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Gravitational potential energy |
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| Dec19-07, 03:06 PM | #1 |
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Gravitational potential energy
1. The problem statement, all variables and given/known data
A 7.26 kg ball hangs from end of 2.5 m rope. The ball is pulled back until rope makes 45 degree angle with vertical. What is the gravitation potential energy? 2. Relevant equations ug=mgh 3. The attempt at a solution I thought I would just multiply mass times gravity times height, but the answer is 52 J. What do I do with the angle I'm given? |
| Dec19-07, 03:09 PM | #2 |
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What are the maximum and minimum vertical heights of the ball?
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| Dec19-07, 03:31 PM | #3 |
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0 m and 2.5 m....?
I provided the entire problem. |
| Dec19-07, 03:42 PM | #4 |
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Gravitational potential energy
In the back of the book it says to find height do
h=2.5m * (1-cos(45)) Then plug that into Ug=mgh. Why 1-cos(45)? |
| Dec19-07, 04:03 PM | #5 |
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Edit: Think of swinging your arm such that it's 45 degrees to vertical. How far has your hand moved vertically? |
| Dec19-07, 05:51 PM | #6 |
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Ok, I think I understand. It's going to be hard to explain, but in other words, when the ball is in the air and has that 45 degree angle, it's going to "lose" height; it's going to be "shorter" than hanging straight down. So that's why 1 is being subtracted from cos(45).
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| Dec19-07, 05:56 PM | #7 |
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I'm not sure I understand your explanation. At 45 degrees the ball is higher than completely vertical- so I'm not sure why you claim it loses height.
Try answering my question. If you swing your arm by 45 degrees from vertical how far has your hand moved vertically (upwards)? (You could take the length of your arm to be e.g. 1.1 meters.) |
| Dec19-07, 06:01 PM | #8 |
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i don't know. i'm flustered and confused.
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| Dec19-07, 06:08 PM | #9 |
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OK, calm down!
This problem involves a little trig. It's best to draw a diagram. Draw a vertical line of length, e.g. 2.5m between points A and B (with A at the top, B at the bottom.) Next- draw a line between points A and C which is of the same length, but at 45 degrees to the AB line. Tell me once you've done this. |
| Dec19-07, 06:21 PM | #10 |
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| Dec19-07, 06:24 PM | #11 |
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kdv: The gravitational potential energy refers to the potential energy that the ball can convert into kinetic energy without cutting the rope. If we allow the rope to be cut- then yes we would have to specify at what height we take as the zero.
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| Dec19-07, 06:30 PM | #12 |
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| Dec19-07, 06:37 PM | #13 |
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| Dec19-07, 06:52 PM | #14 |
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| Dec19-07, 07:04 PM | #15 |
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| Dec19-07, 11:15 PM | #16 |
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christian-as you explained is how i drew my diagram in the first place, so it's done.
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| Dec19-07, 11:27 PM | #17 |
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OK, good. Now for the evil next step.
Draw a line horizontally from C so that it intersects the AB line. Tell me when you're done. |
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