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7-5 The work-energy theorem is also valid for curved paths and varying forces 313

Solve
(a) The force that the athlete exerts is Force exerted by the athlete to F F F co cords on athlete F F F athlete on co
athlete on cords
proportional to the distance x = 47 cm that the hold the cords stretched:

cords are stretched (assuming the end of the F =+ kd

cords when they are not stretched is at =x 0 ). athleteoncords,x  1m 
=  
(86N/m)(47 cm)
 100 cm
1
=
4.0 × 10N

(b) The cords are not stretched at all to Work done by the athlete to stretch the cords:
start (so x = 0) and end up stretched by 1 1

1
2
47cm(so x = 47 cm = 0.47 m). W athleteoncords = kx − kx 2
2
2 2 2 1
1 1
2
= (86N/m)(0.47 m) − (86N/m)(0m) 2
2 2
= 9.5Nm = 9.5 J

(c) When the athlete releases the handle, the cords Work done by the cords on the handle as they relax:
relax from their new starting position ( 1 = 0.47m)  1 1 
x
2
to their final, unstretched position 2 = 0). Unlike W =−  kx 2 2 − kx 1 

(
x
cordsonhandle
an actual spring, the cords don’t push back as the  2 2 
handle passes through the equilibrium position. = −  1 (86N/m)(0m) − 1 (86N/m)(0.47 m) 2 
2
So, the work that the cords do on the handle is   2 2  
given by the negative of Equation 7-13 = −− 9.5 J
.
(9.5 Nm) =
We’ll ignore the force of gravity (that is, we Net work done on handle:
assume the handle flies back horizontally and W W = 9.5J
net =
doesn’t fall). Then the net work done on the cordsonhandle
handle equals the work done by the cords. This Work-energy theorem applied to handle:
is equal to the change in the handle’s kinetic net = K − 1 mv − 1 mv
2
2
energy. Use this to find the handle’s speed when W f i K = 2 f 2 i
the cords are fully relaxed.
Handle is initially at rest, so i v = 0. Solve for final speed of the handle:

1
mv = W net = 9.5J
2
2 f
2W net
2
v =
f
m
2W net 2(9.5 J)
f v = = = 8.7m/s
m 0.25 kg
Reflect
The spring force is not a constant force, so the work done must be less than you would calculate if you multiplied the
force needed to hold the cords fully stretched by the cords’ displacement, because the force exerted is smaller than
that except at the maximum stretch. So we expect our answer to be less than s F d = (4.0 × 10 N)(0.47m) = 19J,
1
which it is.
Extend: Notice that the amount of work that the athlete does to stretch the cords (9.5 J) is the same as the amount
of work that the cords do on the handle when they relax. This suggests that the athlete stores energy in the cords by
stretching them. We’ll explore this idea of storing energy in Section 7-6 as we explore potential energy.
NOW WORK Problem 5 from The Takeaway 7-5.
Like Equation 5-10 Equation 7-13 is also valid when a spring is compressed. If a
,

spring with spring constant k = 1000 N/m is initially relaxed (so x 1 = 0) and you stretch
it by 20cm(so x =+ 20 cm =+ 0.20 m), the work that you do is

2
1 1 1
W = kx − k (0) = (1000 N/m)(0.20m) = 20J
2
2
2
2
2 2 2
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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