453
Answers
701–800
Answers and Explanations
W
kx dx
kx
k
k
k
=
=
= (
)
= (
)
=
∫ 0
0 10
2
0
0 10
2
2
5
2
5 2
0 10
5 2
0 10
5 200
1 0
.
.
( . )
( . )
, 0
00 N m
⋅ = k
Now find the work required to stretch the spring from 30 centimeters to 42 centimeters, or from 0.15 meters from its natural length to 0.27 meters from its natural length:
W
x dx
x
=
=
=
−
(
)
=
∫ 1 000
500
500 0 27
0 15
25
0 15
0 27
2
0 15
0 27
2
2
,
( . ) ( . )
.
.
.
.
.
2 2 J
723.
20
3
ft lb
⋅
Begin by using the equation F(x) = kx to find the value of k, the spring constant. The
force is measured in foot-pounds, so make sure the length is in feet rather than inches.
15
6
12
30
lb
ft
ft lb
= ( )
⋅ =
k
k
Therefore, the force equation is F
 
(x) = 30x.
Because the spring starts at its natural length, x = 0 is the lower limit of integration.
Because the spring is stretched 8 inches, or 8
12
feet, beyond its natural length, the
upper limit of integration is 8
12
2
3
= . Therefore, the integral to find the work is
W
xdx
x
=
=
=
( ) −






=
⋅
∫ 30
15
15 2
3
0
20
3
0
8 12
2
0
2 3
2
2
ft lb
724.
2.25 J
Begin by using the equation F
 
(x) = kx to find the value of the spring constant. Make
sure you convert the centimeters to meters so you can use the units newton-meters,
or joules. If x is the distance the spring is stretched beyond its natural length, then
x = 5
100
m. Therefore, the spring constant is