,\rm m$ in $1\,\rm s$? 

Solution : The power in physics is defined as the ratio of work done on an object to the time taken $P=\frac{W}{t}$. The SI unit of power is the watt ($W$). 

In this problem, first, we must find the amount of work done in lifting the object as much as $1,\rm m$ vertically. The only force involved in this situation is the downward weight force. Thus, \[W=(mg)h=(25\times 10)(1)=250\,{\rm J}\] This amount of work has been done in a time interval of $1\,\rm s$. Hence, the power is calculated as below \[P=\frac{W}{t}=\frac{250}{1}=250\,\rm W\]

Problem (15): A particle having charge $-3.6\,\rm nC$ is released from rest in a uniform electric field $E$ moves a distance of $5\,\rm cm$ through it. The electric potential difference between those two points is $\Delta V=+400\,\rm V$. What work was done by the electric force on the particle? 

Solution : The work done by the electric force on a charged particle is calculated by $W_E=qEd$, where $E$ is the magnitude of the electric field and $d$ is the amount of distance traveled through $E$. But in this case, the electric field strength is not given, and we cannot use this formula. 

We can see this as a problem on electric potential . Recall that the work done by the electric force on a charge to move it between two points with different potentials is given by $W=-q\Delta V$. Substituting the given numerical values into this, we will have \begin{align*} W&=-q\Delta V \\&=-(-3.6)(+400) \\&=\boxed{1440\,\rm J} \end{align*} 

Here, we learned how to calculate the work done by a constant force in physics by solving a couple of example problems. 

Overall, the work done by a constant force is the product of the horizontal component of the force times the displacement between the initial and final points. 

In addition, power, a related quantity to work in physics, is also defined as the rate at which work is done.  

Author : Dr. Ali Nemati Date Published : 9/20/2021

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Course: physics library   >   unit 5.

  • Introduction to work and energy
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  • Work and the work-energy principle
  • Work as the transfer of energy
  • Work example problems
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High School Physics : Calculating Work

Study concepts, example questions & explanations for high school physics, all high school physics resources, example questions, example question #1 : calculating work.

work done physics solved problems

In this case, there is only one force acting upon the object: the force due to gravity. Plug in our given information for the distance to solve for the work done by gravity.

work done physics solved problems

Remember, since the object will be moving downward, the distance should be negative.

work done physics solved problems

The work done is positive because the distance and the force act in the same direction.

Example Question #2 : Calculating Work

work done physics solved problems

Work is a force times a distance:

work done physics solved problems

We know the distance that the book needs to travel, but we need to sovle for the lifting force required to move it.

There are two forces acting upon the book: the lifting force and gravity. Since the book is moving with a constant velocity, that means the net force will be zero. Mathemetically, that would look like this:

work done physics solved problems

We can expand the right side of the equation using Newton's second law:

work done physics solved problems

Use the given mass and value of gravity to solve for the lifting force.

work done physics solved problems

Now that we have the force and the distance, we can solve for the work to lift the book.

work done physics solved problems

This problem can also be solved using energy. Work is equal to the change in potential energy:

work done physics solved problems

Example Question #11 : Work

work done physics solved problems

The relationship between work, force, and distance is:

work done physics solved problems

We are given the force on the toy and the work done. Using these values, we can find the distance. Note that the mass is not relevant for this question.

work done physics solved problems

Example Question #6 : Calculating Work

work done physics solved problems

The relationship between work, force and distance is:

work done physics solved problems

We are given the value for the force and the distance that the toy travels. Using these values, we can find the work done by the cat. Note that the mass of the toy is not relevant for this calculation.

work done physics solved problems

Example Question #7 : Calculating Work

work done physics solved problems

We are given the value for the work done by the cat and the distance that the toy travels. Using these values, we can find the force on the toy. Note that the mass of the toy is not relevant for this calculation.

work done physics solved problems

Example Question #8 : Calculating Work

work done physics solved problems

The formula for work is:

work done physics solved problems

Given the values for force and distance, we can calculate the work done.

work done physics solved problems

Note that no work is done by the force of gravity or the weight of the box, since the vertical position does not change.

work done physics solved problems

Work is the product of force times a distance:

work done physics solved problems

We are given the work and the distance traveled, allowing us to solve for the force. The mass of the cabinet is not necessary information.

work done physics solved problems

Example Question #10 : Calculating Work

work done physics solved problems

None of these

work done physics solved problems

Use the data given to calculate the kinetic energy of the rocket at the two different velocities. Then find the amount of work done using the following equation: 

work done physics solved problems

Kinetic energy of the rocket at the two velocities:

work done physics solved problems

The change in the kinetic energy at the two velocities:

work done physics solved problems

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6.2: Work Done by a Constant Force

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Force in the Direction of Displacement

The work done by a constant force is proportional to the force applied times the displacement of the object.

learning objectives

  • Contrast displacement and distance in constant force situations

Work Done by a Constant Force

When a force acts on an object over a distance, it is said to have done work on the object. Physically, the work done on an object is the change in kinetic energy that that object experiences. We will rigorously prove both of these claims.

The term work was introduced in 1826 by the French mathematician Gaspard-Gustave Coriolis as “weight lifted through a height,” which is based on the use of early steam engines to lift buckets of water out of flooded ore mines. The SI unit of work is the newton-meter or joule (J).

One way to validate if an expression is correct is to perform dimensional analysis. We have claimed that work is the change in kinetic energy of an object and that it is also equal to the force times the distance. The units of these two should agree. Kinetic energy – and all forms of energy – have units of joules (J). Likewise, force has units of newtons (N) and distance has units of meters (m). If the two statements are equivalent they should be equivalent to one another.

\[\mathrm{N⋅m=kg\dfrac{m}{s^2}⋅m=kg\dfrac{m^2}{s^2}=J}\]

Displacement versus Distance

Often times we will be asked to calculate the work done by a force on an object. As we have shown, this is proportional to the force and the distance which the object is displaced, not moved. We will investigate two examples of a box being moved to illustrate this.

Example Problems

Here are a few example problems:

(1.a) Consider a constant force of two newtons (F = 2 N) acting on a box of mass three kilograms (M = 3 kg). Calculate the work done on the box if the box is displaced 5 meters.

(1.b) Since the box is displaced 5 meters and the force is 2 N, we multiply the two quantities together. The object’s mass will dictate how fast it is accelerating under the force, and thus the time it takes to move the object from point a to point b. Regardless of how long it takes, the object will have the same displacement and thus the same work done on it.

(2.a) Consider the same box (M = 3 kg) being pushed by a constant force of four newtons (F = 4 N). It begins at rest and is pushed for five meters (d = 5m). Assuming a frictionless surface, calculate the velocity of the box at 5 meters.

(2.b) We now understand that the work is proportional to the change in kinetic energy, from this we can calculate the final velocity. What do we know so far? We know that the block begins at rest, so the initial kinetic energy must be zero. From this we algebraically isolate and solve for the final velocity.

\[\mathrm{Fd=ΔKE=KE_f−0=\dfrac{1}{2}mv^2_f}\]

\[\mathrm{v_f=\sqrt{2\dfrac{Fd}{m}}=\sqrt{2\dfrac{4N⋅5m}{2kg}}=\sqrt{10}m/s}\]

We see that the final velocity of the block is approximately 3.15 m/s.

Force at an Angle to Displacement

A force does not have to, and rarely does, act on an object parallel to the direction of motion.

  • Infer how to adjust one-dimensional motion for our three-dimensional world

The Fundamentals

Up until now, we have assumed that any force acting on an object has been parallel to the direction of motion. We have considered our motion to be one dimensional, only acting along the x or y axis. To best examine and understand how nature operators in our three-dimensional world, we will first discuss work in two dimensions in order to build our intuition.

A force does not have to, and rarely does, act on an object parallel to the direction of motion. In the past, we derived that \(\mathrm{W = Fd}\); such that the work done on an object is the force acting on the object multiplied by the displacement. But this is not the whole story. This expression contains an assumed cosine term, which we do not consider for forces parallel to the direction of motion. “Why would we do such a thing? ” you may ask. We do this because the two are equivalent. If the angle of the force along the direction of motion is zero, such that the force is parallel to the direction of motion, then the cosine term equals one and does not change the expression. As we increase the force’s angle with respect to the direction of motion, less and less work is done along the direction that we are considering; and more and more work is being done in another, perpendicular, direction of motion. This process continues until we are perpendicular to our original direction of motion, such that the angle is 90, and the cosine term would equal zero; resulting in zero work being done along our original direction. Instead, we are doing work in another direction!

Angle : Recall that both the force and direction of motion are vectors. When the angle is 90 degrees, the cosine term goes to zero. When along the same direction, they equal one.

Let’s show this explicitly and then look at this phenomena in terms of a box moving along the x and y directions.

We have discussed that work is the integral of the force and the dot product respect to x. But in fact, dot product of force and a very small distance is equal to the two terms times cosine of the angle between the two. \(\mathrm{F \times dx = Fd \cos( \theta )}\). Explicitly,

\[\mathrm{\int_{t_2}^{t_1} F⋅dx= \int_{t_2}^{t_1} Fd \cos θ dx=Fd \cos θ}\]

A Box Being Pushed

Consider a coordinate system such that we have x as the abscissa and y as the ordinate. More so, consider a box being pushed along the x direction. What happens in the following three scenarios?

  • The box is being pushed parallel to the x direction?
  • The box is being pushed at an angle of 45 degrees to the x direction?
  • The box is being pushed at an angle of 60 degrees to the x direction?
  • The box is being pushed at an angle of 90 degrees to the x direction?

In the first scenario, we know that all of the force is acting on the box along the x-direction, which means that work will only be done along the x-direction. More so, a vertical perspective the box is not moving – it is unchanged in the y direction. Since the force is acting parallel to the direction of motion, the angle is equal to zero and our total work is simply the force times the displacement in the x-direction.

In the second scenario, the box is being pushed at an angle of 45 degrees to the x-direction; and thus also a 45 degree angle to the y-direction. When evaluated, the cosine of 45 degrees is equal to \(\mathrm{\frac{1}{\sqrt{2}}}\), or approximately 0.71. This means is that 71% of the force is contributing to the work along the x-direction. The other 29% is acting along the y-direction.

