Question

A spring with a constant of 120 N/m stretches by 0.02 m. What is the potential energy of the spring

Answer 1

E = 0.5kx² = 0.5 * 120 * 0.02²

Answer 2

Potential energy of the spring is 0.024 Joules.

Step-by-step explanation:

It is given that,

Spring constant of the spring, k = 120 N/m

The spring is stretched by 0.02 m

We have to find the potential energy of the spring. Mathematically, it is given by :

`PE=(1)/(2)kx^2`

Putting the values of k and x in above equation we get :

`PE=(1)/(2)* 120\ N/m* (0.02\ m)^2`

PE = 0.024 J

Hence, the potential energy of the spring is 0.024 J