## 5 Steps to a 5: AP Physics 1: Algebra-Based 2017 (2016)

### STEP __4__

### Review the Knowledge You Need to Score High

### CHAPTER 13

### Work and Energy

**IN THIS CHAPTER**

**Summary:** An object possesses kinetic energy by moving. Interactions with other objects can create potential energy. Work is done when a force acts over a distance parallel to that force. When work is done on an object (or on a system of objects), kinetic energy can change. This chapter shows you how to recognize the different forms of energy and how to use them to make predictions about the behavior of objects.

**Definitions**

**Kinetic Energy** is possessed by any moving object. It comes in two forms:

is*Translational Kinetic Energy*

It exists when an object”s center of mass is moving.

is*Rotational Kinetic Energy*

It exists when an object rotates.

**Gravitational potential energy** is energy stored in a gravitational field. Near a planet, the formula is

where *h* is the vertical height above a reference position. A long way from a planet, the formula is

where *d* is measured from the planet”s center.

**Elastic potential energy,** also known as spring potential energy, is energy stored by a spring, given by

**Internal Energy** can refer to two somewhat different ideas. Both refer to the concept that multi-object systems can store energy depending on how the objects are arranged in the system.

** Microscopic internal energy **is related to the temperature of the object. As the object warms up, energy can be stored by the vibrations of molecules.

** Internal energy of a two-object system **is just another way of saying “potential energy.”

**Mechanical Energy** refers to the sum of potential and kinetic energies.

**Work** is done when a force acts on something that moves a distance parallel to that force.

**Power** is defined as energy used per second, or work done per second.

**Energy**

The cart pictured above has gravitational potential energy with respect to the location of the motion detector, because the cart is vertically higher than the detector. If the cart were released from rest, it would speed up toward the detector. It would be tempting to try to use the kinematics equations to determine the cart”s maximum speed. However, since the track is curved, the cart”s acceleration will be changing throughout its motion. Whenever acceleration changes, kinematics as studied in __Chapter 10__ are invalid. The methods of energy conservation, as described in this chapter, must be used.

The College Board”s curriculum guide for AP Physics 1 makes much of the difference between **objects** and **systems** . A system is just a collection of objects. In your class, you might well have talked about an object”s kinetic and potential energies. The thing is, the exam development committee doesn”t like that language. Sure, an object can have kinetic energy, just by moving. But a single object technically cannot “have” potential energy.

Why not? Potential energy is always the result of an interaction between objects in a system. For example, gravitational potential energy (equation: *mgh* ) exists only if an object is interacting with the Earth. The Earth-object system stores the potential energy, not just the object itself. Similarly, an object attached to a spring *cannot* store potential energy; the spring-object system stores the energy.

Look, I”m going to talk about objects “having” potential energy. It”s okay with me if you talk about a block on a spring “having” potential energy because the spring is compressed. You”ll still get pretty much everything right on the exam. Just know that the block only “has” potential energy because of its interaction with the spring; and that potential energy is sometimes referred to as the “internal energy of the block-spring system.”

**Work**

**FACT:** Work is done when a force is exerted on an object^{ }^{1}^{ }and that object moves parallel to the direction of the force.

The relevant equation is work = force times parallel displacement:^{ }^{2}

When a force is exerted in the same direction as the object”s motion, the work done is considered to be a positive quantity; when a force is exerted in the opposite direction of the object”s motion, the work done is considered to be a negative quantity. The *net* work on an object is the algebraic sum of the work done by each force.

**Example 1:** A string applies a 10-N force to the right on a 2-kg box, dragging it at constant speed across the floor for a distance of 50 cm.

Let”s calculate the work done by each force acting, and the net work done on the box. Start by sketching a free-body diagram for the box.

** Strategy: **Whenever you are calculating work done by a force with the equation

*W*=

*F*Δ

*x*

_{||}, always sketch the direction of the force and displacement vectors.

Consider each force separately. Start with the 10-N tension in the string. This force acts to the right. The displacement of the box is 50 cm (i.e., 0.50 m) to the right.

Since the force is in the same direction as the displacement, the work done by the tension is just 10 N times 0.50 m = +5 J.

Now consider the friction force. Since the box moves at constant speed, we know the left force (friction) equals the right force (tension); so the friction force has an amount of 10 N.

Since the force is in the *opposite* direction of displacement, the work done by the friction force is 10 N times 0.50, with a negative sign. Thus, the work done by friction is −5 J.

