# Need help deriving work formula

1. Oct 24, 2006

### metalmagik

A body of mass m is in a gravitational field of strength g. The body is moved through a distance h at constant speed v in the opposite direction to the field.

Derive an expression in terms of

m, h and h, for the work done on the body.

Im bad with derivations, I need to get better. So far I have

Work = F x d
Work = ma x h
Work = mgh?

2. Oct 24, 2006

### PhanthomJay

Your answer is correct, however, i don't believe you are clear on the concept. Since the speed is constant, what does that tell you about the magnitude and direction of the force required to move the body upward?

3. Oct 24, 2006

### metalmagik

Right, since the speed is constant, the body is not accelerating, which means force of gravity IS acceleration.

4. Oct 24, 2006

### PhanthomJay

I don't think you quite have it right. Yes, the body is not accelerating, therefore , per Newton's 1st law, there is no net force acting on the body. What is the value of the weight ? And since the weight is acting down, how much pulling force is acting up? Now calculate how much work is done by that force.

5. Oct 24, 2006

### metalmagik

Uh well I Don't have any measurements but I think I understand now...the force acting up is the same as the weight acting down, since it is at constant velocity...correct?

6. Oct 24, 2006

### PhanthomJay

Yes. And since the weight acting down is mg, the force acting up must also be mg, right? So the work done by that force is W = Fh = mgh.
It is important to understand this as you move into problems where the body is accelerating rather than moving at constant speed.

Last edited: Oct 24, 2006
7. Oct 24, 2006

### metalmagik

Ah sweet. Yes I see that. If the body was accelerating this problem would be totally different, meaning, different terms with which to derive, correct?

8. Oct 24, 2006

### PhanthomJay

If the body was accelerating upward at acceleration a, then you'd have to use newton's 2nd law to yield, (denoting the pulling force as F),
F -mg = ma, that is, solving for F,
F = m(a+g), and the work done by the pulling force would be
W = Fh = m(a+g)h. Of course when a= 0, the work becomes just mgh. OK?

9. Oct 24, 2006

### metalmagik

right, makes sense. Just preparing myself incase there's a derivation question on tomorrow's test! Thanks PhanthomJay!