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°

If the leftover energy is 109.8 J and the mass is 2 kg, what speed (in m/s) does the block have at the bottom of its slide? Revisit the definition of KE if needed.

Explanation / Answer

A )

potential energy = mgh

B )

from work energy theorem

mgh -umgdcos36.87 -mgd sin36.87 = KE

KE = 341.387 J

C )

KE = 0.5 mv^2

substituting values we get

v = 10.667 m/sec

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