Calculating Gravitational Potential Energy of a Child on a Swing

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In summary, the gravitational potential energy of the 30.3 N child in the swing attached to ropes 2.14 m long is equal to mgRcos(30.9°), where R is the height above the ground and m is the mass. This can be derived from the formula E_gravitational potential = mgh, where h is the height off the ground. The angle of 30.9° must be taken into account when calculating the height above the ground.
  • #1
nazarip
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A 30.3 N child is in a swing that is attached to ropes 2.14 m long. What is the gravitational potential energy when the ropes make a 30.9° angle with the vertical?

So far I've tried 6 different answers ranging between 0-64.8 J and no luck. Need a clue to get me on the right track. Thanks in advance.
 
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  • #2
You know [itex]E_{gravitational potential}=mgh[/itex], where h is height off the ground and m is mass. I believe you must assume that when the ropes make an angle of 0 with the vertical h=0. Does that help?
 
  • #3
Pretend that its like those incline questions.
 
  • #4
I did assume that when the angle is zero E=0 and when the angle is 90 E=mgh=64.8, so I thought the answer would be (sin30.9)(mgh), but that answer was incorrect. I also tried (sin59.1)(mgh) which was also incorrect, I am not sure where to go from here.
 
  • #5
Well I think when the kid swings up to make a [itex]30.9^\circ[/itex] with the vertical, his height above the ground is increased. By how much? The answer is

[tex]R \cos (30.9^\circ)[/tex]

So the kid's [itex]U=mg(R\cos (30.8^\circ))[/itex]
 

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