In the third scenario, we know that the force is acting at a 60 degree angle to the x-direction; and thus also a 30 degree angle to the y-direction. When evaluated, cosine of 60 degrees is equal to 1/2. This means that the force is equally acting in the x and y-direction! The work done is linear with respect to both x and y.

In the last scenario, the box is being pushed at an angle perpendicular to the x direction. In other words, we are pushing the box in the y-direction! Thus, the box’s position will be unchanged and experience no displacement along the x-axis. The work done in the x direction will be zero.

  • Understanding work is quintessential to understanding systems in terms of their energy, which is necessary for higher level physics.
  • Work is equivalent to the change in kinetic energy of a system.
  • Distance is not the same as displacement. If a box is moved 3 meters forward and then 4 meters to the left, the total displacement is 5 meters, not 7 meters.
  • Work done on an object along a given direction of motion is equal to the force times the displacement times the cosine of the angle.
  • No work is done along a direction of motion if the force is perpendicular.
  • When considering force parallel to the direction of motion, we omit the cosine term because it equals 1 which does not change the expression.
  • work : A measure of energy expended in moving an object; most commonly, force times displacement. No work is done if the object does not move.
  • dot product : A scalar product.

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Gurumuda Networks

Work done by force – problems and solutions

1. A person pulls a block 2 m along a horizontal surface by a constant force F = 20 N. Determine the work done by force F acting on the block.

Work done by a force – problems and solutions 1

Force (F) = 20 N

Displacement (s) = 2 m

Angle (θ ) = 0

Wanted : Work (W)

W = F d cos θ = (20)(2)(cos 0) = (20)(2)(1) = 40 Joule

2. A force F = 10 N acting on a box 1 m along a horizontal surface. The force acts at a 30 o angle as shown in figure below. Determine the work done by force F!

Work done by a force – problems and solutions 2

Force (F) = 10 N

The horizontal force (F x ) = F cos 30 o = (10)(0.5√3) = 5√3 N

Displacement (d) = 1 meter

Wanted : Work (W) ?

W = F x d = (5√3)(1) = 5√3 Joule

3. A body falls freely from rest, from a height of 2 m. If acceleration due to gravity is 10 m/s 2 , determine the work done by the force of gravity !

Object’s mass (m) = 1 kg

Height (h) = 2 m

Acceleration due to gravity (g) = 10 m/s 2

Wanted : Work done by the force of gravity (W)

W = F d = w h = m g h

W = (1)(10)(2) = 20 Joule

W = work, F = force, d = distance, w = weight , h = height, m = mass, g = acceleration due to gravity.

4. An 1-kg object attached to a spring so it is elongated 2 cm. If acceleration due to gravity is 10 m/s 2 , determine (a) the spring constant (b) work done by spring force on object

Mass (m) = 1 kg

Elongation (x) = 2 cm = 0.02 m

Weight (w) = m g = (1 kg)(10 m/s 2 ) = 10 kg m/s 2 = 10 N

Wanted : Spring constant and work done by spring force

(a) Spring constant

Formula of Hooke’s law :

k = F / x = w / x = m g / x

k = (1)(10) / 0.02 = 10 / 0.02

k = 500 N/m

(b) work done by spring force

W = – ½ k x 2

W = – ½ (500)(0.02) 2

W = – (250)(0.0004)

W = -0.1 Joule

The minus sign indicates that the direction of spring force is opposite with the direction of object displacement.

5. A force F = 10 N accelerates a box over a displacement 2 m. The floor is rough and exerts a friction force F k = 2 N. Determine the net work done on the box.

Work done by a force – problems and solutions 3

Force of kinetic friction (F k ) = 2 N

Displacement (d) = 2 m

Wanted : Net work (W net )

Work done by force F :

W 1 = F d cos 0 = (10)(2)(1) = 20 Joule

Work done by force of kinetic friction (F k ) :

W 2 = F k d = (2)(2)(cos 180) = (2)(2)(-1) = -4 Joule

W net = W 1 – W 2

W net = 20 – 4

W net = 16 Joule

6 . What is the work done by force F on the block.

Work done by force – problems and solutions 1

Force (F) = 12 Newton

Displacement (d) = 4 meters

Wanted: Work (W)

W = F d = (12 Newton)(4 meters) = 48 N m = 48 Joule

7 . A block is pushed by a force of 200 N. The block’s displacement is 2 meters. What is the work done on the block?

Force (F) = 200 Newton

Displacement (d) = 2 meters

W = (200 Newton)(2 meters)

W = 400 N m

W = 400 Joule

8 . The driver of the sedan wants to park his car exactly 0.5 m in front of the truck which is at 10 m from the sedan’s position. What is the work required by the sedan?

Work done by force – problems and solutions 2

Displacement (d) = 10 meters – 0.5 meters = 9.5 meters

Force (F) = 50 Newton

W = (50 Newton)(9.5 meters)

W = 475 N m

W = 475 Joule

Work done by force – problems and solutions 3

Work done by Tom and Jerry so the car can move as far as 4 meters. Forces exerted by Tom and Jerry are 50 N and 70 N.

Displacement (s) = 4 meters

Net force (F) = 50 Newton + 70 Newton = 120 Newton

W = F s = (120 Newton)(4 meters) = 480 N m = 480 Joule

10 . A driver pulling a car so the car moves as far as 1000 cm. What is the work done on the car?

Work done by force – problems and solutions 4

Force (F) = 250 Newton

Displacement (s) = 1000 cm = 1000/100 meters = 10 meters

W = F s = (250 Newton)(10 meters) = 2500 N m = 2500 Joule

11. Based on figure below, if work done by net force is 375 Joule, determine object’s displacement.

Work done by force – problems and solutions 11

Work (W) = 375 Joule

Net force ( ΣF) = 40 N + 10 N – 25 N = 25 Newton ( rightward )

Wanted : Displacement ( d )

The equation of work :

Object’s displacement :

d = W / F = 375 Joule / 25 Newton

d = 15 meter s

12. The activities below w hich do not do work is …

A. Push an object as far as 10 meters

B. Push a car until a move

C. Push a wall

D. Pulled a box

W = work , F = force , d = displacement

B ased on the above formula, work done by force and there is a displacement.

The correct answer is C.

1 3 . Andrew pushes an object with force of 20 N so the object moves in circular motion with a radius of 7 meters. Determine the work done by Andrew for two times circular motion .

B. 1400 Joule

C. 1540 Joule

D. 1760 Joule

If the person pushes wheelchair for two times circular motion then the person and wheelchair return to the original position, so the displacement of the person is zero.

Displacement = 0 so work = 0.

The correct answer is A.

14 . Someone push an object on the floor with force of 350 N. The floor exerts a friction force 70 N. Determine the work done by force to move the object as far as 6 meters.

The force of push (F) = 350 Newton

Friction force (F fric ) = 70 Newton

Displacement of object (s) = 6 meters

There are two forces that act on the object, the push force (F) and friction force (F fric ). The push force has the same direction as the displacement of the object because the push force does a positive work. In another hand, the friction force has the opposite direction with a displacement of the object so that the friction force does a negative work.

Work done by push force :

W = F d = (350 Newton)(6 meters) = 2100 Newton-meters = 2100 Joule

Work done by friction force :

W = – (F fric )(s) = – (70 Newton)(6 meters) = – 420 Newton-meters = – 420 Joule

The net work :

W net = 2100 Joule – 420 Joule

W net = 1680 Joule

15 . An object is pushed by a horizontal force of 14 Newton on a rough floor with the friction force of 10 Newton. Determine the net work of move the object as far as 8 meters.

A. 0.5 Joule

C. 32 Joule

D. 192 Joule

Push force (F) = 14 Newton

Friction force (F fric ) = 10 Newton

Displacement of object (d) = 8 meters

There are two forces that act on an object, push force (F) and friction force (F fric ).

The push force has the same direction as the displacement of the object so that the push force does a positive work. In another hand, the friction force has the opposite direction as the displacement of the object so that the friction force does a negative work.

W = F s = (14 Newton)(8 meters) = 112 Newton meters = 112 Joule

W = – (F fric )(s) = – (10 Newton)(8 meters) = – 80 Newton meters = – 80 Joule

W net = 112 Joule – 80 Joule

W net = 32 Joule

16 . Determine the net work based on figure below.

Work

B. 450 Joule

C. 600 Joule

D. 750 Joule

Work = Force (F) x displacement (d)

Work = Area of triangle 1 + area of rectangle + area of triangle 2

Work = 1/2(40-0)(3-0) + (40-0)(9-3) + 1/2(40-0)(12-9)

Work = 1/2(40)(3) + (40)(6) + 1/2(40)(3)

Work = (20)(3) + 240 + (20)(3)

Work = 60 + 240 + 60

Work = 360 Joule

17 . A piece of wood with a length of 60 cm plugged vertically into the ground. Wood hit with a 10-kg hammer from a height of 40 cm above the top of the wood. If the average resistance force of the ground is 2 x 10 3 N and the acceleration due to gravity is 10 m/s 2 , then the wood will enter entirely into the ground after…. hits.

Mass of hammer (m) = 10 kg

Acceleration due to gravity (g) = 10 m/ s 2

Weight of hammer (w) = m g = (10)(10) = 100 kg m/s 2

Displacement of hammer before hits the wood (d) = 40 cm = 0.4 meters

The resistance of wood (F) = 2 x 10 3 N = 2000 N

Length of wood (s) = 60 cm = 0.6 meters

Wanted : T he wood will enter entirely into the ground after…. hits.

Work done on the hammer when hammer moves as far as 0.4 meters is :

W = F d = w s = (100 N)(0.4 m) = 40 Nm = 40 Joule

Work done by the resistance force of the ground :

W = F d = (2000 N)(0.6 m) = 1200 Nm = 1200 Joule

T he wood will enter entirely into the ground after…. hits.

1200 Joule / 40 Joule = 30

The correct answer is D.

[wpdm_package id=’1192′]

  • Work done by force problems and solutions
  • Work-kinetic energy problems and solutions
  • Work-mechanical energy principle problems and solutions
  • Gravitational potential energy problems and solutions
  • The potential energy of elastic spring problems and solutions
  • Power problems and solutions
  • Application of conservation of mechanical energy for free fall motion
  • Application of conservation of mechanical energy for up and down motion in free fall motion
  • Application of conservation of mechanical energy for motion on a curve surface
  • Application of conservation of mechanical energy for motion on an inclined plane
  • Application of conservation of mechanical energy for projectile motion

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StickMan Physics

StickMan Physics

Animated Physics Lessons

Work and Power Example Solutions

Follow along with common work and power example problems and solutions. See how to solve problems when force is applied directly parallel or at an angle.

