a force which does work on an object which is independent of the path taken by the object between its starting point and its ending point

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**Ten Homework Problems**

Chapter 6 Problems 8, 10, 16, 20, 61, 64, 67, 73, 77, 78

**DISCUSSION OF SELECTED SECTIONS**

**6.1 Work Done by a Constant Force**
As you know by now, there are many words in physics that have meanings which are well-defined but not necessarily used in the way they are normally used outside of the context of a physics course, and

*work* is certainly one of these words. In physics, work is defined as the scalar product of force and displacement, that is,

*W = ***F·s =** Fscos**

where *** is the angle between the applied force and the displacement. Here, the force and displacement vectors are multiplied together in such a way that the product yields a scalar. Thus, work is not a vector, and has no direction associated with it. Since work is the product of force and displacement, it has units of newton-meters, or joules (J). A joule is the work done by applying a force of one newton through a displacement of one meter. One joule is about the amount of work you do in lifting your calculator to a height of one
*

*
*meter. For work to be done on an object, the force must have a component in the same direction as the displacement.

**6.2 The Work – Energy Theorem and Kinetic Energy**

*Energy *is the ability to do work, and when work is done, there is always a transfer of energy. Energy can take on many forms, such as potential energy, kinetic energy, and heat energy. The unit for energy is the same as the unit for work, the *joule*. This is because *the amount of work done on a system is exactly equal to the change in energy of the system.* This is called the *work-energy theorem*.

*Kinetic energy* is the energy an object has because it is moving. In order for a mass to gain kinetic energy, work must be done on the mass to push it up to a certain speed, or to slow it down. The work-energy theorem states that the change in kinetic energy of an object is exactly equal to the work done on it, assuming there is no change in the object’s potential energy.

The work done on a system can also be converted into heat energy, and usually some of the work is.

**6.3 Gravitational Potential Energy**

*Potential energy* is the energy a system has because of its position or configuration. When you stretch a rubber band, you store energy in the rubber band as elastic potential energy. When you lift a mass upward against gravity, you do work on the mass and therefore change its energy. The work you do on the mass gives it potential energy relative to the ground.

To lift it, you must apply a force equal to the weight

*mg *of the mass through a displacement height

*h*, and thus the work done in lifting the mass is

*W = Fs = (mg)h *

which must also equal its potential energy

*PE = mgh*

**6.4 Conservative Forces versus Nonconservative Forces, and the Work – Energy Theorem**

A force is said to be conservative if the work done by the force does not depend on the path taken between any two points. The gravitational force and the spring force are two examples of conservative forces. These two forces conserve energy during a round trip. A force is said to be nonconservative if the work done by the force depends on the path taken. Friction is the most common example of a nonconservative force on the AP Physics B exam, since a taking longer path will dissipate more heat energy. Work done by a nonconservative force generally cannot be recovered as usable energy.

**6.5 The Conservation of Mechanical Energy**
When work is done on a system, the energy of that system changes from one form to another, but the *total amount* of energy remains the same. We say that total energy is *conserved*, that is, remains constant during any process. This is also called the *law of conservation of energy.*

We can use the fact that the sum of the potential and kinetic energies remains constant during free fall to solve for quantities such as speed or initial height.

*U*_{top} + K_{top} = U_{bottom} + K_{bottom}
*mgh*_{top} + ½ mv_{top}^{2 }= mgh_{bottom} + ½ mv_{bottom}^{2}
The sum of the kinetic and potential energies of a system is called the

*total mechanical energy* of the system. These same principles can be applied to a block sliding down a frictionless ramp, a pendulum swinging from a height, and many other situations. We could use Newton’s laws and kinematics to solve these types of problems, but usually conservation of energy is easier to apply.

**Example 1**

A small block of mass *m* begins from rest at the top of a curved track at a height *h* and travels around a circular loop of radius *R*. There is negligible friction between the block and the track between points A and D, but the coefficient of kinetic friction on the horizontal surface between points D and E is μ. The distance between points D and E is *x*.

Answer all of the following questions in terms of the given quantities and fundamental constants.

