Answer to Question #319723 in Mechanical Engineering for Nicole Enriques

Question

A spring has a natural length of 8 inches. If a force of 20 lb. stretches the spring 0.5

inches, find the work done in stretching from 8 inches to 11 inches

Answer 1

Firstly needs to calculate spring constant through Hook's law.

F=Kx

20=K*0.50

K=40

Thus F=40x

Work done in stretching 8 to 11 inches= Fdx over range of integration

=40xdx

=40{(11^2/2)-(8^2/2}

=1140 lb- inches