Example Work and Power Problems

1. How much work is done by the stickman that pushes a box 5 meters with a force of 12 Newtons forward?

Since the force is in the same direction as motion you plug numbers directly in and don't have to find the parallel component first.

W = (12)(5) = 60 J

2. What is the power output of the stickman that pushes the box 5 meters in 3 seconds with a constant force of 12 N?

power example 2

3. How much work would be done if 12N of force was applied on an object at an angle of 25° above the horizon to move an object 5 meters horizontally.

A) Find the horizontal component of force:

adj = (cosӨ)(hyp)

adj = (cos(25°))(12)= 10.9 N

B) Find out how much work is done by this component:

W = (10.9)(5) = 54.5 J

Work and Power At Angle

4. What is the power output if 12N of force was applied on an object at an angle of 25° above the horizon to move an object 5 meters horizontally in 3 seconds.

Use the work from the problem above

P = 54.5/3 = 18.2 W 

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How to Calculate Work Done

  • by Abnurlion
  • December 31, 2022 November 2, 2023

Table of Contents

What is Work Done in Physics?

Work is the application of physical or mental effort to overcome an activity. When you press your television remote, work is involved. When you carry a bucket to fetch water, work is also involved. Knocking on your door requires energy to carry out the work (of knocking on the door). Whatever we do, we apply force to carry out the activity (work) successfully. Let us now see how to calculate work done in physics .

How to Calculate Work Done

Therefore, we can define work done in physics as the product of force and the perpendicular distance in the direction where we apply the force.

We use w as the symbol of work done in physics. Work is a scalar quantity, and we measure work in Joules (J).

The formula for calculating work done in physics is

Work (w) = Force (f) x Distance (s)

Which means

w = f x s [where w = work done in joules, f = force in newton, and s = distance in meters]

We can equally say that the formula for work done is

w = ma x s [because f = ma and m = mass of the object, while a = acceleration ]

The formula for work done along an inclined plane is w = f x s x cosθ

Solved Problems: How to Calculate Work Done in Physics

Here are a few examples that will guide you to understanding how to calculate work done in physics.

A drum of oil of a total mass of 100 kilograms rolls down from the floor of a lorry 3.0 meters high. Calculate the work done by gravity on the load.

Mass of the drum, m = 100 kg

height = displacement = distance in a specified direction, s = 3 meters

and the formula for calculating work done is

w = f x s = ma x s

But we are dealing with gravitational force , and this implies that a = g

We can now plugin our data into the above formula

w = 100 x 10 x 3 = 3,000 Joules [Because g = 10ms -2 ]

By converting the above answer to kilojoules, we will now have

w = 3 kJ [Because 1000 Joules = 1 kilojoules]

Therefore, the work done by the gravity on the load is 3 kilojoules.

With the help of a rope inclined at 30 0 to the horizontal, a boy exerted a 25-newton pulling force at a desk and made it move a distance of 5 meters along the horizontal direction. Find the work done by the boy.

Always remember to extract your data from the question. We can see from the above question that

The angle made, θ = 30 0

The force, f = 25 Newton

Displacement, s = 5 meters

work done, w =?

Apply the formula for the work done in an inclined plane, which is w = f x s cosθ

Now, replace the above formula with the data you extracted to obtain

w = 25 x 5 x cos30 0

Therefore, w = 125 x 0.866025

Which implies that

w = 108.25 J

Therefore, the work done by the boy is 108 Joules

The mass of David Johns is 30 kilograms, and he climbed a flight of 20 steps each 15 centimeters high. How much work did David Johns do? [ take g = 10ms -2 ]

Mass of David Johns, m = 30 kg

Displacement covered, s = 20 steps x 15 cm = 300cm

We can now convert displacement into meters to suit our answer

Displacement, s = 300cm = (300/100) m = 3 meters

Work done =?

g = 10 ms -2

By applying the formula for work done, w = mg x s

we will get

w = 30 x 10 x 3 = 900 J

Therefore, the work done by David Johns is 900 Joules.

A constant force of 40-newton acting on a body initially at rest gives it an acceleration of 0.1 ms -2 for 4 seconds. Calculate the work done by the force.

Force, f = 40 N

Initial velocity , u = 0

acceleration, a = 0.1 ms -2

time , t = 4 seconds

we can apply one of the equations of motion ( s = ut + 1/2 at 2 ) to find s

s = 0 x 4 + (1/2) x 0.1 x 4 2

Which becomes s = (1/2) x 0.1 x 16 = 0.8 meters

Now substitute our data above into the formula for work done ( w = f x s )

w = 40 x 0.8 = 32 J

Therefore, the work done by the force is 32 Joules.

A crane lifts a load of 1000 kilograms very slowly through a vertical distance of 150 centimeters. Calculate the work done against gravity if g = 10 ms -2 .

Mass, m = 1000 kg

Distance, s = 150 cm = (150/100) m = 1.5 m

Work done, w =?

gravitational force, g = 10ms -2

Plugin the above data into w = mg x s to get

w = 1000 x 10 x 1.5 = 15,000 Joules = 15 kJ

Therefore, work done against gravity is 15 kilojoules .

A loaded sack of total mass 100 kilograms falls down from the floor of a lorry 2.0 meters high. Calculate the work done by the gravity of the load.

Mass, m = 100 kg

Distance, s (h) = 2 m

The formula for work done is

w = mg x s = mgh

Now, insert your data into the above formula

w = mgh = 100 x 10 x 2 = 2000 J = 2 kJ

Therefore, the work done by gravity is 2 kJ

Nazeef carries a 50-newton bag and climbs five steps of a staircase with a height of 5 meters. The bag was pulled away from his hands (due to its heaviness) with a force of 15 Newton at a constant speed of 0.1m/s through a horizontal distance of 25 meters. Find the work done by Nazeef for his entire motion.

The force on the vertical axis, f v = 50 N

Height along the vertical axis, s v = 5 m

Angle on the vertical axis, θ v = 0

Force on the horizontal axis, f h = 15 N

Height along the horizontal axis, s h = 25 m

The angle on the horizontal axis, θ h = 0

The work done along the vertical axis, w v = f v x s v cosθ

w v = 50 x 5 x cos0 = 250 x 1 = 250 Joules

The work done along the vertical axis, w h = f h x s h cosθ

w h = 15 x 25 x cos0 = 375 x 1 = 375 Joules

Therefore, the work done by Nazeef on his entire motion is

w T = w v + w h = 250 + 375 = 625 Joules

You may also like to read:

What is the work done by the electric force to move a 1 c charge from a to b?

