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

Answer:

Potential energy of the spring is 0.024 Joules.

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 :

$PE=\dfrac{1}{2}kx^2$

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

$PE=\dfrac{1}{2}\times 120\ N/m\times (0.02\ m)^2$

PE = 0.024 J

Hence, the potential energy of the spring is 0.024 J