Motion Problem: Quick Solution | 65 Chars

[metalmagik]

http://img128.imageshack.us/img128/5118/rampproblemec4.png

I did the first 2, I just really really need help with (e). My teacher has the answer down as .12 m...but I do not understand how to achieve height when you don't have an angle with which to achieve x velocity and y velocity. Any hints are greatly appreciated.

 

[berkeman]

Can you use the tradeoff between kinetic energy (related to speed) and potential energy (related to changes in height)?

 

[metalmagik]

GAH. you're right. Forgot about good old conservation of energy. Except, got KE and used that as PE to find h and got .012 as an answer...hmm I used .5 for velocity since that's where it ends off at...what did I do wrong?

 

[berkeman]

Set \Delta KE = \Delta PE

What is the change in KE? What is the change in PE? Note how the mass term cancels out of both sides. Now solve for \Delta H


EDIT -- fixed some LaTex and my typos.

 

[metalmagik]

thanks for the second response :)

I did exactly that and got .012 m rather than .12 m. Here is my equation:

1/2v^2 = gh
1/2(.5m/s)^2 = (-9.8m/s^2)h

 

[berkeman]

0.5m/s is not the change in velocity. The \Delta KE is the initial KE minus the final KE.

 

[metalmagik]

AH I see now. I got the right answer, but why was it necessary to use \Delta KE? I guess just because we needed to find the total height above ground level at which the cart goes? I am just having trouble conceptualizing it.

 

[berkeman]

Because you are using \Delta KE = \Delta PE

The total energy PE + KE is constant because of of the law of conservation of energy. The change in KE is related to the velocity in this problem. The change in PE is related to the change in height in this problem. Does that make sense?


EDIT -- clarified some wording.

 

[metalmagik]

It makes a little sense...Since we need the height relative to the velocity it undergoes in THAT timeframe only, we need \Delta KE?