Question

An ideal spring has a spring constant (force constant) of 2500 N/m, is stretched 4.0 cm, How much elastic potential energy does it possess? (A) 3J (B) 0.00J (C) 1J (D) 2J (E) 4J

Answer 1

Elastic potential energy, E = 2 J

Step-by-step explanation:

It is given that,

Spring constant of the spring, k = 2500 N/m

The spring is stretched to a distance of 4 cm i.e. x = 0.04 m

We have to find the elastic potential energy possessed by the spring. A spring possessed elastic potential energy and it is given by:

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

`E=(1)/(2)* 2500\ N/m* (0.04\ m)^2`

E = 2 Joules.

Hence, the correct option is (d) " 2 Joules ".