problem solving about work in physics

9.1 Work, Power, and the Work–Energy Theorem

Section learning objectives.

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

Describe and apply the work–energy theorem
Describe and calculate work and power

Teacher Support

The learning objectives in this section will help your students master the following standards:

(A) describe and apply the work–energy theorem;
(C) describe and calculate work and power.

In addition, the High School Physics Laboratory Manual addresses the following standards:

Use the lab titled Work and Energy as a supplement to address content in this section.

Section Key Terms

In this section, students learn how work determines changes in kinetic energy and that power is the rate at which work is done.

[BL] [OL] Review understanding of mass, velocity, and acceleration due to gravity. Define the general definitions of the words potential and kinetic .

[AL] [AL] Remind students of the equation W = P E e = f m g W = P E e = f m g . Point out that acceleration due to gravity is a constant, therefore PE e that results from work done by gravity will also be constant. Compare this to acceleration due to other forces, such as applying muscles to lift a rock, which may not be constant.

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

Some things that we typically consider to be work are not work in the scientific sense of the term. Let’s consider a few examples. Think about why each of the following statements is true.

Homework is not work.
Lifting a rock upwards off the ground is work.
Carrying a rock in a straight path across the lawn at a constant speed is not work.

The first two examples are fairly simple. Homework is not work because objects are not being moved over a distance. Lifting a rock up off the ground is work because the rock is moving in the direction that force is applied. The last example is less obvious. Recall from the laws of motion that force is not required to move an object at constant velocity. Therefore, while some force may be applied to keep the rock up off the ground, no net force is applied to keep the rock moving forward at constant velocity.

[BL] [OL] Explain that, when this theorem is applied to an object that is initially at rest and then accelerates, the 1 2 m v 1 2 1 2 m v 1 2 term equals zero.

[OL] [AL] Work is measured in joules and W = f d W = f d . Force is measured in newtons and distance in meters, so joules are equivalent to newton-meters ( N ⋅ m ) ( N ⋅ m )

Work and energy are closely related. When you do work to move an object, you change the object’s energy. You (or an object) also expend energy to do work. In fact, energy can be defined as the ability to do work. Energy can take a variety of different forms, and one form of energy can transform to another. In this chapter we will be concerned with mechanical energy , which comes in two forms: kinetic energy and potential energy .

Kinetic energy is also called energy of motion. A moving object has kinetic energy.
Potential energy, sometimes called stored energy, comes in several forms. Gravitational potential energy is the stored energy an object has as a result of its position above Earth’s surface (or another object in space). A roller coaster car at the top of a hill has gravitational potential energy.

Let’s examine how doing work on an object changes the object’s energy. If we apply force to lift a rock off the ground, we increase the rock’s potential energy, PE . If we drop the rock, the force of gravity increases the rock’s kinetic energy as the rock moves downward until it hits the ground.

The force we exert to lift the rock is equal to its weight, w , which is equal to its mass, m , multiplied by acceleration due to gravity, g .

The work we do on the rock equals the force we exert multiplied by the distance, d , that we lift the rock. The work we do on the rock also equals the rock’s gain in gravitational potential energy, PE e .

Kinetic energy depends on the mass of an object and its velocity, v .

When we drop the rock the force of gravity causes the rock to fall, giving the rock kinetic energy. When work done on an object increases only its kinetic energy, then the net work equals the change in the value of the quantity 1 2 m v 2 1 2 m v 2 . This is a statement of the work–energy theorem , which is expressed mathematically as

The subscripts 2 and 1 indicate the final and initial velocity, respectively. This theorem was proposed and successfully tested by James Joule, shown in Figure 9.2 .

Does the name Joule sound familiar? The joule (J) is the metric unit of measurement for both work and energy. The measurement of work and energy with the same unit reinforces the idea that work and energy are related and can be converted into one another. 1.0 J = 1.0 N∙m, the units of force multiplied by distance. 1.0 N = 1.0 kg∙m/s 2 , so 1.0 J = 1.0 kg∙m 2 /s 2 . Analyzing the units of the term (1/2) m v 2 will produce the same units for joules.

Watch Physics

Work and energy.

This video explains the work energy theorem and discusses how work done on an object increases the object’s KE.

Grasp Check

True or false—The energy increase of an object acted on only by a gravitational force is equal to the product of the object's weight and the distance the object falls.

Repeat the information on kinetic and potential energy discussed earlier in the section. Have the students distinguish between and understand the two ways of increasing the energy of an object (1) applying a horizontal force to increase KE and (2) applying a vertical force to increase PE.

