# Homework Help: Solving a problem using energy and conservative forces concepts

1. Apr 13, 2012

### Hernaner28

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

It asks me to work out the work done by force P and it tells me that the ball m has a CONSTANT speed during its displacement.

That diagram is taken from a book which has already solved the excercise without using the concepts of energy and conservative forces. It used a complicated way of integrating to determinate the work of P. So since all the forces are conservative this could be solved easily, couldn't it?

2. Relevant equations

3. The attempt at a solution
If the speed is constatnt then the ΔK is 0. And we know that ΔK=W so W=0. And we also know that the potential energy ΔU=-W so ΔU=0. But after that I end up writing the integral of P when I don't want to do that. Thanks!

2. Apr 13, 2012

### darkxponent

delta U=-W, are you sure?

3. Apr 13, 2012

### Hernaner28

Yes, that's a definition. The change of potential energy is ΔU=-W - what's wrong with it?

4. Apr 13, 2012

### darkxponent

well this applies only when the object is acted apon by conservative forces only. That is delta U +delta K =0.

5. Apr 13, 2012

### Hernaner28

Yes I know but are you saying that in this system there's a force which is non-conservative? There's the tension, the weight and force P.... oh, force P? hmm...

6. Apr 13, 2012

### darkxponent

You can solve it withput using integral. Just concentrate on work done on the particle by the individual forces

7. Apr 13, 2012

### Hernaner28

OK I'll write that down for each force, but is this a conservative system, right? Now you've said that I am in doubt if P is or not conservative.
Thanks

8. Apr 13, 2012

### darkxponent

p os not conservagive force. Work done by p depends on path taken. You can see from the figure. Only those forces are conservative which do not change the total energy of particle. P does changes the kinetic energy of body and hence it is non-conservagive

9. Apr 13, 2012

### Hernaner28

Hmmm.. I see. So there's no need to use energy concepts either.

$$\begin{array}{l} {W_T} = 0\\ {W_T} = {W_P} + {W_W} + {W_{tension}}\\ 0 = {W_P} - mgh\\ {W_P} = mgh \end{array}$$

Is this right?

Thanks!

10. Apr 13, 2012

### darkxponent

yes that is correct

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