Solve Work Homework: -2.0cm & 4.69cm (Springs & Force 89N)

In summary, the problem involves applying a force of 89 N to hold a block stationary at x = -2.0 cm. By slowly moving the block, a work of +4.0 J is done on the spring-block system, resulting in the block being stationary again. The block's position is found to be either +4.69 cm or -4.69 cm. The spring constant is calculated to be -4450 N/m and the work-energy theorem is used to solve for the final position of the block. The final position is found to be +4.688 cm or -4.688 cm, depending on the direction of the force applied.
  • #1
A_lilah
53
0

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?
 
Last edited:
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  • #2
[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.
 
  • #3




Good job on finding the correct answer for the larger value of x. To find the smaller value, you would need to use the same equation, but instead of subtracting the initial distance, you would add it. This is because the initial distance is in the opposite direction of the final distance, so it would result in a negative value. Therefore, the smaller value of x would be -2.0 cm + 4.69 cm = 2.69 cm. It is important to pay attention to the direction of the initial and final distances when using this equation. Keep up the good work!
 

Related to Solve Work Homework: -2.0cm & 4.69cm (Springs & Force 89N)

1. What is the formula for solving work homework?

The formula for solving work homework is W = F x d, where W represents work, F represents force, and d represents distance.

2. How do I calculate work if both force and distance are given in centimeters?

You will need to convert the measurements to meters before plugging them into the formula. In this case, -2.0cm would be converted to -0.02m and 4.69cm would be converted to 0.0469m.

3. Can you explain how to solve for work using springs?

To solve for work using springs, you will need to use the formula W = 1/2kx^2, where W represents work, k represents the spring constant, and x represents the distance the spring is compressed or stretched.

4. What is the unit of measurement for work?

The unit of measurement for work is joules (J).

5. How do I calculate work if only force and distance are given?

If only force and distance are given, you can use the formula W = F x d to calculate work. Make sure to use the appropriate units of measurement for force and distance.

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