Question

A person pulls on a spring, stretching it 15 cm, which requires a maximum force of 750 N. Find the potential energy stored in the spring when it is stretched by 15 cm and compressed by 5 cm.

Answer 1

Given:

$x_1=\rm 15\: cm=0.15\: m$

$x_2=\rm 5\: cm=0.05\: m$

$F=\rm 750\: N$

The sping constant

$$k=\frac{F}{x_1}=\frac{750\:\rm N}{0.15\:\rm m}=5000\:\rm N/m$$

The potential energy stored in the spring:

(a)

$$E_p=\frac{kx_1^2}{2}=\rm \frac{5000*0.15^2}{2}=56.25\: J$$

(b)

$$E_p=\frac{kx_2^2}{2}=\rm \frac{5000*0.05^2}{2}=6.25\: J$$