A block of mass m begins at rest at the top of a ramp at elevation h with whatev

Question

A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height.

The block slides down the ramp over a distance d until it reaches the bottom of the ramp. How much of its original total energy (in J) survives as KE when it reaches the ground?

(In other words, the acceleration is not zero like it was in lab and friction does not remove 100% of the original PE. How much of that original energy is left over after the friction does work to remove some?)

m = 6.0 kg
h = 10.0 m
d = 5 m
= 0.3
= 36.87°

Explanation / Answer

If the block is initially at rest then the total energy = potential energy = mgh
When the block slides some of the energy get dissipiated in the friction and some get converted in to the kinetic energy.
Frcition force = umgCos(36.87) where u is coefficient of kinetic friction
Work done by friction force = force*distance = umgCos36.87*d
Now by energy balance
Initial total energy = final total energy
mgh = umgdCos36.87 + (1/2)mV2   where V is the velocity at bottom
gh = ugdCo36.87 + 0.5V2
on solving we get
V = 13.1398 m/s

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