GCSE Physics C/W - Speed at bottom of Slope...

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Melanie47
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#1
Report Thread starter 15 years ago
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Okay, the title of this coursework is way too long to be able to fit it in the title of this thread, so I'll put it here:
'How is the speed at the bottom of the slope linked to the height at the top of the slope?'

I don't know many other people who have done this coursework (most people do resistance of a wire), but if there is anyone who ca help me ?
I have obtained all my results, and drawn the graph, but I'm am puzzled over what shape it is supposed to take. I'm in the middle of doing my analysis and then evaluation. For the analysis I know we are supposed to work out by calculations (something do do with PE and KE) what the speed at the bottom of the slope should be for each height, and then compare those calculations with the results we got. What I don't get is how to calculate the speed ?

I really dislike Physics, so I hope that made some kind of sense.
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(Original post by Melanie47)
Okay, the title of this coursework is way too long to be able to fit it in the title of this thread, so I'll put it here:
'How is the speed at the bottom of the slope linked to the height at the top of the slope?'

I don't know many other people who have done this coursework (most people do resistance of a wire), but if there is anyone who ca help me ?
I have obtained all my results, and drawn the graph, but I'm am puzzled over what shape it is supposed to take. I'm in the middle of doing my analysis and then evaluation. For the analysis I know we are supposed to work out by calculations (something do do with PE and KE) what the speed at the bottom of the slope should be for each height, and then compare those calculations with the results we got. What I don't get is how to calculate the speed ?

I really dislike Physics, so I hope that made some kind of sense.
you are using the fact that the potential energy at the top of the slope is equal to the kinetic energy at the bottom. Ie conservation of energy.

so:
PE = mgh
KE=0.5mv^2

so
mgh=0.5mv^2
so rearranging:

v = root(2gh)

so the graph should be a root graph so if you plot v^2 against h you should get a straight line with gradient 2g = 20 metres per sec squared.
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Melanie47
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Report Thread starter 15 years ago
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Okay ... thanks for that, it should help
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Chewwy
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are you going to be varying the angle of the slope, or not?
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