Question

When a belt moves horizontally at a constant speed of $1.5{\text{ms}}^{-1},$ gravels are falling on it at rate of $5$ ${\text{kg s}}^{-1}.$ Then, the extra power required to drive the belt is

• A. $11.25\text{W}$
• B. $37.5\text{W}$
• C. $7.5\text{W}$
• D. $0.75\text{W}$

Answer 1

The correct option is A $11.25\text{W}$

It is given that the velocity of the belt $v=1.5{\text{ms}}^{-1}$

The rate at which the mass of the gravel falls is given by

$\frac{dm}{dt}=5\text{kg}{\text{s}}^{-1}$

By the newtons second law, we know that force is the rate of change of momentum $(p=mv)$

$F=\frac{dp}{dt}=\frac{dv}{dt}m+\frac{dm}{dt}v=5\times 1.5=7.5N$
$($As the velocity $v$ is constant, change in velocity $dv=0)$

$\Rightarrow P=F\times v=7.5\times 1.5=11.25W$

Hence option A is the correct answer