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 ( $g=10 \, m \, s^{- 2}$ )

• A. $2.7m$
• B. $1.8 \, m$
• C. $0.9 \, m$
• D. $1.2m$

Answer 1

Answer: (C)

Acceleration of the suitcase till the slipping continues is
$a=\frac{f_{m a x}}{m}$
$a=\frac{\mu m g}{m}=\mu g=0.5\times 10=5 \, m​s^{- 2}$
Slipping will continue until its velocity also becomes $3ms^{- 1}$ .
$\therefore \, v=u+at$
or or $t=0.6s$
In this time, the displacement of the suitcase
$s_{1}=\frac{1}{2}at^{2}=\frac{1}{2}\times 5\times \left(0.6\right)^{2}=0.9m$
and the displacement of the belt,
$s_{2}=vt=3\times 0.6=1.8m$
Displacement of the suitcase with respect to the belt
$s_{1}-s_{2}=0.9m$