# Energy and mechanics question

1. Nov 2, 2008

### david18

1. The problem statement, all variables and given/known data
http://img120.imageshack.us/img120/3743/physicsquestionnb9.jpg [Broken]
http://g.imageshack.us/img120/physicsquestionnb9.jpg/1/ [Broken]

3. The attempt at a solution

I'm trying this problem and I'm really not sure if my answers are correct at all.

For part a) I get the kinetic energy = Pt (since it is the electrical energy)
I make 0.5m.v^2=Pt and make v the subject, leaving $v=\sqrt \frac{2pt}{m}$

Now I hope what I've done so far is okay. For part b to find the acceleration I differentiated the equation I derived above. To do this I wrote the equation as
$v=(\sqrt \frac{2p}{m})t^\frac{1}{2}$

and so

$dv/dt = \frac{1}{2}(\sqrt \frac{2p}{m})t^\frac{-1}{2}$

While for the distance, by integration, I got

$s= \frac{2}{3}(\sqrt \frac{2p}{m})t^\frac{3}{2}$

Im not very confident in my maths so I'd appreciated it if somebody could check whether I got the right results.

Last edited by a moderator: May 3, 2017
2. Nov 3, 2008

### jcwhitts

As long as your first assumption is correct, that power times t is equal to the kinetic energy, you did the integration and the derivations correctly.