What about the force of gravity? The 20-N weight of the box points downward.

Since no component of the weight is parallel to the displacement, the force of the Earth does zero work on the box.

Similarly, the normal force is straight upward while the displacement is to the right; since no component of the normal force is parallel to the displacement, the normal force does no work on the box.

**Exam Tip from an AP Physics Veteran**

If a force acts at an angle to the displacement, just break that force into components. The component perpendicular to the displacement does no work. The component parallel to the displacement can be multiplied by the displacement to get the work done.

**Finding the Net Work**

You can calculate the net work on the box in two ways:

- First, determine the net force using a free-body diagram, like we showed in
__Chapter 11__. Then, multiply the component of the net force that”s parallel to the displacement by the displacement, just like you would when finding the work done by any force. In Example 1, the net force is zero; so there is no net work done on the box. - First, determine the work done by each force separately. Then, add the work done by each force algebraically (i.e., including negative signs). In Example 1, add the +5 J done by the tension to the −5 J done by friction (and the 0 J done by the normal force and the weight) to get zero net work.

**Conservative Versus Nonconservative Forces**

**FACT:** A “conservative” force converts potential energy to other forms of mechanical energy when it does work. Thus, a conservative force does not change the mechanical energy of a system.

The amount of work done by a conservative force depends only on the starting and ending positions of the object, i.e., it”s “path independent.” The only conservative forces that you need to deal with on the AP Physics 1, Algebra-Based Exam are gravity and springs. When a spring does work on an object, energy is stored in the spring that can be recovered and converted back to kinetic energy. The sum of the potential and kinetic energy of the object-spring system is constant.

Conversely, a “nonconservative” force can change the mechanical energy of a system. Friction is the most common example: Work done by friction on an object becomes microscopic internal energy in the object, raising the object”s temperature. That microscopic internal energy *cannot* be recovered and converted back to kinetic energy. Other nonconservative forces might include, for example, the propeller of an airplane—it does work on the airplane to increase the airplane”s mechanical energy.

**The Work-Energy Theorem**

In your textbook, you”ll see the work-energy theorem written as “net work = change in kinetic energy.” That”s certainly true—*net* work done on an object must change the object”s kinetic energy. The tricky part is, net work must include work done by all forces, conservative and nonconservative.

I think it”s easier to separate conservative and nonconservative forces. Work done by a nonconservative force (*W* _{NC} ) changes the total mechanical energy of a system (*KE* + *PE* ). I write the work-energy theorem as follows:

Generally, the potential energy involved will be either that due to a spring, or due to a gravitational field. The kinetic energy includes *both* translational kinetic energy (½*mv* ^{2} ) and rotational kinetic energy^{ }^{3}^{ }(½*Iω* ^{2} ).

**Example 2:** An archer pulls an arrow of mass 0.10 kg attached to a bowstring back 30 cm by exerting a force that increases uniformly with distance from 0 N to 200 N.

The AP Exam could ask all sorts of questions about this situation. Before you start doing any calculation, categorize the problem. There are only three ways to approach a mechanics problem: kinematics/Newton laws, momentum, and energy. There”s no collision, so momentum is unlikely to be useful. The problem talks about a force; but that force is *changing* . A changing force means a changing acceleration, which means that kinematics equations are not valid. Only the work-energy theorem will be useful.

The only types of potential energy used in AP Physics 1 are due to gravity (*mgh* ) and due to a spring (½*kx* ^{2} ). Which is involved here? The example says that the force of the string varies “uniformly,” which means that the force gets bigger as the distance stretched gets bigger, just like a spring. So treat the bowstring just like a spring.

**Mistake:** An interesting question here might be “how much work does the archer do in pulling back the bowstring?” And you”d be tempted to use the definition of work, *W* = *F* Δ*x* _{||} . But, no, since the force of the archer on the string is changing, this equation for work does not apply. Instead, you must use the entire work-energy theorem.