How to Calculate the Cost of Electricity Per kWh

Also, How to Calculate the Work Function of a Metal

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  • 7.2 Kinetic Energy and the Work-Energy Theorem
  • Introduction to Science and the Realm of Physics, Physical Quantities, and Units
  • 1.1 Physics: An Introduction
  • 1.2 Physical Quantities and Units
  • 1.3 Accuracy, Precision, and Significant Figures
  • 1.4 Approximation
  • Section Summary
  • Conceptual Questions
  • Problems & Exercises
  • Introduction to One-Dimensional Kinematics
  • 2.1 Displacement
  • 2.2 Vectors, Scalars, and Coordinate Systems
  • 2.3 Time, Velocity, and Speed
  • 2.4 Acceleration
  • 2.5 Motion Equations for Constant Acceleration in One Dimension
  • 2.6 Problem-Solving Basics for One-Dimensional Kinematics
  • 2.7 Falling Objects
  • 2.8 Graphical Analysis of One-Dimensional Motion
  • Introduction to Two-Dimensional Kinematics
  • 3.1 Kinematics in Two Dimensions: An Introduction
  • 3.2 Vector Addition and Subtraction: Graphical Methods
  • 3.3 Vector Addition and Subtraction: Analytical Methods
  • 3.4 Projectile Motion
  • 3.5 Addition of Velocities
  • Introduction to Dynamics: Newton’s Laws of Motion
  • 4.1 Development of Force Concept
  • 4.2 Newton’s First Law of Motion: Inertia
  • 4.3 Newton’s Second Law of Motion: Concept of a System
  • 4.4 Newton’s Third Law of Motion: Symmetry in Forces
  • 4.5 Normal, Tension, and Other Examples of Forces
  • 4.6 Problem-Solving Strategies
  • 4.7 Further Applications of Newton’s Laws of Motion
  • 4.8 Extended Topic: The Four Basic Forces—An Introduction
  • Introduction: Further Applications of Newton’s Laws
  • 5.1 Friction
  • 5.2 Drag Forces
  • 5.3 Elasticity: Stress and Strain
  • Introduction to Uniform Circular Motion and Gravitation
  • 6.1 Rotation Angle and Angular Velocity
  • 6.2 Centripetal Acceleration
  • 6.3 Centripetal Force
  • 6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force
  • 6.5 Newton’s Universal Law of Gravitation
  • 6.6 Satellites and Kepler’s Laws: An Argument for Simplicity
  • Introduction to Work, Energy, and Energy Resources
  • 7.1 Work: The Scientific Definition
  • 7.3 Gravitational Potential Energy
  • 7.4 Conservative Forces and Potential Energy
  • 7.5 Nonconservative Forces
  • 7.6 Conservation of Energy
  • 7.8 Work, Energy, and Power in Humans
  • 7.9 World Energy Use
  • Introduction to Linear Momentum and Collisions
  • 8.1 Linear Momentum and Force
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  • 8.4 Elastic Collisions in One Dimension
  • 8.5 Inelastic Collisions in One Dimension
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  • 8.7 Introduction to Rocket Propulsion
  • Introduction to Statics and Torque
  • 9.1 The First Condition for Equilibrium
  • 9.2 The Second Condition for Equilibrium
  • 9.3 Stability
  • 9.4 Applications of Statics, Including Problem-Solving Strategies
  • 9.5 Simple Machines
  • 9.6 Forces and Torques in Muscles and Joints
  • Introduction to Rotational Motion and Angular Momentum
  • 10.1 Angular Acceleration
  • 10.2 Kinematics of Rotational Motion
  • 10.3 Dynamics of Rotational Motion: Rotational Inertia
  • 10.4 Rotational Kinetic Energy: Work and Energy Revisited
  • 10.5 Angular Momentum and Its Conservation
  • 10.6 Collisions of Extended Bodies in Two Dimensions
  • 10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum
  • Introduction to Fluid Statics
  • 11.1 What Is a Fluid?
  • 11.2 Density
  • 11.3 Pressure
  • 11.4 Variation of Pressure with Depth in a Fluid
  • 11.5 Pascal’s Principle
  • 11.6 Gauge Pressure, Absolute Pressure, and Pressure Measurement
  • 11.7 Archimedes’ Principle
  • 11.8 Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action
  • 11.9 Pressures in the Body
  • Introduction to Fluid Dynamics and Its Biological and Medical Applications
  • 12.1 Flow Rate and Its Relation to Velocity
  • 12.2 Bernoulli’s Equation
  • 12.3 The Most General Applications of Bernoulli’s Equation
  • 12.4 Viscosity and Laminar Flow; Poiseuille’s Law
  • 12.5 The Onset of Turbulence
  • 12.6 Motion of an Object in a Viscous Fluid
  • 12.7 Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes
  • Introduction to Temperature, Kinetic Theory, and the Gas Laws
  • 13.1 Temperature
  • 13.2 Thermal Expansion of Solids and Liquids
  • 13.3 The Ideal Gas Law
  • 13.4 Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature
  • 13.5 Phase Changes
  • 13.6 Humidity, Evaporation, and Boiling
  • Introduction to Heat and Heat Transfer Methods
  • 14.2 Temperature Change and Heat Capacity
  • 14.3 Phase Change and Latent Heat
  • 14.4 Heat Transfer Methods
  • 14.5 Conduction
  • 14.6 Convection
  • 14.7 Radiation
  • Introduction to Thermodynamics
  • 15.1 The First Law of Thermodynamics
  • 15.2 The First Law of Thermodynamics and Some Simple Processes
  • 15.3 Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency
  • 15.4 Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated
  • 15.5 Applications of Thermodynamics: Heat Pumps and Refrigerators
  • 15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy
  • 15.7 Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation
  • Introduction to Oscillatory Motion and Waves
  • 16.1 Hooke’s Law: Stress and Strain Revisited
  • 16.2 Period and Frequency in Oscillations
  • 16.3 Simple Harmonic Motion: A Special Periodic Motion
  • 16.4 The Simple Pendulum
  • 16.5 Energy and the Simple Harmonic Oscillator
  • 16.6 Uniform Circular Motion and Simple Harmonic Motion
  • 16.7 Damped Harmonic Motion
  • 16.8 Forced Oscillations and Resonance
  • 16.10 Superposition and Interference
  • 16.11 Energy in Waves: Intensity
  • Introduction to the Physics of Hearing
  • 17.2 Speed of Sound, Frequency, and Wavelength
  • 17.3 Sound Intensity and Sound Level
  • 17.4 Doppler Effect and Sonic Booms
  • 17.5 Sound Interference and Resonance: Standing Waves in Air Columns
  • 17.6 Hearing
  • 17.7 Ultrasound
  • Introduction to Electric Charge and Electric Field
  • 18.1 Static Electricity and Charge: Conservation of Charge
  • 18.2 Conductors and Insulators
  • 18.3 Coulomb’s Law
  • 18.4 Electric Field: Concept of a Field Revisited
  • 18.5 Electric Field Lines: Multiple Charges
  • 18.6 Electric Forces in Biology
  • 18.7 Conductors and Electric Fields in Static Equilibrium
  • 18.8 Applications of Electrostatics
  • Introduction to Electric Potential and Electric Energy
  • 19.1 Electric Potential Energy: Potential Difference
  • 19.2 Electric Potential in a Uniform Electric Field
  • 19.3 Electrical Potential Due to a Point Charge
  • 19.4 Equipotential Lines
  • 19.5 Capacitors and Dielectrics
  • 19.6 Capacitors in Series and Parallel
  • 19.7 Energy Stored in Capacitors
  • Introduction to Electric Current, Resistance, and Ohm's Law
  • 20.1 Current
  • 20.2 Ohm’s Law: Resistance and Simple Circuits
  • 20.3 Resistance and Resistivity
  • 20.4 Electric Power and Energy
  • 20.5 Alternating Current versus Direct Current
  • 20.6 Electric Hazards and the Human Body
  • 20.7 Nerve Conduction–Electrocardiograms
  • Introduction to Circuits and DC Instruments
  • 21.1 Resistors in Series and Parallel
  • 21.2 Electromotive Force: Terminal Voltage
  • 21.3 Kirchhoff’s Rules
  • 21.4 DC Voltmeters and Ammeters
  • 21.5 Null Measurements
  • 21.6 DC Circuits Containing Resistors and Capacitors
  • Introduction to Magnetism
  • 22.1 Magnets
  • 22.2 Ferromagnets and Electromagnets
  • 22.3 Magnetic Fields and Magnetic Field Lines
  • 22.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field
  • 22.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications
  • 22.6 The Hall Effect
  • 22.7 Magnetic Force on a Current-Carrying Conductor
  • 22.8 Torque on a Current Loop: Motors and Meters
  • 22.9 Magnetic Fields Produced by Currents: Ampere’s Law
  • 22.10 Magnetic Force between Two Parallel Conductors
  • 22.11 More Applications of Magnetism
  • Introduction to Electromagnetic Induction, AC Circuits and Electrical Technologies
  • 23.1 Induced Emf and Magnetic Flux
  • 23.2 Faraday’s Law of Induction: Lenz’s Law
  • 23.3 Motional Emf
  • 23.4 Eddy Currents and Magnetic Damping
  • 23.5 Electric Generators
  • 23.6 Back Emf
  • 23.7 Transformers
  • 23.8 Electrical Safety: Systems and Devices
  • 23.9 Inductance
  • 23.10 RL Circuits
  • 23.11 Reactance, Inductive and Capacitive
  • 23.12 RLC Series AC Circuits
  • Introduction to Electromagnetic Waves
  • 24.1 Maxwell’s Equations: Electromagnetic Waves Predicted and Observed
  • 24.2 Production of Electromagnetic Waves
  • 24.3 The Electromagnetic Spectrum
  • 24.4 Energy in Electromagnetic Waves
  • Introduction to Geometric Optics
  • 25.1 The Ray Aspect of Light
  • 25.2 The Law of Reflection
  • 25.3 The Law of Refraction
  • 25.4 Total Internal Reflection
  • 25.5 Dispersion: The Rainbow and Prisms
  • 25.6 Image Formation by Lenses
  • 25.7 Image Formation by Mirrors
  • Introduction to Vision and Optical Instruments
  • 26.1 Physics of the Eye
  • 26.2 Vision Correction
  • 26.3 Color and Color Vision
  • 26.4 Microscopes
  • 26.5 Telescopes
  • 26.6 Aberrations
  • Introduction to Wave Optics
  • 27.1 The Wave Aspect of Light: Interference
  • 27.2 Huygens's Principle: Diffraction
  • 27.3 Young’s Double Slit Experiment
  • 27.4 Multiple Slit Diffraction
  • 27.5 Single Slit Diffraction
  • 27.6 Limits of Resolution: The Rayleigh Criterion
  • 27.7 Thin Film Interference
  • 27.8 Polarization
  • 27.9 *Extended Topic* Microscopy Enhanced by the Wave Characteristics of Light
  • Introduction to Special Relativity
  • 28.1 Einstein’s Postulates
  • 28.2 Simultaneity And Time Dilation
  • 28.3 Length Contraction
  • 28.4 Relativistic Addition of Velocities
  • 28.5 Relativistic Momentum
  • 28.6 Relativistic Energy
  • Introduction to Quantum Physics
  • 29.1 Quantization of Energy
  • 29.2 The Photoelectric Effect
  • 29.3 Photon Energies and the Electromagnetic Spectrum
  • 29.4 Photon Momentum
  • 29.5 The Particle-Wave Duality
  • 29.6 The Wave Nature of Matter
  • 29.7 Probability: The Heisenberg Uncertainty Principle
  • 29.8 The Particle-Wave Duality Reviewed
  • Introduction to Atomic Physics
  • 30.1 Discovery of the Atom
  • 30.2 Discovery of the Parts of the Atom: Electrons and Nuclei
  • 30.3 Bohr’s Theory of the Hydrogen Atom
  • 30.4 X Rays: Atomic Origins and Applications
  • 30.5 Applications of Atomic Excitations and De-Excitations
  • 30.6 The Wave Nature of Matter Causes Quantization
  • 30.7 Patterns in Spectra Reveal More Quantization
  • 30.8 Quantum Numbers and Rules
  • 30.9 The Pauli Exclusion Principle
  • Introduction to Radioactivity and Nuclear Physics
  • 31.1 Nuclear Radioactivity
  • 31.2 Radiation Detection and Detectors
  • 31.3 Substructure of the Nucleus
  • 31.4 Nuclear Decay and Conservation Laws
  • 31.5 Half-Life and Activity
  • 31.6 Binding Energy
  • 31.7 Tunneling
  • Introduction to Applications of Nuclear Physics
  • 32.1 Diagnostics and Medical Imaging
  • 32.2 Biological Effects of Ionizing Radiation
  • 32.3 Therapeutic Uses of Ionizing Radiation
  • 32.4 Food Irradiation
  • 32.5 Fusion
  • 32.6 Fission
  • 32.7 Nuclear Weapons
  • Introduction to Particle Physics
  • 33.1 The Yukawa Particle and the Heisenberg Uncertainty Principle Revisited
  • 33.2 The Four Basic Forces
  • 33.3 Accelerators Create Matter from Energy
  • 33.4 Particles, Patterns, and Conservation Laws
  • 33.5 Quarks: Is That All There Is?
  • 33.6 GUTs: The Unification of Forces
  • Introduction to Frontiers of Physics
  • 34.1 Cosmology and Particle Physics
  • 34.2 General Relativity and Quantum Gravity
  • 34.3 Superstrings
  • 34.4 Dark Matter and Closure
  • 34.5 Complexity and Chaos
  • 34.6 High-temperature Superconductors
  • 34.7 Some Questions We Know to Ask
  • A | Atomic Masses
  • B | Selected Radioactive Isotopes
  • C | Useful Information
  • D | Glossary of Key Symbols and Notation

Learning Objectives

By the end of this section, you will be able to:

  • Explain work as a transfer of energy and net work as the work done by the net force.
  • Explain and apply the work-energy theorem.

