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# Sand is being dropped on a conveyor belt at the rate of $M\, kg/s$. The force necessary to keep the belt moving with a constant velocity of $v\, m/s$ will be:(A) $\dfrac{Mv}{2}$ Newton(B) Zero(C) $Mv$ Newton(D) $2Mv$ Newton

Last updated date: 20th Jun 2024
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Hint:We should know that force is the rate of change of momentum and momentum is defined as a product of mass and acceleration. Substituting the values of rate of change of mass as M kg/s and rate of change of velocity as zero we get the desired value of force. Remember force is a vector quantity and so its direction is also to be taken into account.

Given the rate of change of mass with respect to time is M kg/s.
The constant velocity of the conveyor belt is v m/s.
We know that according to Newton's second law of motion, force applied is directly proportional to the change in the momentum(p) and force is in the direction of the change of momentum.
$F=\dfrac{dp}{dt}$ ---------(1)
Momentum is defined as a product of mass and velocity,$p=mv$.
Therefore equation 1 becomes,
$F=\dfrac{d\left( mv \right)}{dt}$
On differentiating we get,
$F=m\dfrac{dv}{dt}+v\dfrac{dm}{dt}$
The rate of change of velocity with time is zero since the velocity is constant.
Substituting the value of rate of change of mass with respect to time is M kg/s.
Therefore we get force as,
$F=m\times 0+vM \\ \therefore F=Mv$
The force necessary to keep the belt moving with a constant velocity of $v\, m/s$ will be $Mv\,N$.

Hence, option C is correct.

Note:One should know Newton's second law of motion to solve such types of problems, which states that force is directly proportional to the rate of change of momentum, $F=\dfrac{dp}{dt}$. When more than one force acts on any system then we have to add all the forces vectorially using either triangle law of vector addition or parallelogram law of vector addition. If the external force is zero, then the acceleration of the body is zero. Newton’s third law states that forces always occur in pairs.