Losing energy

  • Thread starter Vee9
  • Start date
  • #1
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Homework Statement



Before you turn away from this question: I already did most of the work, so don't be turned off by all of these words, lol. They're just my explanations.

My main point is to get one more idea.


How would you calculate the loss of energy down the ramp, given only the following information:
Mass of ball = 0.0083 kg
and
http://i1097.photobucket.com/albums/...cs_/Energy.jpg [Broken]


2. Homework Equations

E = mgh
E = 0.5mv^2


3. The Attempt at a Solution
I already found one way:
To compare the energy at the top of the ramp, which is E = mgh and the energy at the bottom of the ramp, E2 = mgh + 0.5mv^2 and also compare E to the energy at the bottom of the projectile, E3 = 0.5mv^2.
What I'm confused about is:
when I find E1 = mgh, would I use (0.307+ 0.904) as "h" or would I just use 0.307 as "h?"
And when finding the time of the projectile in E2 using y = 0.5gt^2, would I use (0.019+0.904) as delta y?

But my main question is: What is another way to calculate loss of energy?
 
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Answers and Replies

  • #2
Delphi51
Homework Helper
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The photobucket link says "page not found".
Is it an experiment where you measure the speed at the bottom?
It looks like you have not included the rotational energy.
 
  • #4
Delphi51
Homework Helper
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Okay, I see the picture now. This is a very interesting question that I used with my grade 12 class for many years. It has a bit of a mystery about it that I hope I haven't already spoiled for you.

The experiment has two different parts which really need to be analyzed separately. There is the run down the ramp and the 2D projectile motion through the air. The usual way to analyze it is to do the projectile motion first and figure out the speed of the ball when it is launched. Then you do the run down the ramp and see how fast the ball would have gone had there been no energy loss. This should answer your question about the heights. Keep them separate.
 

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