Question

A spring of stiffness 2N/m is compressed by 20cm Find the elastic potential energy stored in the spring.

Answer 1

Answer:

Explanation: this is the answer of above question

Answer 2

Answer:

The spring of thickness K =2N/m is compressed to a distance of 30m potential energy stored is $0.20J$

Explanation:

• The elastic potential energy is the energy stored when a force applied to deform an elastic object

• A spring with elastic constant K compressed or expanded at a distance x from its mean position, then the potential energy stored is

        $PE=\frac{1}{2}Kx^{2}$

From the question, we have

spring constant $K=2N/m$

compressed distance $x=20cm=0.20m$

substitute these values to get potential energy stored

$PE=\frac{1}{2} (2*0.20)=0.20J$