Anthony T. answered • 06/07/21

Patient Math & Science Tutor

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 ΔX^{2} = 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.