Question

A force of 7 pounds is required to hold a spring stretched 0.4 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.7 feet beyond its natural length?

Answer 1

Answer:

W = 8.575 foot-pounds

Step-by-step explanation:

The equation for work is:  W = F*d

where F is force applied and d distance of the movement

from problem statement we have

d = 0,7 ft

We also know that the spring was stretched 0,4 ft when a force of 7 pounds was applied therefore

K the constant of the spring is

F = K*s      

k = F/s     ⇒   k = 7/0.4         ⇒ k = 17.5  pounds/feet

Now to move from original condition of the spring up to 0,7 feet we need a force of

F = k*s       ⇒  F = 17,5 pounds/feet * 0.7 feet     ⇒ F = 12.25 pounds

And finally the work

W = 12.25 * 0.7   = 8.575 foot-pounds

W = 8.575 foot-pounds