Question

A force of 543 N keeps a certain spring stretched a distance of 0.700 m .
Part A
What is the potential energy of the spring when it is stretched 0.700 m ?
Express your answer with the appropriate units.

Answer 1

k = force/distance = 543/.7

U = (1/2) k x^2 = (1/2) k (.7)^2 = (1/2) (543)(.7^2)/.7
= (1/2)(543)(.7) = average force * distance of course

Answer 2

Newtons * meters = Joules