Question

# 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 $0.5$. The distance that the box will move relative to the belt before coming to rest on it (taking $g=10m/{s}^{2}$), is

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Solution

## Step 1: GivenSpeed of conveyor belt ($u$)=$2m/s$Coefficient of friction ($\mu$)=$0.5$Acceleration due to gravity ($g$)=$10m/{s}^{2}$Final speed =$v$Step 2: Formulae usedThe third equation of motion ${v}^{2}={u}^{2}+2aS$, where v is the final speed, u is the initial speed, a is the acceleration, and S is the distance covered.Step 3: CalculationWe know, that frictional force is the cause of deceleration of the box, thus,$F=\mu mg\left(butF=ma\right)\phantom{\rule{0ex}{0ex}}\therefore -ma=\mu mg\phantom{\rule{0ex}{0ex}}⇒a=-\mu g$where $\mu$ is the coefficient of friction.Here the negative sign indicates the deceleration of the box.Acceleration of the box considering friction,$a=-\mu g=-0.5×10=-5m/{s}^{2}$Applying the third equation of motion${v}^{2}={u}^{2}+2aS$$0={2}^{2}-2×5×S\phantom{\rule{0ex}{0ex}}0=4-10S\phantom{\rule{0ex}{0ex}}S=\frac{4}{10}=\frac{2}{5}=0.4m$Thus, the distance the box will move with respect to the conveyor belt is $0.4$ meters.

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