Question

A force of 570 N keeps a certain ideal spring stretched at a distance of 0.700 m.
A). What is the potential energy of the spring when it is stretched 0.700 m?
B). What is its potential energy when it is compressed 8.00 cm?

Answer 1

Answer:

399j and 456j

Explanation:

(a)spring energy= force * distance

spring energy=570 * 0.700= 399 joules

(b) convert 8.00 cm to m we have 0.08m

E =f * s

E=570 *0.08= 456j