Work Transfers Energy

What happens to the work done on a system? Energy is transferred into the system, but in what form? Does it remain in the system or move on? The answers depend on the situation. For example, if the lawn mower in Figure 7.2 (a) is pushed just hard enough to keep it going at a constant speed, then energy put into the mower by the person is removed continuously by friction, and eventually leaves the system in the form of heat transfer. In contrast, work done on the briefcase by the person carrying it up stairs in Figure 7.2 (d) is stored in the briefcase-Earth system and can be recovered at any time, as shown in Figure 7.2 (e). In fact, the building of the pyramids in ancient Egypt is an example of storing energy in a system by doing work on the system. Some of the energy imparted to the stone blocks in lifting them during construction of the pyramids remains in the stone-Earth system and has the potential to do work.

In this section we begin the study of various types of work and forms of energy. We will find that some types of work leave the energy of a system constant, for example, whereas others change the system in some way, such as making it move. We will also develop definitions of important forms of energy, such as the energy of motion.

Net Work and the Work-Energy Theorem

We know from the study of Newton’s laws in Dynamics: Force and Newton's Laws of Motion that net force causes acceleration. We will see in this section that work done by the net force gives a system energy of motion, and in the process we will also find an expression for the energy of motion.

Let us start by considering the total, or net, work done on a system. Net work is defined to be the sum of work on an object. The net work can be written in terms of the net force on an object. F net F net . In equation form, this is W net = F net d cos θ W net = F net d cos θ where θ θ is the angle between the force vector and the displacement vector.

Figure 7.3 (a) shows a graph of force versus displacement for the component of the force in the direction of the displacement—that is, an F cos θ F cos θ vs. d d graph. In this case, F cos θ F cos θ is constant. You can see that the area under the graph is F d cos θ F d cos θ , or the work done. Figure 7.3 (b) shows a more general process where the force varies. The area under the curve is divided into strips, each having an average force ( F cos θ ) i ( ave ) ( F cos θ ) i ( ave ) . The work done is ( F cos θ ) i ( ave ) d i ( F cos θ ) i ( ave ) d i for each strip, and the total work done is the sum of the W i W i . Thus the total work done is the total area under the curve, a useful property to which we shall refer later.

Net work will be simpler to examine if we consider a one-dimensional situation where a force is used to accelerate an object in a direction parallel to its initial velocity. Such a situation occurs for the package on the roller belt conveyor system shown in Figure 7.4 .

The force of gravity and the normal force acting on the package are perpendicular to the displacement and do no work. Moreover, they are also equal in magnitude and opposite in direction so they cancel in calculating the net force. The net force arises solely from the horizontal applied force F app F app and the horizontal friction force f f . Thus, as expected, the net force is parallel to the displacement, so that θ = 0º θ = 0º and cos θ = 1 cos θ = 1 , and the net work is given by

The effect of the net force F net F net is to accelerate the package from v 0 v 0 to v v . The kinetic energy of the package increases, indicating that the net work done on the system is positive. (See Example 7.2 .) By using Newton’s second law, and doing some algebra, we can reach an interesting conclusion. Substituting F net = ma F net = ma from Newton’s second law gives

To get a relationship between net work and the speed given to a system by the net force acting on it, we take d = x − x 0 d = x − x 0 and use the equation studied in Motion Equations for Constant Acceleration in One Dimension for the change in speed over a distance d d if the acceleration has the constant value a a ; namely, v 2 = v 0 2 + 2 ad v 2 = v 0 2 + 2 ad (note that a a appears in the expression for the net work). Solving for acceleration gives a = v 2 − v 0 2 2 d a = v 2 − v 0 2 2 d . When a a is substituted into the preceding expression for W net W net , we obtain

The d d cancels, and we rearrange this to obtain

This expression is called the work-energy theorem , and it actually applies in general (even for forces that vary in direction and magnitude), although we have derived it for the special case of a constant force parallel to the displacement. The theorem implies that the net work on a system equals the change in the quantity 1 2 mv 2 1 2 mv 2 . This quantity is our first example of a form of energy.

The Work-Energy Theorem

The net work on a system equals the change in the quantity 1 2 mv 2 1 2 mv 2 .

The quantity 1 2 mv 2 1 2 mv 2 in the work-energy theorem is defined to be the translational kinetic energy (KE) of a mass m m moving at a speed v v . ( Translational kinetic energy is distinct from rotational kinetic energy, which is considered later.) In equation form, the translational kinetic energy,

is the energy associated with translational motion. Kinetic energy is a form of energy associated with the motion of a particle, single body, or system of objects moving together.

We are aware that it takes energy to get an object, like a car or the package in Figure 7.4 , up to speed, but it may be a bit surprising that kinetic energy is proportional to speed squared. This proportionality means, for example, that a car traveling at 100 km/h has four times the kinetic energy it has at 50 km/h, helping to explain why high-speed collisions are so devastating. We will now consider a series of examples to illustrate various aspects of work and energy.

Example 7.2

Calculating the kinetic energy of a package.

Suppose a 30.0-kg package on the roller belt conveyor system in Figure 7.4 is moving at 0.500 m/s. What is its kinetic energy?

Because the mass m m and speed v v are given, the kinetic energy can be calculated from its definition as given in the equation KE = 1 2 mv 2 KE = 1 2 mv 2 .

The kinetic energy is given by

Entering known values gives

which yields

Note that the unit of kinetic energy is the joule, the same as the unit of work, as mentioned when work was first defined. It is also interesting that, although this is a fairly massive package, its kinetic energy is not large at this relatively low speed. This fact is consistent with the observation that people can move packages like this without exhausting themselves.

Example 7.3

Determining the work to accelerate a package.

Suppose that you push on the 30.0-kg package in Figure 7.4 with a constant force of 120 N through a distance of 0.800 m, and that the opposing friction force averages 5.00 N.

(a) Calculate the net work done on the package. (b) Solve the same problem as in part (a), this time by finding the work done by each force that contributes to the net force.

Strategy and Concept for (a)

This is a motion in one dimension problem, because the downward force (from the weight of the package) and the normal force have equal magnitude and opposite direction, so that they cancel in calculating the net force, while the applied force, friction, and the displacement are all horizontal. (See Figure 7.4 .) As expected, the net work is the net force times distance.

Solution for (a)

The net force is the push force minus friction, or F net = 120 N – 5 . 00 N = 115 N F net = 120 N – 5 . 00 N = 115 N . Thus the net work is

Discussion for (a)

This value is the net work done on the package. The person actually does more work than this, because friction opposes the motion. Friction does negative work and removes some of the energy the person expends and converts it to thermal energy. The net work equals the sum of the work done by each individual force.

Strategy and Concept for (b)

The forces acting on the package are gravity, the normal force, the force of friction, and the applied force. The normal force and force of gravity are each perpendicular to the displacement, and therefore do no work.

Solution for (b)

The applied force does work.

The friction force and displacement are in opposite directions, so that θ = 180º θ = 180º , and the work done by friction is

So the amounts of work done by gravity, by the normal force, by the applied force, and by friction are, respectively,

The total work done as the sum of the work done by each force is then seen to be

Discussion for (b)

The calculated total work W total W total as the sum of the work by each force agrees, as expected, with the work W net W net done by the net force. The work done by a collection of forces acting on an object can be calculated by either approach.

Example 7.4

Determining speed from work and energy.

Find the speed of the package in Figure 7.4 at the end of the push, using work and energy concepts.

Here the work-energy theorem can be used, because we have just calculated the net work, W net W net , and the initial kinetic energy, 1 2 m v 0 2 1 2 m v 0 2 . These calculations allow us to find the final kinetic energy, 1 2 mv 2 1 2 mv 2 , and thus the final speed v v .

The work-energy theorem in equation form is

Solving for 1 2 mv 2 1 2 mv 2 gives

Solving for the final speed as requested and entering known values gives

Using work and energy, we not only arrive at an answer, we see that the final kinetic energy is the sum of the initial kinetic energy and the net work done on the package. This means that the work indeed adds to the energy of the package.

Example 7.5

Work and energy can reveal distance, too.

How far does the package in Figure 7.4 coast after the push, assuming friction remains constant? Use work and energy considerations.

We know that once the person stops pushing, friction will bring the package to rest. In terms of energy, friction does negative work until it has removed all of the package’s kinetic energy. The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement; hence, this gives us a way of finding the distance traveled after the person stops pushing.

The normal force and force of gravity cancel in calculating the net force. The horizontal friction force is then the net force, and it acts opposite to the displacement, so θ = 180º θ = 180º . To reduce the kinetic energy of the package to zero, the work W fr W fr by friction must be minus the kinetic energy that the package started with plus what the package accumulated due to the pushing. Thus W fr = − 95 . 75 J W fr = − 95 . 75 J . Furthermore, W fr = f d ′ cos θ = – f d ′ W fr = f d ′ cos θ = – f d ′ , where d ′ d ′ is the distance it takes to stop. Thus,

This is a reasonable distance for a package to coast on a relatively friction-free conveyor system. Note that the work done by friction is negative (the force is in the opposite direction of motion), so it removes the kinetic energy.

Some of the examples in this section can be solved without considering energy, but at the expense of missing out on gaining insights about what work and energy are doing in this situation. On the whole, solutions involving energy are generally shorter and easier than those using kinematics and dynamics alone.

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  • Authors: Paul Peter Urone, Roger Hinrichs
  • Publisher/website: OpenStax
  • Book title: College Physics 2e
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Chapter: 11th Physics : UNIT 4 : Work, Energy and Power

Solved example problems for physics: work, energy and power, numerical problems.

1. Calculate the work done by a force of 30 N in lifting a load of 2kg to a height of 10m (g = 10ms -2 )

Force mg = 30 N ; height = 10 m

Work done to lift a load W = ?

W = F.S (or) mgh

W = 300 J 

Ans:  300J

2. A ball with a velocity of 5 m s -1  impinges at angle of 60˚ with the vertical on a smooth horizontal plane. If the coefficient of restitution is 0.5, find the velocity and direction after the impact.

The impluse on the ball acts perpendicular to the smooth plane.

work done physics solved problems

(i) The component of velocity of ball parallel to the surface.