Calculations Involving Work and Power

In applications that involve work, we are often interested in how fast the work is done. For example, in roller coaster design, the amount of time it takes to lift a roller coaster car to the top of the first hill is an important consideration. Taking a half hour on the ascent will surely irritate riders and decrease ticket sales. Let’s take a look at how to calculate the time it takes to do work.

Recall that a rate can be used to describe a quantity, such as work, over a period of time. Power is the rate at which work is done. In this case, rate means per unit of time . Power is calculated by dividing the work done by the time it took to do the work.

Let’s consider an example that can help illustrate the differences among work, force, and power. Suppose the woman in Figure 9.3 lifting the TV with a pulley gets the TV to the fourth floor in two minutes, and the man carrying the TV up the stairs takes five minutes to arrive at the same place. They have done the same amount of work ( f d ) ( f d ) on the TV, because they have moved the same mass over the same vertical distance, which requires the same amount of upward force. However, the woman using the pulley has generated more power. This is because she did the work in a shorter amount of time, so the denominator of the power formula, t , is smaller. (For simplicity’s sake, we will leave aside for now the fact that the man climbing the stairs has also done work on himself.)

Power can be expressed in units of watts (W). This unit can be used to measure power related to any form of energy or work. You have most likely heard the term used in relation to electrical devices, especially light bulbs. Multiplying power by time gives the amount of energy. Electricity is sold in kilowatt-hours because that equals the amount of electrical energy consumed.

The watt unit was named after James Watt (1736–1819) (see Figure 9.4 ). He was a Scottish engineer and inventor who discovered how to coax more power out of steam engines.

[BL] [OL] Review the concept that work changes the energy of an object or system. Review the units of work, energy, force, and distance. Use the equations for mechanical energy and work to show what is work and what is not. Make it clear why holding something off the ground or carrying something over a level surface is not work in the scientific sense.

[OL] Ask the students to use the mechanical energy equations to explain why each of these is or is not work. Ask them to provide more examples until they understand the difference between the scientific term work and a task that is simply difficult but not literally work (in the scientific sense).

[BL] [OL] Stress that power is a rate and that rate means "per unit of time." In the metric system this unit is usually seconds. End the section by clearing up any misconceptions about the distinctions between force, work, and power.

[AL] Explain relationships between the units for force, work, and power. If W = f d W = f d and work can be expressed in J, then P = W t = f d t P = W t = f d t so power can be expressed in units of N ⋅ m s N ⋅ m s

Also explain that we buy electricity in kilowatt-hours because, when power is multiplied by time, the time units cancel, which leaves work or energy.

Links To Physics

Watt’s steam engine.

James Watt did not invent the steam engine, but by the time he was finished tinkering with it, it was more useful. The first steam engines were not only inefficient, they only produced a back and forth, or reciprocal , motion. This was natural because pistons move in and out as the pressure in the chamber changes. This limitation was okay for simple tasks like pumping water or mashing potatoes, but did not work so well for moving a train. Watt was able build a steam engine that converted reciprocal motion to circular motion. With that one innovation, the industrial revolution was off and running. The world would never be the same. One of Watt's steam engines is shown in Figure 9.5 . The video that follows the figure explains the importance of the steam engine in the industrial revolution.

Initiate a discussion on the historical significance of suddenly increasing the amount of power available to industries and transportation. Have students consider the fact that the speed of transportation increased roughly tenfold. Changes in how goods were manufactured were just as great. Ask students how they think the resulting changes in lifestyle compare to more recent changes brought about by innovations such as air travel and the Internet.

Watt's Role in the Industrial Revolution

This video demonstrates how the watts that resulted from Watt's inventions helped make the industrial revolution possible and allowed England to enter a new historical era.

Which form of mechanical energy does the steam engine generate?

Potential energy
Kinetic energy
Nuclear energy
Solar energy

Before proceeding, be sure you understand the distinctions among force, work, energy, and power. Force exerted on an object over a distance does work. Work can increase energy, and energy can do work. Power is the rate at which work is done.

Worked Example

Applying the work–energy theorem.

An ice skater with a mass of 50 kg is gliding across the ice at a speed of 8 m/s when her friend comes up from behind and gives her a push, causing her speed to increase to 12 m/s. How much work did the friend do on the skater?

The work–energy theorem can be applied to the problem. Write the equation for the theorem and simplify it if possible.

Identify the variables. m = 50 kg,

Substitute.

Work done on an object or system increases its energy. In this case, the increase is to the skater’s kinetic energy. It follows that the increase in energy must be the difference in KE before and after the push.

