Asked by Mary
A mass sits at the top of a 2 meter long ramp inclined at an angle of 30 degrees (to the horizontal). When the mass is released, it slides down the ramp and then onto a level floor. If both the ramp and the board have a coefficient of .25 (Friction), how far out onto the floor will the box slide before coming to a stop?
Answers
Answered by
drwls
Then mass drops H = L sin 30 = 1 meter in elevation, and then slides an additional distance X.
P.E. loss = friction work done
M g L sin30 = 0.25 M g cos 30 * L + 0.25 M g X
M g cancels out
L sin30 = 0.25 cos30 L + 0.25 X
X = (4 L sin30 - L cos 30)
= L*(2 - sqrt3) = 0.536 m
P.E. loss = friction work done
M g L sin30 = 0.25 M g cos 30 * L + 0.25 M g X
M g cancels out
L sin30 = 0.25 cos30 L + 0.25 X
X = (4 L sin30 - L cos 30)
= L*(2 - sqrt3) = 0.536 m
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