Question

A block is gently placed on a conveyor belt moving horizontally at a constant speed. After $t=4\text{s}$, the velocity of the block becomes equal to the velocity of the belt. If the coefficient of static friction between the block and the belt is $\mu =0.2$, then what is the velocity of the conveyor belt ?

• A. $2\text{m/s}$
• B. $4\text{m/s}$
• C. $6\text{m/s}$
• D. $8\text{m/s}$

Answer 1

The correct option is D $8\text{m/s}$

Initially block is at rest while conveyer belt is moving with velocity $v$, due to relative motion between the block and belt, limiting friction will act on the block in the direction of motion of belt.
After time $t$ when block reaches velocity equal to that of belt relative motion ceases.

$f=\mu \times N$
$f=\mu \times mg...\left(i\right)$
applying newton's $2^{nd}$ law on block
$f=ma...\left(ii\right)$
$\Rightarrow \mu \times mg=ma$
$\therefore a=0.2\times 10=2{\text{m/s}}^2$

Applying kinematic equation on the block $(u=0$), its velocity $\left(v\right)$ at time $t=4\text{s}$ is given by :
$v=u+at$
$\therefore v=0+2\times 4=8\text{m/s}$
At time $t=4\text{s}$ both block and belt have same velocity, hence velocity of conveyor belt is $8\text{m/s}$.