Question

A suitcase is gently dropped on a conveyor belt moving at a velocity of $3m{s}^{-1}$ . 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 ( $g=10m{s}^{-2}$ )

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

Answer 1

Answer: (C)

Acceleration of the suitcase till the slipping continues is
$a=\frac{f_{max}}{m}$
$a=\frac{\mu mg}{m}=\mu g=0.5\times 10=5m{s}^{-2}$
Slipping will continue until its velocity also becomes $3m{s}^{-1}$ .
$\therefore v=u+at$
or or $t=0.6s$
In this time, the displacement of the suitcase
$s_1=\frac{1}{2}a{t}^2=\frac{1}{2}\times 5\times {\left(0.6\right)}^2=0.9m$
and the displacement of the belt,
$s_2=vt=3\times 0.6=1.8m$
Displacement of the suitcase with respect to the belt
$s_1-s_2=0.9m$