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

Answer 1

Step 1: Given

Speed 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 used

The 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: Calculation

We know, that frictional force is the cause of deceleration of the box, thus,

$$\begin{gathered} F=\mu mg\left(butF=ma\right) \\ \therefore -ma=\mu mg \\ \Rightarrow a=-\mu g \end{gathered}$$

where $\mu$ is the coefficient of friction.

Here the negative sign indicates the deceleration of the box.

Acceleration of the box considering friction,$a=-\mu g=-0.5\times 10=-5m/s^2$

Applying the third equation of motion

$v^2=u^2+2aS$

$$\begin{gathered} 0=2^2-2\times 5\times S \\ 0=4-10S \\ S=\frac{4}{10}=\frac{2}{5}=0.4m \end{gathered}$$

Thus, the distance the box will move with respect to the conveyor belt is $0.4$ meters.