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# A box is gently dropped on a conveyor belt in an airport moving of 3m/s. If the coeffiecient of friction between belt and box is 0.5, how far will the box move on the belt (relative to belt) before coming to rest ?

$(a)\;2\; m \quad (b)\; -0.918 \;m \quad (c)\; 0.863 \;m \quad (d)\;0.918\;m$

The initial velocity of box relative to the belt is 3m/s in backward direction. (opposite to motion of belt) ie 3 m/s. The box experience maximum frictional force $F=\mu R=\mu mg$
Therefore $a=\large\frac{Force}{mass}$
$\qquad\qquad=\large\frac{\mu mg}{m}$
$\qquad\qquad=\mu g$
$\qquad\qquad=0.5 \times 9.7 =4.9 m/s^2$
in forward direction
Let the box come to rest in distance S
$v^2=a^2+2as$
$0=(-3)^2+2 \times 4.9 \times s$
$-9=2 \times 4.9 \times S$
$S=-0.918\;m$
We sign shows the box moves in opposite direction of motion of belt
Hence b is the correct answer

edited Jan 26, 2014 by meena.p