A key idea to use in this question is that the integral of force (in respect to position or "x") equals the work done. If you think about the force vs position graph, you can find the work done by calculating the area underneath, right? It's the same logic.
Hi,
Thank you for your question.
A key idea to use in this question is that the integral of force (in respect to position or "x") equals the work done. If you think about the force vs position graph, you can find the work done by calculating the area underneath, right? It's the same logic.
So, here is my solution:
Does this answer your question?