Answer to Question #186222 in Classical Mechanics for Jethro

Question

An upstretched spring is 10 ft. long. A pull of 40 lb stretches the spring by 1/2 ft. Find the work done in stretching from 10 ft. to 14 ft.

Answer 1

Spring constant

$K=\frac{40}{\frac{1}{2}}=80lb/ft$

Work done will be store as potential energy of spring so

$W=\frac{1}{2}kx^2$

$W=\frac{1}{2}(80)(4)^2=640 lbft$