To find the work done by the archer in pulling back the bowstring, write the work-energy theorem, considering the time from when he starts pulling until the maximum extension. Since the arrow is at rest before the archer starts pulling, and is *still* at rest when the string is pulled all the way back, the change in kinetic energy is zero. The potential energy of a spring is zero at the equilibrium position and is ½*kx* ^{2} at full extension. The work done by the archer is a nonconservative force, since it changes the mechanical energy of the string-arrow system. We get

*W* _{NC} = (0) + (½*kx* ^{2} − 0)

But this isn”t solvable yet—we know the distance *x* the archer pulled to be 0.30 m (i.e., 30 cm). But what is *k* , the spring constant of the “spring”? Use the equation *F* = *kx* when the bow is fully extended. The problem says the maximum force the archer pulls with is 200 N when the string is extended 0.30. Plugging into *F* = *kx* and solving for *k* gives *k* = 670 N/m.^{ }^{4}

Now the work done by the archer is ½(670 N/m)(0.30 m)^{2} = 30 J.

The AP Exam could certainly ask for this calculation. Or, the exam might ask, “If the archer instead pulls back 60 cm, what will happen to the work done by the archer?” The work-energy theorem still applies, the kinetic energy terms still go away, and the work done is still ½*kx* ^{2} . The spring constant is a property of a spring (or in this case, a bowstring). So *k* doesn”t change. We doubled the displacement from equilibrium, *x* . Since the variable *x* is squared, then we don”t multiply the work done by a factor of two when we double *x* ; we multiply the work done by a factor of 2^{2} (i.e., by a factor of four). The archer has to do four times as much work.

How about finding out how fast the arrow would be traveling when the archer shoots it? We cannot use kinematics with a varying net force or a varying acceleration. Use the work-energy theorem again; except this time, let the problem start when the archer released the bowstring, and end when the string gets back to the equilibrium position and the arrow is released. Now, since no nonconservative forces act,^{ }^{5}^{ }the equation becomes

0 = (0 − ½*mv* ^{2} ) + (½*kx* ^{2} − 0)

The kinetic energy goes from zero to something; while we don”t know the value of the arrow”s final kinetic energy, we know the equation for that kinetic energy is ½*mv* ^{2} . The potential energy goes from ½*kx* ^{2} to zero, because for a spring *x* = 0 is by definition the equilibrium position.

Whenever all the forces acting are conservative, mechanical energy is conserved. This means that energy can be changed from potential to kinetic or back, but the total mechanical energy must remain the same always. Here the initial potential energy was 30 J—that”s the same ½*kx* ^{2} calculation that we did above. These 30 J of potential energy are entirely converted to kinetic energy. So set 30 J = ½*mv* ^{2} and solve for *v* . (Use the 0.10-kg mass of the arrow for *m* .) The speed is 24 m/s.

When you are asked to do a calculation, it”s worth asking: Is the answer reasonable? One m/s is about 2 miles per hour.^{ }^{6}^{ }This arrow was traveling in the neighborhood of 50 mph—about the speed of a car, but less than the speed of a professional baseball pitch.^{ }^{7}

**Exam Tip from an AP Physics Veteran**

If the spring were instead hanging vertically instead of vibrating horizontally, this problem would be *solved the same way* . When you have a vertical spring, define the equilibrium position *x* = 0 as the place where the mass would hang at permanent rest. Then, *F* = *kx* gives the *net* force on the hanging object, rather than just the force applied by the spring; and you can use *PE* = ½*kx* ^{2} (not *mgh* ) to calculate the object”s potential energy.

**Power**

Whether you walk up a mountain or whether a car drives you up the mountain, the same amount of work has to be done on you. (You weigh a certain number of newtons, and you have to be lifted up the same distance either way!) But clearly there”s something different about walking up over the course of several hours and driving up over several minutes. That difference is power.

Power is, thus, measured in units of joules/second, also known as watts. A car engine puts out hundreds of horsepower, equivalent to maybe 100 kilowatts; whereas, you”d work hard just to put out a power of a few hundred watts.

** Practice Problems**

** Note: **An additional drill involving graphical analysis of a mass on a spring is available in

__Chapter 18__.

** 1 .** A 0.5-kg cart released from rest at the top of a smooth incline has gravitational energy of 6 J relative to the base of the incline.

(a) Calculate the cart”s speed at the bottom of the incline.

(b) When the cart has rolled halfway down the incline, the cart”s gravitational potential energy will be:

(A) Greater than 3 J

(B) Less than 3 J

(C) Equal to 3 J

Justify your answer.

(c) When the cart has rolled halfway down the incline, the cart”s kinetic energy will be

(A) Greater than 3 J

(B) Less than 3 J

(C) Equal to 3 J

(D) Unknown without knowledge of the cart”s speed

Justify your answer.

