## 5 Steps to a 5: AP Physics C (2016)

### STEP __4__

### Review the Knowledge You Need to Score High

### CHAPTER 14

### Energy Conservation

**IN THIS CHAPTER**

**Summary:** While kinematics can be used to predict the speeds of objects with constant acceleration, energy conservation is a more powerful tool that can predict how objects will move even with a changing acceleration.

**Key Ideas**

Work is related to kinetic energy through the work–energy theorem.

There are many types of potential energy. Two (due to gravity and due to a spring) are discussed in this chapter.

To use conservation of energy, add potential + kinetic energy at two positions in an object”s motion. This sum must be the same everywhere.

A potential energy function can be derived for any conservative force.

**Relevant Equations**

The definition of work:

*W* = *F* ·*d* _{||}

The work–energy theorem:

*W* _{net} = ΔK

The force of a spring:

*F* = −*kx*

Two different types of potential energy:

Power:

Relationship between a conservative force *F* and the potential energy *U* it creates:

As with momentum, the energy of an isolated system is always conserved. It might change form—potential energy can be converted to kinetic energy, or kinetic energy can be converted to heat—but it”ll never simply disappear.

Conservation of energy is one of the most important, fundamental concepts in all of physics … translation: it”s going to show up all over the AP exam. So read this chapter carefully.

**Kinetic Energy and the Work-Energy Theorem**

We”ll start with some definitions.

What this second definition means is that work equals the product of the distance an object travels and the component of the force acting on that object directed parallel to the object”s direction of motion. That sounds more complicated than it really is: an example will help.

A box is pulled along the floor, as shown in __Figure 14.1__ . It is pulled a distance of 10 m, and the force pulling it has a magnitude of 5 N and is directed 30° above the horizontal. So, the force component that is PARALLEL to the 10 m displacement is (5 N)(cos 30°).

**Figure 14.1 Box is pulled along the floor.**

One newton·meter is called a joule, abbreviated as 1 J.

- Work is a scalar. So is energy.
- The units of work and of energy are joules.
- Work can be negative … this just means that the force is applied in the direction opposite displacement.

This means that the kinetic energy of an object equals one-half the object”s mass times its speed squared.

The net work done on an object is equal to that object”s change in kinetic energy. Here”s an application:

A train car with a mass of 200 kg is traveling at 20 m/s. How much force must the brakes exert in oder to stop the train car in a distance of 10 m?

Here, because the only horizontal force is the force of the brakes, the work done by this force is *W* _{net} .

Let”s pause for a minute to think about what this value means. We”ve just calculated the change in kinetic energy of the train car, which is equal to the net work done on the train car. The negative sign simply means that the net force was opposite the train”s displacement.

To find the force:

**Potential Energy**

Potential energy comes in many forms: there”s gravitational potential energy, spring potential energy, electrical potential energy, and so on. For starters, we”ll concern ourselves with gravitational potential energy.

Gravitational PE is described by the following equation:

*U = mgh*

In this equation, *m* is the mass of an object, *g* is the gravitational field of 10 N/kg on Earth, and *h* is the height of an object above a certain point (called “the zero of potential”).^{ }^{1}^{ }That point can be wherever you want it to be, depending on the problem. For example, let”s say a pencil is sitting on a table. If you define the zero of potential to be the table, then the pencil has no gravitational PE. If you define the floor to be the zero of potential, then the pencil has PE equal to *mgh* , where *h* is the height of the pencil above the floor. Your choice of the *zero* of potential in a problem should be made by determining how the problem can most easily be solved.

REMINDER: *h* in the potential energy equation stands for *vertical* height above the zero of potential.

**Conservation of Energy: Problem-Solving Approach**

Solving energy-conservation problems is relatively simple, as long as you approach them methodically. The general approach is this: write out all the terms for the initial energy of the system, and set the sum of those terms equal to the sum of all the terms for the final energy of the system. Let”s practice.

A block of mass *m* is placed on a frictionless plane inclined at a 30° angle above the horizontal. It is released from rest and allowed to slide 5 m down the plane. What is its final velocity?

If we were to approach this problem using kinematics equations (which we could), it would take about a page of work to solve. Instead, observe how quickly it can be solved using conservation of energy.