(ii) For the component of velocity of ball perpendicular to the surface, apply law of restitution.

The component of velocity parallel to the surface will be changed.

v cos α = u cos 60°

v cos α = 5 × 1/2 = 5/2       ….(1)

According to law of restitution

v sin α = e u sin 60°

v sin α = 1/2 × 5 × √3/2 =  5 (√3/4)        ….(2)

Squaring and adding (1) and (2)

v 2 (sin 2 α + cos 2 α) = 

work done physics solved problems

v = 3.3 ms -1

Ans:  v = 0.3 m  s -1

3. A bob of mass m is attached to one end of the rod of negligible mass and length r, the other end of which is pivoted freely at a fixed center O as shown in the figure. What initial speed must be given to the object to reach the top of the circle? (Hint: Use law of conservation of energy). Is this speed less or greater than speed obtained in the section 4.2.9?

Ans: √ 4gr ms-1

work done physics solved problems

The horizontal distance, draw the point of projection to the point where the ball returns to the same level

work done physics solved problems

AC = OA-OC = r - rcosθ

minimum velocity = at v L = √[5 gr ]

work done physics solved problems

v 1 2 = v L 2 - 2gr (1 - cos θ)

v 1 2 = 5 gr - 2gr (l-cosθ)

v 1 2 = 5gr-2gr(l – 1/2)

v 1 2 = 5gr - gr => v 1 2 = 4gr

v 1 = √[4 gr ] ms -1 .

4. Two different unknown masses A and B collide. A is initially at rest when B has a speed v. After collision B has a speed v/2 and moves at right angles to its original direction of motion. Find the direction in which A moves after collision.

Momentium is censerved in both × and y direction.

In x - direction

M B V B = O + M A V A ' cosɸ ...(l)

In y - direction

O = M B V B ' - M A V A 'sinɸ  ...(2)

(2)/(1) tanɸ = V B ’/V B = 1/2

ɸ = 26.6° (or) 26° 36' [1° = 60']

Ans:  θ   = 26° 33 ′

5. A bullet of mass 20 g strikes a pendulum of mass 5 kg. The centre of mass of pendulum rises a vertical distance of 10 cm. If the bullet gets embedded into the pendulum, calculate its initial speed.

Mass of the bullet m 1 = 20 g = 0.02 kg.

Mass of the pendulum m 2 = 5 kg

Centre of mass of pendulum rises to a height = h = 10 cm = 0.1 m

Speed of the bullet = u 1

Pendulum is at rest .:. u 2 = 0

Common velocity of the bullet and the pendulum after the bullet is embeded into the object = v

work done physics solved problems

From II equation of motion

v = √[2gh] = √[2x9.8x0.1] = √[1.96] = 1.4 ms -1

Substitute the value of v in equation (1)

1.4 = 0.02 u 1 /5.02

u 1 = 5.02x1.4 / 0.02

u 1 = 351.4 ms -1

Ans:  v = 351.4m  s -1

Conceptual Questions

1. A spring which in initially in un-stretched condition, is first stretched by a length x and again by a further length x. The work done in the first case W 1  is one third of the work done in second case W 2 . True or false?

work done physics solved problems

2. Which is conserved in inelastic collision? Total energy (or) Kinetic energy?

Total energy is always conserved.

But K.E. is not conserved.

3. Is there any net work done by external forces on a car moving with a constant speed along a straight road?

If a car is moving at a constant speed, then external force will be zero.

Because a = [v - u] / t

work done physics solved problems

For constant speed v = u , then a =0. ( a -acceleration)

F = ma   .'. F = zero. i.e.. no external force.

W = F.S. = 0. So net work done is zero.

4. A car starts from rest and moves on a surface with uniform acceleration.

Draw  the  graph  of  kinetic  energy versus displacement. What information you can get from that graph?

work done physics solved problems

5. A charged particle moves towards another charged particle. Under what conditions the total momentum and the total energy of the system conserved?

(i) Both charged particles shall be dissimilar charge, (i.e. positive and negative)

(ii) After collision the charged particles; stick together permanent.

(iii) They should move with common velocity

SOLVED EXAMPLE

work done physics solved problems

2. A particle moves along X- axis from x=0 to x=8 under the influence of a force given by F= 3 x   2  - 4 x  + 5. Find the work done in the process.

work done physics solved problems

3. A body of mass 10kg at rest is subjected to a force of 16N. Find the kinetic energy at the end of 10 s.

Mass m = 10 kg

Force F = 16 N

time t = 10 s

work done physics solved problems

4. A body of mass 5kg is thrown up vertically with a kinetic energy of 1000 J. If acceleration due to gravity is 10 ms -2 , find the height at which the kinetic energy becomes half of the original value.

Mass m = 5kg

K.E          E = 1000J

g = 10 m s -2

work done physics solved problems

5. Two bodies of mass 60 kg and 30 kg move in the same direction along straight line with velocity 40 cm s -1  and 30 cm   s -1   respectively suffer one dimensional elastic collision. Find their velocities after collision.

Mass m 1  = 60 kg

Mass m 2  = 30 kg

V 1  =  40cm s -1

V 2  =  30c m  s -1

work done physics solved problems

Likewise, 

work done physics solved problems

6. A particle of mass 70 g moving at 50 cm s -1  is acted upon by a variable force as shown in the figure. What will be its speed once the force stops?

work done physics solved problems

The area under the graph gives the impulse.

Impulse I = area of  ∆ OAE+ area of rectangle ABDE+ area of  ∆ DBC

work done physics solved problems

But Impulse = 2 × initial momentum of the particle= 2 × m × u

work done physics solved problems

Hence the particle will reverse its direction and move with its initial speed.

7. A particle strikes a horizontal frictionless floor with a speed u at an angle θ with the vertical and rebounds with the speed v at an angle Φ  with an vertical. The coefficient of restitution between the particle and floor is e. What is the magnitude of v?

work done physics solved problems

Applying component of velocities,

work done physics solved problems

The x - component of velocity is

work done physics solved problems

The magnitude of y – component of velocity is not same, therefore, using coefficient of restitution,

work done physics solved problems

8. A particle of mass m is fixed to one end of a light spring of force constant k and un-stretched length l. It is rotated with an angular velocity  ω   in horizontal circle. What will be the length increase in the spring?

Mass spring = m Force

constant = k

Un-stretched length = l

Angular velocity = ω

work done physics solved problems

Let ‘x’ be the increase in the length of the spring.

The new length = (l+x) = r

When the spring is rotated in a horizontal circle,

Spring force = centripetal force.

work done physics solved problems

9. A gun fires 8 bullets per second into a target X. If the mass of each bullet is 3 g and its speed 600 s -1 . Then, calculate the power delivered by the bullets.

Power = work done per second = total kinetic energy of 8 bullets per second

work done physics solved problems

Solved Example Problems for Work

Example 4.1.

A box is pulled with a force of 25 N to produce a displacement of 15 m. If the angle between the force and displacement is 30 o , find the work done by the force.

work done physics solved problems

Force, F = 25 N

Displacement, dr = 15 m

Angle between F and dr, θ = 30 o

Work done ,  W  =  Fdr cos θ

work done physics solved problems

Solved Example Problems for Work done by a constant force

Example 4.2

An object of mass 2 kg falls from a height of 5 m to the ground. What is the work done by the gravitational force on the object? (Neglect air resistance; Take g = 10 m s -2 )

work done physics solved problems

Work done by gravitational force is

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The work done by the gravitational force on the object is positive.

Example 4.3

An object of mass  m = 1 kg is sliding from top to bottom in the frictionless inclined plane of inclination angle θ = 30 o  and the length of inclined plane is 10 m as shown in the figure. Calculate the work done by gravitational force and normal force on the object. Assume acceleration due to gravity, g = 10 m s -2

work done physics solved problems

We calculated in the previous chapter that the acceleration experienced by the object in the inclined plane as g sin θ  .

According to Newton’s second law, the force acting on the mass along the inclined plane F  =  mg sin θ . Note that this force is constant throughout the motion of the mass.

The work done by the parallel component of gravitational force  (  mg sin θ )  is given by

work done physics solved problems

Example 4.4

If an object of mass 2 kg is thrown up from the ground reaches a height of 5 m and falls back to the Earth (neglect the air resistance). Calculate

a) The work done by gravity when the  object reaches 5 m height

b) The work done by gravity when the   object comes back to Earth

c) Total work done by gravity both in  upward and downward motion and mention the physical significance of the result.

When the object goes up, the displacement points in the upward direction whereas the gravitational force acting on the object points in downward direction. Therefore, the angle between gravitational force and displacement of the object is 180°.

a. The work done by gravitational force in the upward motion.

Given that ∆r =5 m and F mg

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b. When the object falls back, both the gravitational force and displacement of the object are in the same direction. This implies that the angle between gravitational force and displacement of the object is 0°.

work done physics solved problems

c. The total work done by gravity in the entire trip (upward and downward motion)

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It implies that the gravity does not transfer any energy to the object. When the object is thrown upwards, the energy is transferred to the object by the external agency, which means that the object gains some energy. As soon as it comes back and hits the Earth, the energy gained by the object is transferred to the surface of the Earth (i.e., dissipated to the Earth).

Example 4.5

A weight lifter lifts a mass of 250 kg with a force 5000 N to the height of 5 m.

a. What is the workdone by the weight lifter?

b. What is the workdone by the gravity?

c. What is the net workdone on the object?

a. When the weight lifter lifts the mass, force and displacement are in the same direction, which means that the angle between them θ = 0 0 . Therefore, the work done by the weight lifter,

work done physics solved problems

b. When the weight lifter lifts the mass, the gravity acts downwards which means that the force and displacement are in opposite direction. Therefore, the angle between them θ = 180 0

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c. The net workdone (or total work done) on the object

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Solved Example Problems for Work done by a variable force

Example 4.6.

A variable force  F   =  k x 2  acts on a particle which is initially at rest. Calculate the work done by the force during the displacement of the particle from  x   =  0 m to  x   =  4 m. (Assume the constant  k   = 1 N m -2 )

work done physics solved problems

Solved Example Problems for  Kinetic energy

Example 4.7.

