Question

A spring of force constant 10n/m has an initial stretch 0.20m in changing the stretch to 0.25m the increase in potential energy is

Answer 1

Potential energy = 1/2kx^2 where the parameters are defined as follows.

K = The spring constant.

x = the extension of the spring

We need to get the potential energy before and after.

We get this as follows :

Initial potential energy :

= 1/2 × 10 × 0.20 × 0.20 = 0.2 joules

The potential energy when the stretch changes to 0.25m is :

It is given as :

1/2 × 0.25 × 0.25 × 10 = 0.3125 Joules

The increase in potential energy is given by :

0.3125 - 0.2 = 0.1125 joules