We will define our zero of potential to be the height of the box after it has slid the 5 m down the plane. By defining it this way, the PE term on the right side of the equation will cancel out. Furthermore, because the box starts from rest, its initial KE also equals zero.

The initial height can be found using trigonometry: *h* _{i}_{ }= (5m) (sin 30°) = 2.5 m.

In general, the principle of energy conservation can be stated mathematically like this:

The term *W* in this equation stands for work done on an object. For example, if there had been friction between the box and the plane in the previous example, the work done by friction would be the *W* term. When it comes to the AP exam, *you will include this W* term only when there is friction (or some other exteral force) involved. When friction is involved, *W* = *F* _{f}_{ }*d* , where *F* _{f}_{ }is the force of friction on the object, and *d* is the distance the object travels.

Let”s say that there was friction between the box and the inclined plane.

A box of mass *m* is placed on a plane inclined at a 30° angle above the horizontal. The coefficient of friction between the box and the plane is 0.20. The box is released from rest and allowed to slide 5.0 m down the plane. What is its final velocity?

We start by writing the general equation for energy conservation:

*K _{i} + U_{i} + W = E_{f} + U_{f}*

*W* equals *F* _{f}_{ }*d* , where *F* _{f}_{ }is the force of friction, and *d* is 5 m.^{ }^{2}

The value for *W* is negative because friction acts opposite displacement. You may want to draw a free-body diagram to understand how we derived this value for *F* _{N}_{ }.

Now, plugging in values we have

We rearrange some terms and cancel out *m* from each side to get

*v* _{f}_{ }= 5.7 m/s

This answer makes sense—friction on the plane *reduces* the box”s speed at the bottom.

**Springs**

Gravitational potential energy isn”t the only kind of PE around. Another frequently encountered form is spring potential energy.

The force exerted by a spring is directly proportional to the amount that the spring is compressed. That is,

In this equation, *k* is a constant (called the spring constant), and *x* is the distance that the spring has been compressed or extended from its equilibrium state. The negative sign is simply a reminder that the force of a spring always acts opposite to displacement—an extended spring pulls back toward the equilibrium position, while a compressed spring pushes toward the equilibrium position. We call this type of force a restoring force, and we discuss it more in __Chapter 17__ on simple harmonic motion. However, we can ignore this sign unless we are doing calculus.

When a spring is either compressed or extended, it stores potential energy. The amount of energy stored is given as follows.

Similarly, the work done by a spring is given by *W* _{spring} = ½*kx* ^{2} . Here”s an example problem.

A block with a mass of 2 kg is attached to a spring with *k* = 1 N/m. The spring is compressed 10 cm from equlibrium and than released. How fast is the block traveling when it passes through the equilibrium point? Neglect friction.

It”s important to recognize that we CANNOT use kinematics to solve this problem! Because the force of a spring changes as it stretches, the block”s acceleration is not constant. When acceleration isn”t constant, try using energy conservation.

We begin by writing our statement for conservation of energy.

Now we fill in values for each term. PE here is just in the form of spring potential energy, and there”s no friction, so we can ignore the *W* term. Be sure to plug in all values in meters!

Plugging in values for *k* and *m* , we have

**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.

**Potential Energy vs. Displacement Graphs**

A different potential energy function can actually be derived for ANY conservative force. (A conservative force means that energy is conserved when this force acts … examples are gravity, spring, and electromagnetic forces; friction and air resistance are the most common nonconservative forces.) The potential energy *U* for a force is given by the following integral:

Note that this equation works for the gravitational force *F* = −*mg* (where − is the down direction) and the spring force *F* = −*kx* ; the potential energy attributable to gravity integrates to *mgh* , and the spring potential energy becomes ½*kx* ^{2} .

**Chris on a Skateboard**

Once a potential energy of an object is found as a function of position, making a *U* vs. *x* graph tells a lot about the long-term motion of the object. Consider the potential energy of a spring, ½*kx* ^{2} . A graph of this function looks like a parabola, as shown in __Figure 14.2__ .

