A conveyor belt is moving at a constant speed of 2m/s. A box
is gently dropped on it. The coefficient of friction between
them is μ = 0.5. The distance that the box will move relative to
belt before coming to rest on it taking g = 10 ms⁻², is
(a) 1.2 m (b) 0.6 m
(c) zero (d) 0.4 m
Answers
Answered by
0
Answer:
zero, I guess
Explanation:
Since box dosent have any horizontal velocity in the first place the only way it will get a horizontal force is from the conveyor belt.
if box eventually comes to rest on the conveyor belt means that friction between it is sufficient to keep it on the belt and not slide off causing a varition in the force experienced.
so both belt n box experiences same net force so relative displacement is 0
Ps usually u need mass to calculate the friction force and thereby diatance.
Since its not given its most likely a juke question and u can assume the answer I gave.
If I m wrong and this is an IIT lvl question pls correct me.
Similar questions