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

Set Up F F F n n
Again we’ll use the work-energy theorem for an object, Work done by a constant force,
v of the bobsleigh linear displacement:

this time to determine the final speed f

(which starts at rest, so its initial speed is i =v 0 ). The

normal force and gravitational force do no work on the W = Fd cos θ (7-2) d
sleigh because they are exerted on the sleigh perpendicular Kinetic energy: F F F k k
to its displacement. The four men of the crew do positive 20.0°
work, while the kinetic friction and drag forces do nega- K = 1 mv (7-8) F F F cr crew
2
F ). 2

tive work (we will just refer to this combined force as k
The new wrinkle is that the force exerted by the crew
points at an angle to the sleigh’s displacement. We’ll deal Work-energy theorem for an object:
with this using Equation 7-2 Note that the horizontal W K − i K (7-9)
.
net =

and vertical forces are drawn to different scales. f mg
Solve
The sleigh starts at rest, so its initial kinetic energy is Work-energy theorem d = 50.0 m
= 50.0
d d
zero. From Equation 7-9 the net work done on the sleigh for an object with
is therefore equal to its final kinetic energy, which is initial kinetic energy
related to the final speed .v equal to zero: v v v = 0 v v v f f
i i
f
1
2
net =
W K − 0 = mv
f
f
2
.
The force of the crew is exerted at θ = 20 0° to the network = work done by the crew plus work done by

displacement, and the friction force is at θ = 180° Use friction

.

these in Equation 7-2 to find the net work done on the W net = W crew + W
sleigh. friction
= F crew d cos 20.0° + Fd cos 180°
k
= (285 N)(50.0m)(0.940)
+ (60.0N)(50.0m)( 1)
−
4
= 1.04 × 10 J
1
2

W
Substitute net into the work-energy theorem for an Work-energy theorem for an object: net = mv
W
f
v .
object and solve for the final speed f 2
4
2W 2(1.04 × 10J)
v v = net =

:
Solve for f f
m 210 kg
= 9.95 m/s
Reflect
The sleigh gains speed and kinetic energy during the motion, so the net work done on the sleigh by the crew and

v is

friction must be positive (even though individually one was negative), which is what we got! The answer for f
relatively close to the speed of world-class sprinters, so it seems reasonable. In reality, the start of an Olympic four-man
bobsleigh race is slightly downhill, so the sleigh’s speed at the end of the starting stretch is typically even faster (11 or
12m/s ) because a component of the force of gravity exerted on the sleigh by Earth is in the direction of motion, adding

to the work done.
NOW WORK Problems 2 and 3 from The Takeaway 7-4.
EXAMPLE 7-5 Find the Work Done by an Unknown Force
An adventurous parachutist of mass 70.0 kg drops from the top of Angel Falls in Venezuela, the world’s highest waterfall.
The waterfall is 979 m (3212 ft) tall and the parachutist deploys her chute after falling 295 m, at which point her speed is
54.0m/s During the 295-m drop, (a) what was the net work done on her and (b) what was the work done on her by the

.
force of air resistance?
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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