Question

Sand drops fall vertically at the rate of $3\,kg{s^{ - 1}}$ on the conveyor belt moving horizontally with a speed of $0.25\,m{s^{ - 1}}$ the extra power required to keep it moving is?

Answer 1

In order to solve this question, we will first calculate the force acting on the conveyor belt using the general formula of force and given parameters and then will use the general formula of power to figure out the net power required to keep the conveyor belt moving.

Formula used:
If $m, v$ be the mass and velocity of a body then according to newton’s second law of motion, force $F$ is calculated as
$F = \dfrac{d}{{dt}}(mv)$
which shows the rate of change of product of mass and velocity of the body.
Power of a body is calculated as,
$P = Fv$
where, $F, v$ be the force acting on the body, velocity of the body.

Complete step by step answer:
According to the question, we have given that if m represents mass of the sand and v represent velocity of conveyor belt then,
$\dfrac{{dm}}{{dt}} = 3kg{s^{ - 1}}$
Rate of mass of sand falling on the belt.
$v = 0.25\,m{s^{ - 1}}$ constant speed of the belt.
Force $F$ acting on the belt is calculated using formula,
$F = \dfrac{d}{{dt}}(mv)$

Since, $v$ is constant here so it can be written as,
$F = v\dfrac{{dm}}{{dt}}$
On putting the value of parameters we get,
$F = 0.25 \times 3$
$\Rightarrow F = 0.75N$
Now, power can be calculated by using the formula, $P = Fv$. We get,
$P = 0.75 \times 0.25$
$\therefore P = 0.1875W$

Hence, the power required to keep the conveyor belt moving is $P = 0.1875\,W$.

Note: It should be remembered that, most of the times body has the change in its velocity which is called acceleration of the body instead of change in its mass in classical mechanics but in relativistic mechanics its mass of the body which always changes while moving at the speed comparable to the speed of light.