a force of 1250 pounds compresses a spring 5 inches from its natural length. find the work done in compressing the spring 8 additional inches.

f=kd
1250=k5
k=250

then what do i do.

the answer choices are:
a.3250
b.21,125
c.18000
d.2000
e. none of these

Work = (1/2)* F * deflection
After 8 additional inches of deflection, the Force will increase to
(13/5)*1250 = 3250 lb
The total work done (from zero deflection) will be
(1/2)*3250 *13 = 21,125 in-lb
The original work done (for the first 5 inches of deflection) was
(1/2)*1250*5 = 3125 in-lb
The additional work done is
21,125 - 3125 = 18,000 in-lb

Another way to do this would be to calculate the spring constant k = 250 lb/in, as you have done. Then subtract
(1/2) k 5^2 from (1/2) k (13^2).
The difference is (1/2)*k * 144 = ?

1 answer

  1. work= force * distance, so you can set up an integral to solve for work

    F=k*x, k=250

    force =250x
    distance =dx

    so then you integrate(250x)dx from 0->(5+8)=21,125

    idk if its right but its an answer choice

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