Work on Springs using Hooke's Law

Question

I'm currently stuck on parts c and d of this problem. The problem says Suppose a force of 20 N is required to stretch and hold a spring 0.4 m from its equilibrium position (0). I found k constant to be 50 and part b required using integration on 0 to -0.3 for spring compression on the and I found the work to be 2.25 J. Here's part c and d...

c) does it take twice as much work to compress the spring 0.6 meters from equilibrium as it does to compress 0.3 meters from equilibrium?

d)Does it take twice as much force to compress the spring 0.6 meters from equilibrium as it does to compress 0.3 meters from equilibrium?

I'm assuming you just solve the work for 0.6 compression and compare it to 2.25 J for c but im not sure about d. Please let know your thoughts and ways to approach this.

Answer 1

Hint: From Hooke's Law $$ F=kx $$ we can derive the formula for spring potential energy by integrating with respect to position $$ E=\frac12kx^2 $$

Answer 2

For any future people who'd like to know the answer to this question.

For c) you can just calculate the work for compressing the spring 0.6 meters and compare it to the work for 0.3.

And for d) all you have to do is plug in -0.3 and -0.6 and compare the result. Hooke's law assumes a linear spring ( F(x) = kx ) which is an odd function so stretching or compressing x amount either direction takes the same amount of force and since F(x) = kx doubling or tripling the x-value will do the same to the result, as shown below.

F(x) = kx

F(2x) = k(2x) = 2kx

or

F(-x) = -kx

F(-2x) = k(-2x) = -2kx