Aspring with a constant of 120 n/m stretches by 0.02 m. what is the potential energy of the spring

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 :

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

PE = 0.024 J

Hence, the potential energy of the spring is 0.024 J

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