Two objects of masses 2 kg and 4 kg are moving with the same momentum of 20 kg m s -1 .

a. Will they have same kinetic energy?

b. Will they have same speed?

a. The kinetic energy of the mass is given by

work done physics solved problems

Note that  KE 1  ≠  KE 2  i.e., even though both are having the same momentum, the kinetic energy of both masses is not the same. The kinetic energy of the heavier object has lesser kinetic energy than smaller mass. It is because the kinetic energy is inversely proportional to the mass (KE   ∝  1/m) for a given momentum.

b. As the momentum,  p  =  mv , the two objects will not have same speed.

Solved Example Problems for  Potential Energy

Example 4.8.

An object of mass 2 kg is taken to a height 5 m from the ground  g =  10ms -2  .

a. Calculate the potential energy stored in the object.

b. Where does this potential energy come from?

c. What external force must act to bring the mass to that height?

d. What is the net force that acts on the object while the object is taken to the height ‘h’?

a. The  potential energy  U   = m g h  = 2  ×  10  ×  5 = 100 J

Here the positive sign implies that the energy is stored on the mass.

b. This potential energy is transferred from external agency which applies the force on the mass.

work done physics solved problems

d. From the definition of potential energy, the object must be moved at constant velocity. So the net force acting on the object is zero.

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Solved Example Problems for  Elastic Potential Energy

Example 4.9

Let the two springs A and B be such that k A >k B . On which spring will more work has to be done if they are stretched by the same force?

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The work done on the springs are stored as potential energy in the springs.

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k A >k B  implies that U B >U A  . Thus, more work is done on B than A.

Example 4.10

A body of mass m is attached to the spring which is elongated to 25 cm by an applied force from its equilibrium position.

a. Calculate the potential energy stored in the spring-mass system?

b. What is the work done by the spring force in this elongation?

c. Suppose the spring is compressed to the same 25 cm, calculate the potential energy stored and also the work done by the spring force during compression. (The spring constant, k = 0.1 N m -1 ).

The spring constant, k = 0.1 N m -1

The displacement,  x  = 25 cm = 0.25 m

a. The potential energy stored in the spring is given by

work done physics solved problems

Note that the potential energy is defined through the work done by the external agency. The positive sign in the potential energy implies that the energy is transferred from the agency to the object. But the work done by the restoring force in this case is negative since restoring force is in the opposite direction to the displacement direction.

c. During compression also the potential energy stored in the object is the same.

work done physics solved problems

Work done by the restoring spring force during compression is given by

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In the case of compression, the restoring spring force acts towards positive  x -axis and displacement is along negative  x  direction.

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Solved Example Problems for  Conservative and nonconservative forces

Example 4.11.

Compute the work done by the gravitational force for the following cases

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(As the displacement is in two dimension; unit vectors and are used)

a. Since the motion is only vertical, horizontal displacement component d x  is zero. Hence, work done by the force along path 1 (of distance h).

work done physics solved problems

Therefore, the total work done by the force along the path 2 is

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Note that the work done by the conservative force is independent of the path.

Example 4.12

Consider an object of mass 2 kg moved by an external force 20 N in a surface having coefficient of kinetic friction 0.9 to a distance 10 m. What is the work done by the external force and kinetic friction ? Comment on the result. (Assume g = 10 ms - 2 )

m  = 2 kg,  d  = 10 m,  F ext   = 20 N,    k   = 0.9.   When an object is in motion on the horizontal surface, it experiences two forces.

a. External force,  F ext   =  20 N

b. Kinetic friction,

f k  =μ k mg  = 0.9x(2)x10=18N.

The work done by the external force   W ext  = Fs = 20x20 =200J

The work done by the force of kinetic  friction  W k  = f k d  = (-18) x10=-180J  Here the negative sign implies that the  force of kinetic friction is  pposite to the  direction of displacement.

The total work done on the object 

W total  = Wext + W k  = 200 J – 180 J = 20 J .

Since the friction is a non-conservative force, out of 200 J given by the external force, the 180 J is lost and it can not be recovered.

Solved Example Problems for  Law of conservation of energy

Example 4.13.

An object of mass 1 kg is falling from the height  h  = 10 m. Calculate

a. The total energy of an object at  h   = 10 m

b. Potential energy of the object when it is at  h   =  4 m

c. Kinetic energy of the object when it is at  h   =  4 m

d. What will be the speed of the object when it hits the ground?

(Assume  g   = 10 m s -2 )

a. The gravitational force is a conservative force. So the total energy remains constant throughout the motion. At  h  = 10   m, the total energy E  is entirely   potential energy.

work done physics solved problems

b. The potential energy of the object at  h  =   4   m is

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c. Since the total energy is constant throughout the motion, the kinetic energy at  h   =  4 m must be  KE  =  E  -  U  =   100   -   40   =   60J

Alternatively, the kinetic energy could also be found from velocity of the object at 4 m. At the height 4 m, the object has fallen through a height of 6 m.

The velocity after falling 6 m is calculated from the equation of motion,

work done physics solved problems

d. When the object is just about to hit the ground, the total energy is completely kinetic and the potential energy,  U   =  0.

work done physics solved problems

Example 4.14

A body of mass 100 kg is lifted to a height 10 m from the ground in two different ways as shown in the figure. What is the work done by the gravity in both the cases? Why is it easier to take the object through a ramp?

work done physics solved problems

m = 100 kg, h = 10 m

Along path (1):

The minimum force  F 1  required to move the object to the height of 10 m should be equal to the gravitational force, F 1  mg = 100 x 10 = 1000 N

The distance moved along path (1) is,  = 10   m

The work done on the object along path (1) is

W = Fh = 1000 x 10 = 10,000 J

Along path (2):

In the case of the ramp, the minimum force  F 2   that we apply on the object to take it   up is not equal to  mg , it is rather equal to  mg  sin θ   . ( mg sin <  mg)  .

Here, angle θ = 30 o

Therefore, F 2   = mg sinθ = 100 × 10 ×  sin30 o  = 100 × 10 × 0.5 = 500N

Hence,  (mg sinθ < mg)

The path covered along the ramp is,

l = h/sin30 = 10/0.5 =20m

The work done on the object along path  (2) is, W = F2 l = 500 × 20 = 10,000 J

Since the gravitational force is a conservative force, the work done by gravity on the object is independent of the path taken.

In both the paths the work done by the gravitational force is 10,000 J

Along path (1): more force needs to be applied against gravity to cover lesser distance .

Along path (2): lesser force needs to be applied against the gravity to cover more distance.

As the force needs to be applied along the ramp is less, it is easier to move the object along the ramp.

Example 4.15

An object of mass m is projected from the ground with initial speed v 0 .

Find the speed at height h.

Since the gravitational force is conservative; the total energy is conserved throughout the motion.

work done physics solved problems

Final values of potential energy, kinetic energy and total energy are measured at the height  h .

By law of conservation of energy, the initial and final total energies are the same.

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Note that in section (2.11.2) similar result is obtained using kinematic equation based on calculus method. However, calculation through energy conservation method is much easier than calculus method.

Example 4.16

An object of mass 2 kg attached to a spring is moved to a distance  x   = 10 m from its equilibrium position. The spring constant  k  = 1   N m -1 and assume that the surface is   frictionless.

a. When the mass crosses the equilibrium position, what is the speed of the mass?

b. What is the force that acts on the object when the mass crosses the equilibrium position and extremum position  x  =  ±  10 m.

a. Since the spring force is a conservative force, the total energy is constant. At x  = 10   m, the total energy is purely   potential.

work done physics solved problems

When the mass crosses the equilibrium position  x  = 0 , the potential energy

work done physics solved problems

The entire energy is purely kinetic energy at this position.

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b. Since the restoring spring force is F = - kx, when the object crosses the equilibrium position, it experiences no force. Note that at equilibrium position, the object moves very fast. When the object is at  x  = +10 m (elongation), the force F = - k  x

F = - (1) (10) = - 10 N. Here the negative sign implies that the force is towards equilibrium i.e., towards negative  x -axis and when the object is at  x  = - 10 (compression), it experiences a forces F = - (1) (- 10) = +10 N. Here the positive sign implies that the force points towards positive  x -axis.

The object comes to momentary rest at  x  =  ± 10 m  even though it experiences a maximum force at both these points.

Solved Example Problems for  Motion in a vertical circle

Example 4.17.

Water in a bucket tied with rope is whirled around in a vertical circle of radius 0.5 m. Calculate the minimum velocity at the lowest point so that the water does not spill from it in the course of motion. (g = 10 ms -2 )

work done physics solved problems

Solved Example Problems for Unit of power

Example 4.18.

Calculate the energy consumed in electrical units when a 75 W fan is used for 8 hours daily for one month (30 days).

Power, P = 75 W

Time of usage, t = 8 hour × 30 days = 240 hours

Electrical energy consumed is the product of power and time of usage.

Electrical energy = power × time of usage = P × t

work done physics solved problems

Solved Example Problems for Relation between power and velocity

Example 4.19.

A vehicle of mass 1250 kg is driven with an acceleration 0.2 ms - 2  along a straight level road against an external resistive force 500 N . Calculate the power delivered by the vehicle’s engine if the velocity of the vehicle is 30 m  s - 1  .

The vehicle’s engine has to do work against resistive force and make vechile to move with an acceleration. Therefore, power delivered by the vehicle engine is

Solved Example Problems for collision

Solved Example Problems for Elastic collisions in one dimension

Example 4.20.

A lighter particle moving with a speed of 10 m s -1  collides with an object of double its mass moving in the same direction with half its speed. Assume that the collision is a one dimensional elastic collision. What will be the speed of both particles after the collision?

work done physics solved problems

Let the mass of the fi rst body be m which  moves with an initial velocity, u 1  = 10 m s -1 .

Therefore, the mass of second body is 2m and its initial velocity is u 2  = ½ u 1  = ½(10ms -1 )

Then, the fi nal velocities of the bodies can be calculated from the equation (4.53) and equation (4.54)

work done physics solved problems

As the two speeds v 1 and v 2  are positive, they move in the same direction with the velocities, 3.33 m s −1  and 8.33 m s −1  respectively.

Solved Example Problems for Perfect inelastic collision

Example 4.21.

A bullet of mass 50 g is fired from below into a suspended object of mass 450 g. The object rises through a height of 1.8 m with bullet remaining inside the object. Find the speed of the bullet. Take g = 10 ms -2 .

m 1  = 50 g = 0.05 kg; m 2  = 450 g = 0.45kg

work done physics solved problems

The speed of the bullet is u 1 . The second body is at rest u 2  = 0 . Let the common velocity of the bullet and the object after the bullet is embedded into the object is v.

work done physics solved problems

The combined velocity is the initial velocity for the vertical upward motion of the combined bullet and the object. From second equation of motion,

work done physics solved problems

Substituting this in the above equation, the value of u 1  is

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Solved Example Problems for Coefficient of restitution (e)

Example 4.22.

