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292 Chapter 7 Conservation of Energy and an Introduction to Energy and Work

If you again hold the object in one hand, but now rest your elbow on a table,
AP ® Exam Tip
you’ll be able to hold the object for a longer time with far less fatigue. That’s because
Chemical energy is not a topic in this case you’re using only the muscles of your hand and forearm, not those of your
on the AP® Physics 1 exam. upper arm. Fewer muscles are involved, so you need to convert less energy. The table
does not have to use chemical energy to support you or the book.

Work by Forces Not Parallel to Displacement,
and Negative Work
How can we calculate the work done by a constant force that is not exerted in the direc-
tion of the object’s motion? As an example, in Figure 7-4a a groundskeeper is using a rope

to pull a screen across a baseball diamond to smooth out the dirt. The force F that the

man exerts on the screen is at an angle θ with respect to the displacement d of the screen,
so only the component of the force along the displacement does work on the screen
(Figure 7-4b). This component is cosF θ, so the amount of work done by the force is

Work done on an object by a constant force F that
points at an angle q to the object’s displacement d Magnitude of the constant force F
EQUATION IN WORDS
Calculating the work done
by a constant force exerted at (7-2) W = Fd cos q
an angle θ to the displacement
Magnitude of the object’s displacement d Angle between the directions of F and d

To understand Equation 7-2, let’s look at some special cases.

• If F is in the same direction as the motion, θ = 0 and cos θ = cos0 = 1. This is
AP ® Exam Tip
the same situation as shown in Figure 7-2, and in this case Equation 7-2 gives
F cos θ is the component of the same result as Equation 7-1: W = Fd.
the force that is parallel to the • If the angle θ between the force and displacement is more than 0 but less than
°
displacement, so is also called 90 , as in Figure 7-4b, then cos θ is less than 1 but still positive. The work W

F on the equation sheet. done by the force F on the object is positive but less than Fd.

• If F is perpendicular to the direction of motion, θ = 90° and cos θ = cos90 ° = 0.

In this case Equation 7-2 tells us that force F does zero work on the object.
An example is the force exerted by a lazy dog lying on top of the screen from
Figure 7-4a (Figure 7-5). The dog exerts a downward force on the screen as the
screen moves horizontally, so θ = 90 ° and the lazy dog does no work at all.

(a) (b)
F
The man exerts a force F on the q
screen. The direction of the force Screen
is not parallel to the ground. F cos q
d

Only the component of force parallel to the
displacement contributes to the work that
the man does on the screen.
F
d

As he exerts the force, the
screen undergoes a displacement
d along the ground.

Figure 7-4 When force is not aligned with displacement (a) A man exerts a force on a screen
to pull it across a baseball field. (b) Finding the work that the man does on the screen (seen from

the side). If a constant force F is exerted on an object at an angle θ to the object’s displacement

d the work done on the object by the force equals the force component F cos θ parallel to the
displacement multiplied by the object’s displacement d: W = Fd cos θ.

Uncorrected proofs have been used in this sample. Copyright © Bedford, Freeman & Worth Publishers.
Distributed by Bedford, Freeman & Worth Publishers. For review purposes only. Not for redistribution.

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