A force of 4 pounds is required to hold a spring stretched 0.1 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.6 feet beyond its natural length?

Answer 1

work=INT force*dx from 0 to .6
= int(k x dx)= 1/2 k x^2 over limits
where k= 4lbs/.1f= 20 lbs/ft
work= 1/2 k x^2 from zero to x=.6
work= 1/2 (20lbs/ft)*.6^2 ft^2 - 0
= 36 ft-lbs work