How Far Will a Mass Travel on a Frictionless Incline After Going Through a Loop?

AI Thread Summary
A mass on a 10-meter incline at a 36-degree angle experiences kinetic friction with a coefficient of 0.26 before entering a frictionless loop with a 2-meter radius. The problem involves calculating how far the mass will travel up a second incline inclined at 33 degrees after completing the loop. Using conservation of energy principles, the initial energy and speeds at various points are determined, leading to the conclusion that the mass will travel 9.18 meters up the second incline. Key equations include energy conservation and work done against friction. The analysis emphasizes the role of kinetic energy and friction in determining the mass's final position.
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Homework Statement


A mass starts at top of incline with a length of 10 meters with an angle of 36 degrees. Incline has friction with a coefficient of kinetic friction of .26. At bottom of incline is a horizontal, frictionless surface which has a loop of radius 2 meters. After loop is a frictionless incline which is inclined at an angle of 33 degrees. If mass is given enough velocity at top of first incline with friction such that it just makes it around loop, how far up second incline will mass travel in meters? Answer is 9.18.

Homework Equations


\DeltaE=Wnc

The Attempt at a Solution


h=Lsin\theta
Wnc=Wfk+WFn=-fkL
Ef-Eo=-fkL
Eo=uo+KE
Ef=uf+KEf
 
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So using conservation of energy, what is the initial energy? At the bottom of the incline (just before the loop) what is the speed?
 
vo=4.9m/s
vf=9.9m/s
 

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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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