Tips For Success

This problem illustrates a general technique for approaching problems that require you to apply formulas: Identify the unknown and the known variables, express the unknown variables in terms of the known variables, and then enter all the known values.

Identify the three variables and choose the relevant equation. Distinguish between initial and final velocity and pay attention to the minus sign.

Practice Problems

Identify which of the following actions generates more power. Show your work.

carrying a 100 N TV to the second floor in 50 s or
carrying a 24 N watermelon to the second floor in 10 s ?
Carrying a 100 N TV generates more power than carrying a 24 N watermelon to the same height because power is defined as work done times the time interval.
Carrying a 100 N TV generates more power than carrying a 24 N watermelon to the same height because power is defined as the ratio of work done to the time interval.
Carrying a 24 N watermelon generates more power than carrying a 100 N TV to the same height because power is defined as work done times the time interval.
Carrying a 24 N watermelon generates more power than carrying a 100 N TV to the same height because power is defined as the ratio of work done and the time interval.

Check Your Understanding

work and force
energy and weight
work and energy
weight and force

When a coconut falls from a tree, work W is done on it as it falls to the beach. This work is described by the equation

Identify the quantities F , d , m , v 1 , and v 2 in this event.

F is the force of gravity, which is equal to the weight of the coconut, d is the distance the nut falls, m is the mass of the earth, v 1 is the initial velocity, and v 2 is the velocity with which it hits the beach.
F is the force of gravity, which is equal to the weight of the coconut, d is the distance the nut falls, m is the mass of the coconut, v 1 is the initial velocity, and v 2 is the velocity with which it hits the beach.
F is the force of gravity, which is equal to the weight of the coconut, d is the distance the nut falls, m is the mass of the earth, v 1 is the velocity with which it hits the beach, and v 2 is the initial velocity.
F is the force of gravity, which is equal to the weight of the coconut, d is the distance the nut falls, m is the mass of the coconut, v 1 is the velocity with which it hits the beach, and v 2 is the initial velocity.

Use Check Your Understanding questions to assess students’ achievement of the section’s learning objectives. If students are struggling with a specific objective, the Check Your Understanding will help identify which one and direct students to the relevant content.

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Work and Energy

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.

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

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 $200\,{\rm m}$?

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, $\theta=30^\circ$.

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

Write something.
Write something else.
Write something different.
Write something completely different.
the acceleration of the bullet in the rifle
the force of the propellant on the bullet in the rifle
the work done on the bullet while it is in the barrel
the horizontal acceleration of the bullet as it flies through the air
the force of aerodynamic drag on the bullet
the work done by aerodynamic drag on the bullet
How far did the block move during this part of the experiment?
How much work was done on the block during this part of the experiment?
coefficient of static friction for wood on wood
coefficient of kinetic friction for wood on wood
Draw a free body diagram showing all the forces acting on the model rocket.
the weight of the rocket
the net force on the rocket while the engine was running
the acceleration of the rocket while the engine was running
the distance traveled by the rocket while the engine was running
the speed of the rocket when the engine stopped
the work done by the engine on the rocket
Draw a free body diagram showing all the forces acting on the model rocket after the engine shut down.
What is the acceleration of the rocket after the engine shut down?
What maximum height above the ground did the rocket reach?
How much work did gravity do on the rocket from launch until it reached its maximum height?
If your answers to part g. and part k. are not equal (to within 2 or 3 significant digits), you've made a mistake somewhere. If they are equal, you've probably done it correctly (probably).
the weight of the car
the force of the tires pushing the car up the ramp
the distance the car traveled up the ramp
the increase in height of the car
the work done by the engine pushing the car up the ramp
Draw a free body diagram showing all the forces acting on the lawnmower. Do not resolve any of the forces into components. Do indicate their directions, however.
the gravitational force of the Earth
the normal force of the ground
the force applied by the homeowner
the force of friction from the ground
What does the area under this curve represent?
Calculate its cumulative value at 200 m intervals. Compile your results in a table like the one below.
Sketch a graph of this quantity with respect to displacement.

statistical

Use the given data to create a force-displacement graph.
Determine the work done on the projectile as a function of its displacement.
Compute the launch speed of the projectile.

Data adapted from Kampen, Kaczmarczik, and Rath; 2006 .

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Physics library

Course: physics library > unit 5.

Introduction to work and energy
Work and energy (part 2)
Conservation of energy
What are energy and work?
What is kinetic energy?
What is gravitational potential energy?
What is conservation of energy?
Work and the work-energy principle
Work as the transfer of energy
Work example problems
Work as area under curve
Thermal energy from friction
What is thermal energy?

Work/energy problem with friction

Conservative forces
What is power?

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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.