How much work is done in stretching this spring from a length of 2.5 m to a length of 3.2 m?

Question

A spring with a natural length of 1.8 meters is held at a length of 2.4 meters by a force of 12 N. How much work is done in stretching this spring from a length of 2.5 m to a length of 3.2 m?

Answer 1

As the force constant wasn't given, we will have to calculate it. By Hooke's law, F = KΔX. K = F/ΔX =

12N / (2.4 - 1.8)m = 20 Nm.

The potential energy at 2.5 m then is 1/2 K ΔX² = 1/2 * 20 * (2.5 - 1.8) = 7.0 joules. At a stretch of 3.2 m the potential energy is

1/2 * 20 * (3.2 - 1.8) = 14 joules.

The work done in stretching the spring form 2.5 m to 3.2 m is the PE difference = 7 joules.