**Figure 14.2 Potential energy vs. displacement graph for a spring.**

You can get a general feel for how the mass on a spring moves by imagining that Chris is riding a skateboard on a ramp shaped like the graph. A ramp shaped like this looks like a half-pipe. If he starts from some height above the bottom, Chris will oscillate back and forth, going fastest in the middle, and turning around when he runs out of energy at the right or left end. Although this is not precisely how a mass on a spring moves—the mass only moves back and forth, for example—the long-term properties of Chris”s motion and the motion of the mass on a spring are the same. The mass oscillates back and forth, with its fastest speed in the middle, just like Chris does.

Thinking about Chris on a skateboard works for all *U* vs. *x* graphs. Consider a model of the energy between two atoms that looks like the graph in __Figure 14.3__ .

**Figure 14.3 Potential energy vs. displacement graph for two atoms.**

If Chris on his skateboard released himself from rest near position *x* _{1} , he”d just oscillate back and forth, much like in the mass on a spring problem. But if he were to let go near the position labeled *x* _{2} , he”d have enough energy to keep going to the right as far as he wants; in fact, he”d make it off the page, never coming back. This is what happens to the atoms in molecules, too. If a second atom is placed pretty close to a distance *x* _{1} from the first atom, it will just oscillate back and forth about that position. However, if the second atom is placed very close to the first atom, it will gain enough energy to escape to a faraway place.

** Practice Problems**

**Multiple Choice:**

Questions 1 and 2

A block of weight *mg* = 100 N slides a distance of 5.0 m down a 30-degree incline, as shown above.

** 1 .** How much work is done on the block by gravity?

(A) 500 J

(B) 430 J

(C) 100 J

(D) 50 J

(E) 250 J

** 2 .** If the block experiences a constant friction force of 10 N, how much work is done by the friction force?

(A) −43 J

(B) −25 J

(C) −500 J

(D) −100 J

(E) −50 J

** 3 .** A mass experiences a potential energy

*U*that varies with distance

*x*as shown in the graph above. The mass is released from position

*x*= 0 with 10 J of kinetic energy. Which of the following describes the long-term motion of the mass?

(A) The mass eventually comes to rest at *x* = 0.

(B) The mass slows down with constant acceleration, stopping at *x* = 5 cm.

(C) The mass speeds up with constant acceleration.

(D) The mass oscillates, never getting farther than 5 cm from *x* = 0.

(E) The mass oscillates, never getting farther than 10 cm from *x* = 0.

** 4 .** Two identical balls of mass

*m*= 1.0 kg are moving towards each other, as shown above. What is the initial kinetic energy of the system consisting of the two balls?

(A) 0 joules

(B) 1 joules

(C) 12 joules

(D) 18 joules

(E) 36 joules

**Free Response:**

** 5 .** A 1500-kg car moves north according to the velocity–time graph shown.

(a) Determine the change in the car”s kinetic energy during the first 7 s.

(b) To determine how far the car traveled in these 7 s, the three basic kinematics equations can not be used. Explain why not.

(c) Use the velocity–time graph to estimate the distance the car traveled in 7 s.

(d) What was the net work done on the car in these 7 s?

(e) Determine the average power necessary for the car to perform this motion.

** Solutions to Practice Problems**

** 1 . E—** The force of gravity is straight down and equal to 100 N. The displacement parallel to this force is the

*vertical*displacement, 2.5 m. Work equals force times parallel displacement, 250 J.

** 2 . E—** The force of friction acts up the plane, and displacement is down the plane, so just multiply force times distance to get 50 J. The negative sign indicates that force is opposite displacement.

** 3 . D—** Think of Chris on a skateboard—on this graph, he will oscillate back and forth about

*x*= 0. Because he starts with a KE of 10 J, he can, at most, have a potential energy of 10 J, which corresponds on the graph to a maximum displacement of 5 cm. (The mass cannot have constant acceleration because constant acceleration only occurs for a constant force; a constant force produces an energy graph that is linear. The mass will not come to rest because we are assuming a conservative force, for which KE can be converted to and from PE freely.)

** 4 . E—** Kinetic energy is a scalar, so even though the balls move in opposite directions, the KEs cannot cancel. Instead, kinetic energy ½(1 kg)(6 m/s)

^{2}attributable to different objects adds together algebraically, giving 36 J total.

