Question

A force of 20 lb is required to hold a spring stretched 1 ft. beyond its natural length. How much work is done in stretching

the spring from 1 ft. beyond its natural length to 3 ft. beyond its
natural length?
the answer is supposed to be 80 but I'm really not sure how to get there.

Answer 1

Explanation:

Work of stretching a spring = 1/2 k (xf^2 - xo^2)

k = spring constant = 20 lbs/ft

xf =3 ft xo = 1ft

1/2 ( 20 ) ( 3^2 - 1^2) 2) = 80 ft lbs