(a) Determine the speed of the block at point B, at the bottom of the loop.

(b) Determine the kinetic energy of the block at point C, at the top of the loop.

After the block slides down the loop from point C to point D, it enters the rough portion of the track. The speed of the block at point E is half the speed of the block at point D.

(c) Determine the speed of the block at point D, just before it enters the rough portion of

the track.

(d) Determine the amount of work done by friction between points D and E.

(e) Find an expression for the coefficient of kinetic friction μ.

**Solution**
Since there is no friction on the track between points A and B, there is no loss of energy. Thus,

Solving for *v*_{B}* ,*

(b) Conservation of energy:

(c) There is no energy lost on the track between points A and D, so the speed of the block at point D is the same as the speed at point B:

(d) The work done by friction is the product of the frictional force and the displacement through which it acts.

(e) The frictional force causes the block to have a negative acceleration according to Newton’s second law:

Using a kinematic equation,

**6.7 Power**
Work can be done slowly or quickly, but the time taken to perform the work doesn’t affect the amount of work which is done, since there is no element of time in the definition for work. However, if you do the work quickly, you are operating at a higher power level than if you do the work slowly. *Power is defined as the rate at which work is done. *Oftentimes we think of electricity when we think of power, but it can be applied to mechanical work and energy as easily as it is applied to electrical energy. The equation for power is

and has units of joules/second or

*watts* (W). A machine is producing one watt of power if it is doing one joule of work every second. A

75-watt light bulb uses 75 joules of energy each second.

**Example 2**
A motor raises a mass of 3 kg to a height *h* at a constant speed of 0.05 m/s. The battery (not shown) which provides energy to the motor originally stores 4 J of energy, all of which can be used to lift the mass.

(a) What is the power developed in the motor?

(b) To what maximum height can the motor lift the mass using its stored energy?

**Solution**

(a)

(b)

**6.9 Work Done by a Variable Force**

Graphically, the work done on an object or system is equal to the area under a *Force vs. displacement* graph:

The area under the graph from zero to 20 meters is 1450 N m. Thus, the force represented by the graph does 1450 J of work. This work is also a measure of the *energy* which was transferred while the force was being applied.

**Supplemental Section: Potential Energy vs. Displacement Graphs**
Since work done by a force is equal to the product of the force and the displacement through which it acts, the area under a force versus displacement graph is equal to the work done. Doing work on a system changes the energy of that system. Consider the potential energy versus displacement graph below:

This graph could represent the potential energy of

a mass oscillating on a spring, as the work done on the spring is equal to the potential energy at any point on the graph. That potential energy in turn is converted into kinetic energy as the mass is accelerated by the spring. The total energy of the system remains constant.

**Example 3** Let’s say the total energy of the object represented by the

*U *vs.

*x* graph above is

20 J. This means that the sum of the potential and kinetic energy of the object at any displacement is 20 J, even though the potential energy on the graph goes higher than 20 J.

(a) What is the relationship between the potential energy and the displacement of this object?

(b) What is the potential energy of the object when it is at a displacement of – 2 m?

(c) What is the maximum displacement of the object?

(d) What is the kinetic energy of the object when it is at a displacement of 3 m?

(e) What is the speed of the object when it is at a displacement of 3 m?

**Solution**

(a) Since the *U* vs. *x* graph is a parabola, the potential energy must be directly proportional to the square of the displacement, that is, .

(b) If we go over to x = - 2, and go up to meet the graph, the potential energy is 4 J.

(c) The maximum displacement of the object corresponds to the maximum potential energy, which is 20 J. If we look up 20 J on the graph, we find that the corresponding displacement is approximately ± 4.4 m.

(d) The potential energy that corresponds to x = 3 m is 9 J. Since the total energy is 20 J, the kinetic energy at this displacement must be 20 J – 9 J = 11 J.

(e)

**CHAPTER 6 REVIEW QUESTIONS**

*For each of the multiple choice questions below, choose the best answer. *

*Unless otherwise noted, use g = 10 m/s*^{2} and neglect air resistance.