Gravitational potential energy.

AI Thread Summary
The discussion focuses on calculating gravitational potential energy for a sled sliding down a hill. The boy on the sled has a mass of 72 kg and starts from a height of 35 m, with a friction force of 250 N. The correct formula for gravitational potential energy is identified as PE = mgh, leading to a calculated value of 24,696 J. The angle of the hill (37 degrees) is noted but deemed unnecessary for the potential energy calculation. The conversation shifts to finding the sled's speed at the bottom, prompting further questions about the work done by friction and the distance traveled.
cruisx
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Homework Statement


Hey guys, just wondering if you guys could help me out a bit.

A boy on a sled with a total mass of 72kg slides from rest down a hill, which is 35m high. The friction force between sled and snow has a magnitude of 250N. If the slope of the hill makes an angle of 37 degrees with the horizontal, find:
a) the gravitational potential energy of the sled at the top of the hill relative to the base.

Homework Equations



If i am not mistaken the equation for Gav.pot energy is
U= GMe * m
...Re2

and the value of g can be found as g = G * Me/ (R + h)2

The Attempt at a Solution



I know i have the following from the question:

m= 72kg
h1=35m
Ff= 250N

now what i don't get is where the 37 degrees is supposed to come in and i am not quite sure that that is the equation that i am supposed to use. Could someone please explain to me how i would begin this question? I want to solve it my slef but can someone point me into a direction. thank you.edit: wait a second, i think i was doing it wrong. Would it be like E1g = mgh = (72kg)(9.8m/s2)(35m) = 24696 J ?
 
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cruisx said:

Homework Equations



If i am not mistaken the equation for Gav.pot energy is
U= GMe * m
...Re2

and the value of g can be found as g = G * Me/ (R + h)2
Much simpler would be to use PE = mgh, which is perfectly good near the Earth's surface.


now what i don't get is where the 37 degrees is supposed to come in
Maybe you don't need it for part a. :wink:
 
cruisx said:
edit: wait a second, i think i was doing it wrong. Would it be like E1g = mgh = (72kg)(9.8m/s2)(35m) = 24696 J ?

Now you've got it.
 
Doc Al said:
Now you've got it.


Ok cool thanks guys, i think i can handle it form here ^_^. look like i will be studying all day for that physics test tomorrow :frown:
 
Sorry for posting again but now i am stuck at part c of the question which is : the speed of the sled at the bottom of the hill.

How would i find this out? I know i have to use the 37 degrees somehow but i am stuck, I now have the following variables.

m=72kg
h1 = 35m
Ff = 250N
gav.pot energy : 24696J
Work: 8750J

Its been a while since i have had to find speed so a hint would be helpful =)
 
What's the work done by friction? Hint: What's the distance traveled by the sled?

Set up an energy equation.
 

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