# 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 = ?