Block on a ramp problem - energy

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SUMMARY

The discussion focuses on solving a physics problem involving a package projected up a 15º incline with a coefficient of kinetic friction (μk) of 0.12. The height at the top of the ramp is calculated to be 2.588 meters. The user initially misapplies the work-energy principle by incorrectly accounting for potential energy and neglecting the distance in the work terms. The correct approach emphasizes that potential energy should not be double-counted when calculating work done by gravity.

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Marioqwe
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Homework Statement



Package is projected 10 m up a 15º incline so that it justreaches the top of the incline with zero velocity. μk =0.12

a) Determine initial velocity of package at A

Homework Equations



Energy and friction equations

The Attempt at a Solution



Height at the top of ramp h = 2.588m

Work = change in energy

W = -μmgcos(15) - mgsin(15) = mgh - 0.5mv^2

But I get the wrong answer. Am I supposed to use potential energy? and if not why not?

Thanks
 
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Marioqwe said:
W = -μmgcos(15) - mgsin(15) = mgh - 0.5mv^2

But I get the wrong answer. Am I supposed to use potential energy? and if not why not?
(1) If you include potential energy, that automatically takes care of the work done by gravity. Don't count it twice.
(2) The work terms must be of a form force*distance. You left out the distance.
 
I see my mistake. many thanks Doc Al.
 

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