A suitcase is gently dropped on a conveyor belt moving at a velocity of 3 m s - 1 . If the coefficient of friction between the belt and the suitcase is 0 . 5 , the displacement of the suitcase relativ

Question

A suitcase is gently dropped on a conveyor belt moving at a velocity of 3 m s - 1 . If the coefficient of friction between the belt and the suitcase is 0 . 5 , the displacement of the suitcase relative to conveyor belt before the slipping between the two is stopped, is ( g = 10 m s - 2 )

TestHub · Accepted Answer

Acceleration of the suitcase till the slipping continues is
$a = \frac{f_{m a x}}{m}$
$a = \frac{μ m g}{m} = μ g = 0.5 \times 10 = 5 m s^{- 2}$
Slipping will continue until its velocity also becomes $3 m s^{- 1}$ .
$∴ v = u + a t$
or or $t = 0.6 s$
In this time, the displacement of the suitcase
$s_{1} = \frac{1}{2} a t^{2} = \frac{1}{2} \times 5 \times {(0.6)}^{2} = 0. 9 m$
and the displacement of the belt,
$s_{2} = v t = 3 \times 0.6 = 1 .8 m$
Displacement of the suitcase with respect to the belt
$s_{1} - s_{2} = 0.9 m$

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