(d) When the cart has rolled halfway down the incline, the cart”s speed will be:

(A) Half of its speed at the bottom

(B) Greater than half of its speed at the bottom

(C) Less than half of its speed at the bottom

(e) A 1.0-kg cart is released from rest at the top of the same incline. At the bottom, it will be moving

(A) Faster than the 0.5-kg cart

(B) Slower than the 0.5-kg cart

(C) The same speed as the 0.5-kg cart

** 2 .** A 0.1 kg stone sits at rest on top of a compressed vertical spring. The potential energy stored in the spring-earth-stone system is 40 J. The spring is released, throwing the stone straight up into the air.

(a) How much kinetic energy will the stone have when it first leaves the spring?

(b) How much gravitational energy, relative to the spot where the stone was released, will the stone have when it reaches the peak of its flight?

(c) Calculate the height above the release point to which the stone travels.

(d) Suggest something we could change about this situation that would cause the stone to reach a height double that calculated in Part (c).

** 3 .** In the laboratory, a motion detector records the speed of a cart as a function of time, stopping its reading when the cart is 15 cm in front of the detector at the line marked on the track. The cart is released from rest at the position shown.

(a) The kinetic energy of the cart at the line marked on the track is equal to the gravitational energy *mgh* of the cart at its initial position. On the preceding diagram, draw and label the distance you would measure for the height *h*of the cart.

(b) Explain in some detail how commonly available laboratory equipment could be used to measure the labeled height *h* .

(c) If the height *h* were doubled in a second trial, the motion detector would read

(A) The same speed as in the first trial

(B) Two times the speed in the first trial

(C) Four times the speed in the first trial

(D) times the speed in the first trial

Justify your answer.

** 4 .** Student A lifts a 50-N box from the floor straight up to a height of 40 cm in 2 s. Student B lifts a 40-N box straight up from the floor to a height of 50 cm in 1 s.

(a) Compared to Student A, Student B does

(A) The same work but develops more power

(B) The same work but develops less power

(C) More work but develops less power

(D) Less work but develops more power

Justify your answer

(b) Now Student A instead lifts the 50-N box from the floor diagonally, moving the box 40 cm to the right and 40 cm upward in the same 2 s.

(A) Compared to the work he did originally, does Student A do more, less, or the same work?

(B) Compared to the power he developed originally, does Student A develop more, less, or the same power?

** Solutions to Practice Problems**

** 1 .** (a) Here gravitational energy is converted to kinetic energy. (“Smooth” generally means friction is negligible.) Kinetic energy at the bottom will be 6 J, which is equal to ½

*mv*

^{2}. Plug in the 0.5-kg mass of the cart and solve for

*v*to get 4.9 m/s.

(b) Gravitational potential energy is *mgh* . Since the *h* term is in the numerator and not squared or square rooted, cutting *h* in half cuts the whole equation in half as well; so the cart”s gravitational potential energy will be 3 J.

(c) Any gravitational potential energy lost by the cart must be converted to kinetic energy. The cart lost 3 J of gravitational energy, so the cart now has 3 J of kinetic energy.

(d) At the bottom, to find the speed we set 6 J = ½*mv* ^{2} . Solving for the cart”s speed, we get . We”re going to cut that 6 J term in half. Since the 6 J is under the square root, though, we don”t cut the speed in half; instead, we multiply the speed by . If you don”t see why that gives a speed greater than half the speed at the bottom, try carrying out the entire calculation—you should get 3.5 m/s.

(e) The energy conversion is the same—we”re setting gravitational energy (*mgh* ) at the top equal to kinetic energy (½*mv* ^{2} ) at the bottom. Notice the mass on both sides—the mass cancels, and so doesn”t affect the result. Thus, the speed is the same for either cart. Again, feel free to do the calculation with *m* = 1.0 kg to get 4.9 m/s again.

** 2 .** (a) No forces external to the spring-earth-stone system act on the block. Therefore, mechanical energy is conserved. The 40 J of potential energy is converted to the stone”s kinetic energy, or 40 J.

(b) Again, mechanical energy is conserved. The 40 J of kinetic energy are now converted entirely to 40 J of gravitational energy.

(c) That 40 J of gravitational energy at the peak can be set equal to *mgh* . Solve for *h* to get 40 m.

