# Work/force with a magnet problem

1. Mar 11, 2008

### the7joker7

The strength of magnetic force varies inversely with the square of the distance between the magnets. In other words,

Force = k/(distance^2)

Suppose that when two magnets are 4 cm apart, there is a force of 4 newtons. Find the work required to move the magnets from a distance of 3 cm apart to a distance of 11 cm apart.

4N = k/(0.04$$^{2}$$)m

4N = k/(0.0016m)

k = 2500

One thing that's thrown me for a loop though, is...

The distance between the magnets is 4 CM, or 0.04M.

So do I take the square of 4 CM to get 16 CM, or .16M.

Or do I square 0.04M to get 0.0016M, or .16CM.

I think it's the latter, but I'm not totally sure and I don't get second chances on this question.

2. Mar 11, 2008

### Hootenanny

Staff Emeritus
It is in fact the former, when raising a quantity to a power it's always a good idea to convert it to it's base units.

$$\left(4 cm\right)^2 = \left(4\cdot10^{-2}m\right)^2 = 16\cdot10^{-4}m^2 = 1.6\cdot10^{-3}m^2$$

Last edited: Mar 11, 2008
3. Mar 11, 2008

### the7joker7

So k is indeed 2500, eh?

Alright, lemme see...

As for finding work...

Work = Integral of ((2500/x^2) * (x)) from 3 to 11.

4. Mar 11, 2008

### Hootenanny

Staff Emeritus
No, $k\neq2500$ you may want to check you arithmetic.

For the integral, I'm not sure why your multiplying by x here, perhaps it should be dx? Also take care of your units, your best using meters, i.e. integrate from 0.03 to 0.11.

Perhaps a minor point, but don't forget to square your units, i.e. m2.

5. Mar 11, 2008

### the7joker7

Wait, does k = .0064?

I multiply by x because Work is force times distance. I calculate force in the first part. Then I calculate distance with the x. And yes, at the end I forgot to add a (with respect to dx).

6. Mar 11, 2008

### the7joker7

So now I have...

Integral of (((.0064/x^2) * (x))dx from 0.03 to 0.11.

Gives 0.0083154 joules.

7. Mar 11, 2008

### Hootenanny

Staff Emeritus
Sounds much better
Ahh, but the definition of rectilinear [1D] mechanical work is,

$$W = \int^{b}_{a}F(x)dx$$

8. Mar 11, 2008

### the7joker7

Ah, okay. Thanks.

Now I have...

Integral of (((.0064/x^2))dx from .03 to .11

Gives .1551515152 joules.

Sound accurate?

9. Mar 11, 2008

### Hootenanny

Staff Emeritus
Sounds good to me