# Work done by a spring

1. Feb 11, 2012

### Supernejihh

1. The problem statement, all variables and given/known data

We must apply a force of magnitude 80N to hold the block stationary at x = -2.0cm. From that position, we then slowly move the block so that our force does +4.0J of work on the spring-block system; the block is then again stationary. What is the block's position? (Two answers.)

2. Relevant equations

F = -kx

W = (1/2)*k*xi^2 - (1/2)*k*xf^2

3. The attempt at a solution

80 = -k(-.02m)
k = 4000 N/m

With the spring constant, i plug it into the Work equation to find xf

4 = .5*4000*(-.02^2) - .5*4000*xf^2

3.2 = -2000xf^2
-.0016 = xf^2

And this is where I am stuck. The answer cannot be an imaginary number, but a real one. I know the equations are correct, so what am I doing wrong? Please explain.

2. Feb 11, 2012

### Staff: Mentor

You want final minus initial, not the other way around.

3. Feb 11, 2012

### Supernejihh

The problem is that, that is how the book presented us with the equation. It was initial minus final. I understand where you are coming from, but that's how it is.

4. Feb 11, 2012

### Staff: Mentor

Then what does W stand for in your book? The work done by the spring? The work done by you will be opposite to that.

5. Feb 11, 2012

### Supernejihh

Doc Al. So what you are telling me is that the Work of +4.0J is caused by me. And its positive because the direction and force are directed towards one way. But the force of a spring is always opposite the displacement. So the work of the spring is -4.0J because its moving one way, while the force is directed towards the opposite?

6. Feb 11, 2012

### Staff: Mentor

Right.

7. Feb 11, 2012

### Supernejihh

Thanks for your help Doc. I really appreciate it.