## SAT Physics Subject Test

## Chapter 4 Work, Energy, and Power

It”s difficult to give a precise definition of energy. Loosely speaking, energy is a quantity which gives an object or system the ability to accomplish something (what we will define as **work**). There are different forms of energy partly because there are different kinds of forces. There”s kinetic energy (a train zooming at high speeds), gravitational energy (a meteor crashing into the earth), elastic energy (a stretched rubber band), thermal energy (an oven), radiant energy (sunlight), electrical energy (a lamp plugged into a wall socket), nuclear energy (nuclear power plants), and mass energy (the heart of Einstein”s equation *E = mc*^{2}).

Energy can come into a system or leave it via various interactions that produce changes. For the SAT Physics Subject Test, you should think of **force** as the agent of change, **energy** as the measure of change, and **work** as the way of transferring energy from one system to another. And one of the most important laws in physics—the **law of conservation of energy**, equivalent to the **first law of thermodynamics**—says that the total amount of energy in a given process will stay constant—that is, it will be **conserved**. For example, electrical energy can be converted into light and heat (this is how a light bulb works), but the amount of electrical energy coming in to the lightbulb equals the total amount of light and heat given off. Energy cannot be created or destroyed; it can only be transferred (from one system to another) or transformed (from one form to another).

### WORK

When you lift a book from the floor, you exert a force on it over a distance, and when you push a crate across a floor, you also exert a force on it over a distance. The application of force over a distance and the resulting change in energy of the system give rise to the concept of **work**. When you hold a book in your hand, you exert a force (normal force) on the book, but since the book is at rest, the force does not act through a distance, so you do no work on the book. Although you did work on the book as you lifted it from the floor, once it”s at rest in your hand, you are no longer doing work on it.

In short, if a constant force **F** acts over a distance *d*, and **F** is parallel to **d**, then the work done by **F** is the product of force and distance. If a constant force **F** acts over a distance **d**, and *θ* is the angle between **F** and **d**, then the work done by **F** is the product of the component of force in the direction of the motion and the distance.

*W* = **F***d* cos*θ*

Notice that, although work depends on two vectors (**F** and **d** where **d** points in the direction of motion), work itself is *not* a vector. Work is a scalar quantity.

1. You slowly lift a book of mass 2 kg at constant velocity a distance of 3 m. How much work did you do on the book?

Here”s How to Crack It

In this case, the force you exert must balance the weight of the book (otherwise the velocity of the book wouldn”t be constant), so *F = mg* = (2 kg)(10 m/s^{2}) = 20 N. Since this force is straight upward and the displacement of the book is also straight upward, **F** and **d** are parallel, so the work done by your lifting force is *W = Fd* = (20 N)(3 m) = 60 N·m. The unit for work, the newton-meter (N·m) is renamed a **joule**, and abbreviated J. So the work done is 60 J.

**When the formula forwork works**

*W* = *Fd* cos*θ* only works

when the Force does

not change as the object

moves.

2. A 15 kg crate is moved along a horizontal floor by a warehouse worker who”s pulling on it with a rope that makes a 60° angle with the horizontal. The tension in the rope is 200 N and the crate slides a distance of 10 m. How much work is done on the crate by the worker?

Here”s How to Crack It

The figure below shows that **F**_{T} and **d** are not parallel. It”s only the component of the force acting along the direction of motion, **F**_{T} cos*θ*, that does any work.

Therefore

*W* = (*F*_{T} cos *θ*)*d* = (200 N • cos 60°)(10 m) = 1,000 J

3. In question 2, assume that the coefficient of kinetic friction between the crate and the floor is 0.4.

(a) How much work is done by the normal force?

(b) How much work is done by the friction force?

Here”s How to Crack It

(a) Clearly, the normal force is not parallel to the motion, so we use the general definition of work. Since the angle between **F**_{N} and **d** is 90° (by the definition of *normal*) and cos 90° = 0, the normal force does zero work.

(b) The friction force, **F**_{f}, is also not parallel to the motion; it”s *antiparallel*. That is, the angle between **F**_{f} and **d** is 180°. Since cos 180° = –1, and the strength of the normal force is *F _{N}*

*= F*=

_{f}*mg*= (15 kg)(10 m/s

^{2}) = 150 N, the work done by the friction force is

*W* = –**F**_{f} *d* = –µ_{k}**F**_{N}*d* = –(0.4)(150 N)(10 m) = –600 J

The two previous examples show that work—which, as we said, is a scalar quantity—may be positive, negative, or zero. If the angle between **F** and **d** (*θ*) is less than 90°, then the work is positive (because cos*θ* is positive in this case); if *θ* = 90°, the work is zero (because cos 90° = 0); and if*θ* > 90°, then the work is negative (because cos *θ* is negative). In other words, if a force helps the motion, the work done by the force is positive, but if the force opposes the motion, then the work done by the force is negative.

For situations where *q* is something other than 0°, 90°, or 180°, it is sometimes useful to break the force into components, *F*_{┴} and *F*_{║}

Where *F*_{║} is the component in the direction of motion (or opposite if it is negative) and *F*_{┴} is the component perpendicular to the direction of motion. We can now write this formula as *W* = *F*_{║}*d*.

**Questions 4-7**

A box slides down an inclined plane (incline angle = 40°). The mass of the block, *m*, is 40 kg, the coefficient of kinetic friction between the box and the ramp, *μ*_{k}, is 0.3, and the length of the ramp, *d*, is 10 m. (Use: sin 40° ≈ 0.6 and cos 40° ≈ 0.8.)

4. How much work is done by gravity?

5. How much work is done by the normal force?

6. How much work is done by friction?

7. What is the total work done?

Here”s How to Crack It

4. Remember that the force that”s directly responsible for pulling the box down the plane(*F*_{║}) is the component of the gravitational force that”s parallel to the ramp: **F**_{w} sin *θ* = *mg* sin *θ* (where *θ* is the incline angle). This component is parallel to the motion, so the work done by gravity is:*W*_{by gravity} = (*mg* sin*θ*)*d* = (40 kg)(10 N/kg)(sin 40°)(10 m) = 2,400 J

Notice that the work done by gravity is positive, as we would expect it to be, since gravity is helping the motion. Also, be careful with the angle *θ*. The general definition of work reads *W* = (*F* cos *θ*)*d*, where *θ* is the angle between **F** and **d**. However, the angle between **F**_{w} and **d** is *not*40° here, so the work done by gravity is not (*mg* cos 40°)*d*. The angle *θ* used in the calculation above is the incline angle. This is why *W* = *F*_{║}*d* is a useful way of writing the formula.

5. Since the normal force is perpendicular to the motion, the work done by this force is zero.

6. The strength of the normal force is **F**_{w} cos *θ* (where *θ* is the incline angle), so the strength of the friction force is **F**_{f} = µ_{k}**F**_{N} = µ_{k}**F**_{w} cos *θ* = µ_{k}*mg* cos *θ*. Since **F**_{f} is antiparallel to **d**, the cosine of the angle between these vectors (180°) is –1, so the work done by friction is

Notice that the work done by friction is negative, as we expect it to be, since friction is opposing the motion.

7. Since work is a scalar, we can find the total work done simply by adding the values of the work done by each of the forces acting on the box: