Springs and Work

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Homework Statement


In the figure below, we must apply a force of magnitude 89 N to hold the block stationary at x = -2.0 cm. From that position we then slowly move the block so that our force does +4.0 J of work on the spring-block system; the block is then again stationary. What is the block's position (x)? (There are two answers.)
---cm (smaller value)
---cm (larger value)




Homework Equations



W = .5 * [-k(xfinal^2) + k(xinitial^2)]

k = spring constant = force / distance stretched (or compressed)

The Attempt at a Solution



1. Find the spring constant:

K = 89N / -.02m = -4450 N/m

2. Plug values in

W = 4 J = .5 * -4450N.m * [ -xf^2 - xi^2]
The initial distance from 0 was -.02m so I plugged that in for the initial distance and solved for the final distance, which = .0469 m = 4.69cm.
This was the right answer, but I'm not sure in which direction it was (although it was the larger value) and or how to find the smaller one. Would I add the initial distance instead of subtract it?
 
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Answers and Replies

  • #2
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[tex]x_{i}[/tex] = 0.02 m
F = 89 N
W = 4 J
[tex]\vec{F_{spring}}[/tex] = -kx

The positive work done on the system means that energy from compressing the block into the spring more was converted into the potential (spring) energy, and the potential energy increased. There is negative work done on the block, so [tex]\Delta[/tex]KE is negative.
+[tex]\Delta[/tex]PE = 4 J = -[tex]\Delta[/tex]KE = -[tex]\int[/tex][tex]\vec{F}[/tex] [tex]\bullet[/tex] d[tex]\vec{r}[/tex] = -[tex]\int[/tex][tex]\vec{F_{spring}}[/tex] [tex]\bullet[/tex] d[tex]\vec{r}[/tex] = -[tex]\int^{x_{f}}_{x_{i}}[/tex](-kx)dx = k[tex]\int^{x_{f}}_{x_{i}}xdx[/tex] = (1/2)k([tex]x_{f}^{2}[/tex] - [tex]x_{f}^{2}[/tex]).

So +[tex]\Delta[/tex]PE = 4 J = (1/2)k([tex]x_{f}^{2}[/tex] - [tex]x_{i}^{2}[/tex]).

To solve for k, we know that at a displacement of [tex]x_{i}[/tex] = 0.02 m, the block is at static equilibrium, and net force is zero. So,
F + [tex]F_{spring}[/tex] = 0
F + (-kx) = 0
k = F/[tex]x_{i}[/tex].

4 = (1/2)k([tex]x_{f}^{2}[/tex] - [tex]x_{i}^{2}[/tex])
[tex]x_{f}^{2}[/tex] = 4/(1/2)k + [tex]x_{f}^{2}[/tex]
[tex]x_{f}[/tex] = [tex]\pm[/tex][tex]\sqrt{8/k + x_{f}^{2}[/tex] = [tex]\pm[/tex]0.046880196 m
I think the negative number reflects that if the compressing force is released, the spring will push the block past [tex]x_{i}[/tex] toward zero so the displacement is in the other direction, but I'm not sure. The spring constant k is never negative. Feel free to correct any mistakes I've made.
 

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