Solving a problem using energy and conservative forces concepts

[Hernaner28]

Homework Statement

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?

Homework Equations

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!

 

[darkxponent]

delta U=-W, are you sure?

 

[Hernaner28]

delta U=-W, are you sure?

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

 

[darkxponent]

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

 

[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...

 

[darkxponent]

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

 

[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

 

[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

 

[Hernaner28]

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

\begin{array}{l}<br /> {W_T} = 0\\<br /> {W_T} = {W_P} + {W_W} + {W_{tension}}\\<br /> 0 = {W_P} - mgh\\<br /> {W_P} = mgh<br /> \end{array}

Is this right?

Thanks!

 

[darkxponent]

yes that is correct