459
Answers
701–800
Answers and Explanations
W
k x dx
kx
k
L
L
L
L
L
=
=
=
−
−
−
−
−
−
∫ 0 40
0 60
2
0 40
0 60
2
25
2
25 2
0 60
040
.
.
.
.
( .
) ( . − −
=
−
[
]
L
k
L
)
.
.
2
50
0 2 0 4
Likewise, expanding the second integral gives you
W
k x dx
kx
k
L
L
L
L
L
=
=
=
−
−
−
−
−
−
∫ 0 60
0 80
2
0 60
0 80
2
40
2
40 2
0 80
060
.
.
.
.
( .
) ( . − −
=
−
[
]
L
k
L
)
.
.
2
80
0 28 0 4
Now solve for the spring constant k. By subtracting 50 = k[0.2 – 0.4L] from 80 = k[0.28 –
0.4L], you’re left with 30 = k(0.08) so that k = 375.
Finally, find the natural length of the spring. Substituting the value of k into the equation 50 = k[0.2 – 0.4L] gives you 50 = 375[0.2 – 0.4L]; solving for L yields L = 1
6
meters, so
the natural length of the spring is approximately 16.67 centimeters.
733.
615 752
,
J
Let n be the number of subintervals of length Δ x and let x i
* be a sample point in the ith
subinterval.
Here’s a view of the tank underground:
And here’s a cross-section of the tank:
Answers
701–800
Answers and Explanations
W
k x dx
kx
k
L
L
L
L
L
=
=
=
−
−
−
−
−
−
∫ 0 40
0 60
2
0 40
0 60
2
25
2
25 2
0 60
040
.
.
.
.
( .
) ( . − −
=
−
[
]
L
k
L
)
.
.
2
50
0 2 0 4
Likewise, expanding the second integral gives you
W
k x dx
kx
k
L
L
L
L
L
=
=
=
−
−
−
−
−
−
∫ 0 60
0 80
2
0 60
0 80
2
40
2
40 2
0 80
060
.
.
.
.
( .
) ( . − −
=
−
[
]
L
k
L
)
.
.
2
80
0 28 0 4
Now solve for the spring constant k. By subtracting 50 = k[0.2 – 0.4L] from 80 = k[0.28 –
0.4L], you’re left with 30 = k(0.08) so that k = 375.
Finally, find the natural length of the spring. Substituting the value of k into the equation 50 = k[0.2 – 0.4L] gives you 50 = 375[0.2 – 0.4L]; solving for L yields L = 1
6
meters, so
the natural length of the spring is approximately 16.67 centimeters.
733.
615 752
,
J
Let n be the number of subintervals of length Δ x and let x i
* be a sample point in the ith
subinterval.
Here’s a view of the tank underground:
And here’s a cross-section of the tank: