Question

the spring of thickness K =20n/m is compressed to distance of 30m find potential energy stored in it.
please answer ​

Answer 1

Answer:

The spring of thickness K =20N/m is compressed to a distance of 30m potential energy stored is $300J$

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 $K=20N/m$

compressed distance $x=30m$

substitute these values to get potential energy stored

⇒

$PE=\frac{1}{2} (20*30)=300J$