Question

A spiral spring is compressed by 0.1cm. calculate the elastic potential energy in the spring if the stiffness of the spring is 100Nm^1

Answer 1

`E=5* 10^(-5)\ J`

Step-by-step explanation:

Given that,

A spring is compressed by 0.1 cm or 0.001 m

The spring constant of the spring, k = 100 N/m

The elastic potential energy in the spring is given by :

`E=(1)/(2)kx^2\\\\E=(1)/(2)* 100* 0.001^2\\\\E=5* 10^(-5)\ J`

So, the elastic potential energy of the spring is equal to
`5* 10^(-5)\ J`.