Show that the ratio of velocities of equal masses in an inelastic collision when one of the masses is stationary is

v 1 /v 2  = 1-e/1+e

work done physics solved problems

From the law of conservation of linear momentum,

work done physics solved problems

Using the equation (2) for u 1  in (1), we get

work done physics solved problems

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Mechanics: Work, Energy and Power

Power and Work Done

A series of free GCSE/IGCSE Physics Notes and Lessons. In these lessons, we will

  • Describe what is meant by power.
  • Calculate power using either energy or work done.

Related Pages IGCSE Physics Lessons

Power - Physics

The following diagram gives the formula for power and work done. Scroll down the page for more examples and solutions on how to use the formula.

Power and Work Done

Example: When a car stops, 40000J of work is done by the brakes in a time of 5s. Calculate the power of the brakes.

(a) How much work is done in lifting an object that weighs 250N to a height of 4m? (b) At what rate is work done if the 250N object is lifted to a height 4m in 4s?

A block of stone weighs 500 newtons. It is 20m pushed up a slope. (a) Neglecting friction, how much force is needed to push it up the incline at constant velocity? (b)(i) How much work is done to push it up the incline? (ii) How much work is done in lifting the block vertically 4m?

Someone takes 11 minutes to walk up a hill 120m high. His weight is 550N. (a) How much work did he do in climbing the hill? (b) How many seconds did he take to climb the hill? (c) What power is developed by his muscles as he climbs? (d) If he runs up the hill his muscles can exert 500 Watts. How long will it take him to get to the top of the hill?

An electric motor is 75% efficient. It is used to raise a weight of 1000N a vertical distance of 5m in 10s. (a) How much work muset the motor do in raising the weight? (b) What is the power developed by the motor as it raises the weight? (c) Given its efficiency, at what rate must power be supplied to the motor. (d) What happens to the energy used to raise the weight?

The escalator at Highgate underground station transports people up and sown a vertical height of 30m in one minute. (a) How much work was done by the motor of the escalator in taking someone with weight 750N from the bottom to the top? (b) Calculate the power exerted by the motor in raising each traveler. At busy times, the escalator moves more than 120 people up the stairs every minute. (c) What is the minimum amount of power required from the electric motors that drive the escalators. (Assume that each passenger has an average weight of 750N)

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IMAGES

  1. Solved Example Problems for Physics: Work, Energy and Power

    work done physics solved problems

  2. GCSE Physics Work done and energy transfer (AQA 9-1)

    work done physics solved problems

  3. Formula For Net Work Physics

    work done physics solved problems

  4. Work Done by a Constant Force

    work done physics solved problems

  5. Solved A 31 lb block sits on a friction-free inclined plane

    work done physics solved problems

  6. work done physics solved problems

    work done physics solved problems

VIDEO

  1. Solve* Any PHYSICS Numerical In 3 Easy STEPS😎| NEET 2024

  2. Physics 71 Day 14B [Part 1/8]

  3. WORK DONE|| PHYSICS||LECTURE 2|| XI

  4. A level Physics Worksheet on Work done, Energy and power

  5. Work, Energy & Power L4

  6. Work to be done#construction

COMMENTS

  1. Physics Work Problems for High Schools

    (a) The applied force (b) The normal force exerted by the table (c) The force of gravity Solution: In this problem, the force makes an angle with the displacement. In such cases, we should use the work formula W=Fd\cos\theta W = F dcosθ where \theta θ is the angle between force F F and displacement.

  2. Work example problems (video)

    Work example problems Google Classroom About Transcript David goes through some example problems on the concept of work. By reviewing these, you'll have a better knowledge of how to calculate work done by individual forces on an object in motion.

  3. Work, Energy, and Power Problem Sets

    Problem 1: Renatta Gass is out with her friends. Misfortune occurs and Renatta and her friends find themselves getting a work out. They apply a cumulative force of 1080 N to push the car 218 m to the nearest fuel station. Determine the work done on the car. Audio Guided Solution Show Answer Problem 2:

  4. Work and Power: Problems

    The magnitude of the force is given by F = ma = (10) (5) = 50 N. It acts over a distance of 20 m, in the same direction as the displacement of the object, implying that the total work done by the force is given by W = Fx = (50) (20) = 1000 Joules. Problem : A ball is connected to a rope and swung around in uniform circular motion.

  5. 7.3 Work-Energy Theorem

    W net = W grav = − m g ( y f − y i), where y is positive up. The work-energy theorem says that this equals the change in kinetic energy: − m g ( y f − y i) = 1 2 m ( v f 2 − v i 2). Using a right triangle, we can see that ( y f − y i) = ( s f − s i) sin θ, so the result for the final speed is the same.

  6. 7.1 Work

    The total work done by gravity is zero [20 J + 0 J + (− 20 J) = 0]. [20 J + 0 J + (− 20 J) = 0]. Unlike friction or other dissipative forces, described in Example 7.2, the total work done against gravity, over any closed path, is zero. Positive work is done against gravity on the upward parts of a closed path, but an equal amount of ...

  7. 9.1 Work, Power, and the Work-Energy Theorem

    Section Key Terms The Work-Energy Theorem In physics, the term work has a very specific definition. Work is application of force, f f, to move an object over a distance, d, in the direction that the force is applied. Work, W, is described by the equation W = fd. W = f d.

  8. Work/energy problem with friction (video)

    Since friction is always an opposing force you subtract this from the 38.5KJ and get the 8455J mentioned. This is the kinetic energy so 1/2mv^2 and you then multiply both sides by 2 and get 16910 = mv^2. The mass is 90kg so divide both sides by 90 and get v^2=187.8889. Square root this and you end up with 13.7m/s.

  9. Work Physics Problems with Solutions

    Work Physics Problems with Solutions Work is done when an object moves in the same direction, while the force is applied and also remains constant. Refer the below work physics problems with solutions and learn how to calculate force, work and distance. Work Example Problems Example 1:

  10. Calculating Work

    Correct answer: Explanation: The formula for work is , work equals force times distance. In this case, there is only one force acting upon the object: the force due to gravity. Plug in our given information for the distance to solve for the work done by gravity. Remember, since the object will be moving downward, the distance should be negative.

  11. 6.2: Work Done by a Constant Force

    Example Problems. Here are a few example problems: (1.a) Consider a constant force of two newtons (F = 2 N) acting on a box of mass three kilograms (M = 3 kg). Calculate the work done on the box if the box is displaced 5 meters. (1.b) Since the box is displaced 5 meters and the force is 2 N, we multiply the two quantities together.

  12. Work done by force

    Solution : W = Fx d = (5√3) (1) = 5√3 Joule 3. A body falls freely from rest, from a height of 2 m. If acceleration due to gravity is 10 m/s2, determine the work done by the force of gravity! Known : Object's mass (m) = 1 kg Height (h) = 2 m Acceleration due to gravity (g) = 10 m/s2 Wanted : Work done by the force of gravity (W) Solution :

  13. Work, Energy, and Power Problem Sets

    The problems below are provided as additional problems to the original set of 32 Work and Energy problems.Just like problems in the original set, these problems consist of a problem, an answer (revealed by clicking a link), an audio guided solution, and links to an Overview page of formulas and to The Physics Classroom Tutorial pages. The Extra Problems pertain to various aspects of the topic ...

  14. Solved Example Problems for Work and Work done by a force

    Physics : Work, Energy and Power : Work and Work done by a force Solved Example Problems for Work Example 4.1 A box is pulled with a force of 25 N to produce a displacement of 15 m. If the angle between the force and displacement is 30o, find the work done by the force. Solution Force, F = 25 N Displacement, dr = 15 m

  15. Work and Power Example Solutions

    4. What is the power output if 12N of force was applied on an object at an angle of 25° above the horizon to move an object 5 meters horizontally in 3 seconds. Use the work from the problem above. P = W/t. P = 54.5/3 = 18.2 W.

  16. PDF Physics

    Yes, it does feel like you are putting work in just by holding a heavy box, but in physics the definition of work requires that a force causes a displacement in order for work to be done. Our bodies need to use chemical energy to hold the box up, and we are producing a force to keep it in the air.

  17. How to Calculate Work Done

    Problem 1 A drum of oil of a total mass of 100 kilograms rolls down from the floor of a lorry 3.0 meters high. Calculate the work done by gravity on the load. Solution Data Mass of the drum, m = 100 kg

  18. 7.2 Kinetic Energy and the Work-Energy Theorem

    7.11. The quantity 1 2 mv 2 in the work-energy theorem is defined to be the translational kinetic energy (KE) of a mass m moving at a speed v. ( Translational kinetic energy is distinct from rotational kinetic energy, which is considered later.) In equation form, the translational kinetic energy, KE = 1 2 mv 2 , 7.12.

  19. Solved Example Problems for Physics: Work, Energy and Power

    1. Calculate the work done by a force of 30 N in lifting a load of 2kg to a height of 10m (g = 10ms-2) Answer: Given : Force mg = 30 N ; height = 10 m Work done to lift a load W = ? W = F.S (or) mgh = 30x10 W = 300 J Ans: 300J 2. A ball with a velocity of 5 m s-1 impinges at angle of 60˚ with the vertical on a smooth horizontal plane.

  20. Work, Energy, and Power Problem Sets

    Use the work equation to calculate the work done, a force value, or a displacement value. Includes 8 problems. Problem Set WE2: Work 2 Use the work equation to calculate the work done, a force value, or a displacement value. Includes 6 problems. Problem Set WE3: Work and Power 1

  21. How To Calculate Work On An Inclined Plane

    In this video, I tackle a problem involving pulling a block up a ramp and determining the net work acting on it. I explain the concept of work and emphasize ... AP Physics 1: Algebra-Based

  22. Power and Work Done (examples, solutions, videos, notes)

    Power - Physics. The following diagram gives the formula for power and work done. Scroll down the page for more examples and solutions on how to use the formula. Example: When a car stops, 40000J of work is done by the brakes in a time of 5s. Calculate the power of the brakes.