Answer to Question #295510 in Physics for Hanum

Question

A person pulls on a spring, stretching it 20 cm, which requires a maximum force of

800 N. Find the potential energy stored in the spring when it is stretched 20 cm and

compressed 5 cm.

Answer 1

Given:

$x_1=\rm 20\: cm=0.2\: m$

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

$F=\rm 800\: N$

The sping constant

$$k=\frac{F}{x_1}=\frac{800\:\rm N}{0.2\:\rm m}=4000\:\rm N/m$$

The potential energy stored in the spring:

(a)

$$E_p=\frac{kx_1^2}{2}=\rm \frac{4000*0.2^2}{2}=80\: J$$

(b)

$$E_p=\frac{kx_2^2}{2}=\rm \frac{4000*0.05^2}{2}=5\: J$$