(d) The energy conversion here is spring energy → gravitational energy. Mathematically, that”s ½*kx* ^{2} = *mgh* . Solving for *h* , . To double the height, we could use a spring with double the spring constant of the original spring, because *k* is unexponented^{ }^{8}^{ }in the numerator. We could compress the spring 1.4 times its original compression, since when the *x* in the numerator is squared that would multiply the whole expression by 2. We could use a rock of mass 0.05 kg; with *m* in the denominator, halving the mass doubles the entire expression. (Okay, I suppose we could go to some new planet where *g* is 5 N/kg. If you will fund that trip, I”ll give you credit for that answer.)

** 3 .** (a) In the equation

*mgh, h*represents the vertical distance above the lowest position or some reference point. Here the reference point is the line on the track. The motion detector reads the front of the cart, so

*h*must be measured to the front of the cart, not the middle or back. See above for the answer.

(b) Use a meterstick, obviously, but it”s not an easy measurement to make. First, measure the vertical distance from the desk to the line on the track 15 cm in front of the detector. Then measure the vertical distance from the desk to the track directly under the front of the cart; then subtract the two distance measurements. You can get more accurate measurements if you use a bubble level and plumb bob to ensure the table is horizontal and the measurements are vertical. (If you want to measure along the track the distance from the front of the cart to the line, use an angle measurer to get the angle of the track, then use trigonometry.)

(c) The energy conversion here is gravitational energy → kinetic energy. In equations, that”s *mgh* = ½*mv* ^{2} . Solve for *v* to get . The variable *h* is in the numerator but under the square root, so doubling *h*multiplies the speed by the square root of 2, choice D.

** 4 .** (a) The work done by the student is equal to the change in the box”s gravitational potential energy—that”s

*mgh*. The time it takes the student to lift the box doesn”t depend on time at all. Plugging in, we find that Student A does (5 kg)(10 N/kg)(0.40 m) = 20 J of work on the box. Student B does (4 kg)(10 N/kg)(0.50 m) = 20 J of work, also. Now, power is work divided by the time it takes to do that work. Since they do the same amount of work, whoever takes less time to do the work develops more power. That”s Student B. So the answer is choice A.

(b) (A) As above, the work done by Student A on the box is *mgh* . Here *h* represents the vertical height above the lowest position. Since that vertical height is still 40 cm, Student A has done the same work. (You could also recognize that the horizontal displacement is not parallel to the box”s weight, or to the force Student A applied to lift the box.^{ }^{9}^{ })

(B) Since the work done by Student A is the same as before, and it took the same amount of time, the power (= work/time) is the same.

** Rapid Review**

- Work is done when a force is exerted on an object, and that object moves parallel to the direction of the force.
- A “conservative” force converts potential energy to other forms of mechanical energy when it does work. Thus, a conservative force does not change the mechanical energy of a system.
- The only types of potential energy used in AP Physics 1 are due to gravity (
*mgh*) and due to a spring (½*kx*^{2}). - The work-energy theorem can be written as
*W*_{NC}= (Δ*KE*) + (Δ*PE*), where*W*_{NC}is the work done by a nonconservative force. - Whenever the force on an object is not steady, energy conservation methods must be used to solve the problem. The most common of these situations are curved tracks, springs, and pendulums.

__ ^{1}__ Or on a system of objects—if a force is exerted on a system of objects and the system”s center of mass moves parallel to the force, work was done.

__ ^{2}__ Or, equivalently, the equation is displacement times parallel force, which may sometimes be a more convenient expression.

__ ^{3}__ See

__Chapter 14__for further discussion of rotational kinetic energy.

__ ^{4}__ No,

*not*666.6666666666 N/m. Use two significant figures, unless you want your 10th-grade chemistry teacher to have heart palpitations.

__ ^{5}__ The force of the bowstring is a conservative force, because the potential energy it stores is part of the mechanical energy of the bowstring-arrow system.

__ ^{6}__ For those of you who didn”t grow up in America, 1 m/s is a bit less than 4 km/hr.

__ ^{7}__ For those of you who are uncomfortable with the World”s Greatest Sport, this is less than the speed of a tennis player”s serve.

__ ^{8}__ If “unexponented” is a word…

__ ^{9}__ The force Student A applies on the box is straight up, at least while the box is moving at constant speed. Once the box starts moving horizontally, no force in that direction is necessary to continue its motion—that”s Newton”s first law.