Question

It takes 130 j of work to compress a certain spring 0.10m. (a) what is the force constant of this spring? To compress the spring an additional 0.10m, does it take 130

Answer 1

Answer:

Explanation:

The PE equation for a mass/spring system is

$PE=\frac{1}{2}k$Δx² and filling in:

$130=\frac{1}{2}k(.10)^2$ and

$k=\frac{2(130)}{(.10)^2}$ so

k = 26000 N/m

If the displacement from equilibrium changes more, the PE needed to compress it will also change.

$PE=\frac{1}{2}(26000)(.20)^2$ gives us that

PE = 520J