Question

A box is gently placed on a horizontal conveyer belt moving with a speed of $4m{s}^{-1}$. If the coefficient of friction between the box and the belt is 0.8, through what distance will the block slide without slipping? Take g = $10m{s}^{-2}$

Answer 1

Step 1: Given

Speed of the conveyor belt = $4m{s}^{-1}$

Coefficient of friction = 0.8

Step 2: Formula used

Frictional force , $f=\mu mg$

$v^2=u^2+2as$

Step 3: Calculation

Frictional force is responsible for the motion of the block.

Therefore,

$ma=\mu mg$

$\Rightarrow a=\mu g$

$\Rightarrow a=0.8\times 10$

$\Rightarrow a=8m{s}^{-2}$

When the block is sliding without slipping, then the velocity of the block is equal to velocity of the conveyor belt i.e.

$4m{s}^{-1}$

Now, by applying 3rd equation of motion

$v^2=u^2+2as$

${\Rightarrow 4}^2=0+2\times 8\times s$

$\Rightarrow s=\frac{16}{16}$

$\Rightarrow s=1m$

Therefore, block will move 1 m without slipping.