** 5 .** (a) The car started from rest, or zero KE. The car ended up with ½(1500 kg)(40 m/s)

^{2}= 1.2 × 10

^{6}J of kinetic energy. So its change in KE is 1.2 × 10

^{6}J.

(b) The acceleration is not constant. We know that because the velocity–time graph is not linear.

(c) The distance traveled is found from the area under the graph. It is easiest to approximate by counting boxes, where one of the big boxes is 10 m. There are, give-or-take, 19 boxes underneath the curve, so the car went 190 m.

(d) We cannot use work = force × distance here, because the net force is continually changing (because acceleration is changing). But W_{net} = ΔKE is always valid. In part (a) the car”s change in KE was found to be 1.2 × 10^{6} J; so the net work done on the car is also 1.2 × 10^{6} J.

(e) Power is work divided by time, or 1.2 × 10^{6} J/7 s = 170 kW. This can be compared to the power of a car, 220 horsepower.

** Rapid Review**

- Energy is the ability to do work. Both energy and work are scalars.
- The work done on an object (or by an object) is equal to that object”s change in kinetic energy.
- Potential energy is energy of position, and it comes in a variety of forms; for example, there”s gravitational potential energy and spring potential energy.
- The energy of a closed system is conserved. To solve a conservation of energy problem, start by writing K
_{i }+*U*+_{i}*W*= K_{ f }+*U*, where “_{f}*i*” means “initial,” “*f*” means “final,” and*W*is the work done by friction or an externally applied force. Think about what type of*U*you”re dealing with; there might even be more than one form of*U*in a single problem! - Power is the rate at which work is done, measured in watts. Power is equal to
*work/time*, which is equivalent to force multiplied by velocity. - If the functional form of a conservative force is known, then the potential energy attributable to that force is given by

When this *U* is graphed against displacement, the motion of an object can be predicted by imagining “Chris on a skateboard” skating on the graph.

^{1}^{ }Note that 10 N/kg is exactly the same as 10 m/s^{2} .

^{2}^{ }Note this difference carefully. Although potential energy involves only a vertical height, work done by friction includes the *entire* distance the box travels.

**CHAPTER 14**

**Energy**

** 1 .** A 2 kg mass experiences a potential energy that varies with

*x*as shown in the graph. Which statement correctly describes the behavior of the mass?

(A) When at a location of 1 m, the mass will experience a force in the negative *x* direction.

(B) When at a location of 1 m, the mass will experience no force.

(C) If released from rest at a location of 3 m, the mass will reach a maximum speed of .

(D) If released from rest at a location of 3 m, the mass will oscillate about an equilibrium at location of 2 m on the *x* -axis.

(E) If released from rest at a location of 5 m, the mass will reach a maximum speed of 20 m/s.

** 2 .** An 80 kg roller-coaster car traveling 10 m/s passes over the crest of a 40 m hill, as shown in the figure. The speed of the car at the bottom of the hill is 24 m/s. The energy lost to friction is most nearly

(A) 60 J

(B) 1,000 J

(C) 5,000 J

(D) 13,000 J

(E) 22,500 J

** 3 .** A block of mass (

*m*) is released from the top of a frictionless incline at a distance of

*l*from the bottom. The block slides down the incline and then across a frictionless horizontal plane. At the far end of the horizontal plane is a fixed spring with force constant

*k*. What will be the maximum compression of the spring?

(A)

(B)

(C)

(D)

(E)

** 4 .** The following graphs represent the force (

*F*) applied to an experimental car of mass

*m*. The engine of the car can supply a maximum force (

*F*

_{max}). Which force distribution will cause the greatest change in kinetic energy of the car from

*x*= 0 to

*x*

_{f}?

(A)

(B)

(C)

(D)

(E)

** Answers**

__1__ .**A** — . This can be written in differential form as , which shows that the force on the object will be the negative of the slope of the potential energy graph. At 1 m, this indicates that the force will be in the negative *x* -direction. *Note:* The peak of the potential energy graph at 2 m is not a stable equilibrium point, and the forces will push the 2 kg mass away from this point.

__2__ .**C** —Using conservation of energy and *g* = 10 m/s^{2} , we get

__3__ .**C** —Using conservation of energy,

__4__ .**D** —The net work done on the car is the integration (area) of the *F-x* graph. The graph with the most area is choice D.