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April 15th, 2016, 01:14 PM   #1
kkg
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Conservation of Energy Problem

Each tile has a mass of 2kg and the coefficient of friction between the tile and roof is 0.75. The roof is inclined 30 degrees to the horizontal and the bottom of the roof is 6m above the ground.

A tile is kicked from point A directly down the roof. The distance from A to the edge of the roof is 4m. When it reaches point B which is a distance x from the edge of the roof as measured along its slope, it's speed is 4m/s. It subsequently hits the ground travelling at 9m/s. In the motion of the time from B to the ground, the work done against sliding and other resistances is 90J. Find x using an appropriate method.

I've tried solving this but I can't seem to get the answer. I don't understand how you can solve for x using conservation of energy.
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April 15th, 2016, 01:52 PM   #2
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What conservation of energy equation are you using?
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April 15th, 2016, 05:52 PM   #3
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Seems that what happens from A to B is inconsequential since the question asks to determine the distance x from B to the roof's edge. Also, the coefficient of friction is not needed since the energy lost from B to the ground from work done by all non-conservative forces is given.

Starting at point B ...

Total mechanical energy at B - 90J = kinetic energy at ground level

$E_B-90=81 \implies E_B=171 \, J$

Total mechanical energy at B = kinetic energy at B + gravitational potential energy at B

$171=16+2g\left(6+\frac{x}{2}\right)$

$x=\dfrac{155}{g}-12$
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April 16th, 2016, 08:19 AM   #4
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Quote:
Originally Posted by skeeter View Post
Seems that what happens from A to B is inconsequential since the question asks to determine the distance x from B to the roof's edge. Also, the coefficient of friction is not needed since the energy lost from B to the ground from work done by all non-conservative forces is given.

Starting at point B ...

Total mechanical energy at B - 90J = kinetic energy at ground level

$E_B-90=81 \implies E_B=171 \, J$

Total mechanical energy at B = kinetic energy at B + gravitational potential energy at B

$171=16+2g\left(6+\frac{x}{2}\right)$

$x=\dfrac{155}{g}-12$
Thanks.

Also, how did you work out the GPE at B to be 2g(6+x/2)? I don't understand where the x/2 comes from?
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April 16th, 2016, 08:55 AM   #5
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Quote:
Also, how did you work out the GPE at B to be 2g(6+x/2)? I don't understand where the x/2 comes from?
sin30 = 1/2
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April 16th, 2016, 01:51 PM   #6
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30-60-90 triangle ...
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File Type: jpg 30_60_90.jpg (8.9 KB, 1 views)
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