A 10 kg sandbag drops a negligibly small distance and lands upon a horizontally

Question

A 10 kg sandbag drops a negligibly small distance and lands upon a horizontally moving conveyor belt. The belt moves with a velocity of 2.5 m/s +x with respect to the earth’s surface. The kinetic friction coefficient between the bag and belt is 0.4. Draw clearly labeled free body diagrams to support your answer to the questions.

a) How long does it take for the sandbag to stop slipping on the conveyor belt?
b) Calculate the displacement of the bag - relative to the point where it first made

contact with the belt - once slipping ceases.
c) Calculate the displacement of the bag relative to the earth once slipping ceases.

Explanation / Answer

a) problem a can be visualized as relative slipping where back is moving at 2.5m/s backwards with respect to conveyor belt at rest.

So for the bag to be stopped,

v = u -at

So = 0 = 2.5 - mu*g * t

t = 2.5 / mu * g ==> 2.5 / (0.4*9.8) = 0.63s

Displacement of bag =

s = u^2/2a = 2.5^2 / (2*0.4*9.8) = 0.797 m

Displacement relative to earth = Displacement of belt - displacement of bag

SO 2.5*t = 2.5 * 0.63 = 1.575 - 0.797 m = 0.778 m

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