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7-4 The work-energy theorem can simplify many physics problems 305

Set Up
We are given the parachutist’s mass and the distance Work done by a constant force, linear F F F F air
air
that she falls, so we can find the gravitational force that displacement:
Earth exerts on her and the work done by that force using

Equation 7-2 Air resistance also exerts a force on the W = Fd cos θ (7-2)
.
parachutist. We don’t know its magnitude, so we can’t Work-energy theorem for an object: d
directly calculate the work done by this force. Instead, we’ll m g
net =

use the work-energy theorem for an object. We’ll assume W K − i K (7-9)
f
that the parachutist starts at rest, so her initial kinetic Kinetic energy:
energy is zero.
1
2
K = mv (7-8)
2
Solve

(a) We can model the parachutist as an object well, W net = K = 1 mv f 2
f
because the relative motion of her limbs does not affect 2 v v v = 0
i i
what we are trying to find, which is just how much work 1 (70.0kg)(54.0m/s)
2
is done by the various forces in this situation. (How the = 2
= 295
d d
parachutist holds herself does slightly impact the effects d = 295 m
5
of air resistance, but we aren’t trying to change the size = 1.02 × 10 J
of that.) Because i K = 0, Equation 7-9 states that the net

work done on the parachutist is just equal to her final
kinetic energy.
v v v = 54.0 m/s
= 54.0 m/s
f f f f
(b) We know that total energy must be conserved. This Work done by the gravitational force:

does not mean that total energy is constant (that is only Displacement is in the same direction
true for a closed, isolated system, which our parachutist (downward) as the force, so
is not), but that any change in the energy of the system
must be equal to transfers of energy into or out of W grav = mgd
2
the system by external interactions. Any time we use = (70.0kg)(9.80m/s )(295 m)
5
work to determine changes in energy, we are building = 2.02 × 10 J
from this conservation principle. The net work on the
parachutist is the sum of the work done by the forces Substitute and solve: The work done
exerted on her; the work done by the gravitational by the force of air resistance is
force, grav ; and the work done by the force of air W air = W net − W

W
.
W
resistance, air Hence air is the difference between grav 5

W
5
net and grav .
W W = 1.02 × 10 J − 2.02 × 10J
5
=− 1.00 × 10 J
Reflect
The work done by the force of air resistance is negative Average upward force of air resistance:
because the force of air resistance is directed upward,
air =
opposite to the downward displacement of the parachutist. W F air d cos 180° =− F air d
We can use the calculated value of air to find the average Solving:
W

value of the force of air resistance. This is only an average 5
−
value because this force is not constant: The faster the F air =− W air =− (1.00 × 10J)
parachutist falls, the greater the force of air resistance. d 295 m
This is less than the gravitational force on the parachutist, = 339 J/m = 339 N
as it needs to be, in order for the parachutist to fall and
gain speed.
NOW WORK Problem 4 from The Takeaway 7-4.
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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