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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
Detailed Solution
Take the belt together with the sand lying on it as the system; its mass $m$ increases with time as sand drops on it.
Newton's second law: $F_{ext} = \dfrac{dp}{dt} = \dfrac{d(mv)}{dt}$
$F_{ext} = m\dfrac{dv}{dt} + v\dfrac{dm}{dt}$
The belt moves with constant velocity, so $\dfrac{dv}{dt} = 0$.
$\Rightarrow F_{ext} = v\dfrac{dm}{dt}$
Given $\dfrac{dm}{dt} = M$ kg/s, so $F_{ext} = Mv$ newton.
Alternative view: the sand falls with zero horizontal velocity and has to be given a horizontal velocity $v$. Every second a mass $M$ gains horizontal momentum $Mv$, so the reaction on the belt is $Mv$ backward, and a forward force $Mv$ is needed to keep the velocity constant.
This force acts in the direction of the velocity of the belt and equals $Mv$ newton.
Newton's second law: $F_{ext} = \dfrac{dp}{dt} = \dfrac{d(mv)}{dt}$
$F_{ext} = m\dfrac{dv}{dt} + v\dfrac{dm}{dt}$
The belt moves with constant velocity, so $\dfrac{dv}{dt} = 0$.
$\Rightarrow F_{ext} = v\dfrac{dm}{dt}$
Given $\dfrac{dm}{dt} = M$ kg/s, so $F_{ext} = Mv$ newton.
Alternative view: the sand falls with zero horizontal velocity and has to be given a horizontal velocity $v$. Every second a mass $M$ gains horizontal momentum $Mv$, so the reaction on the belt is $Mv$ backward, and a forward force $Mv$ is needed to keep the velocity constant.
This force acts in the direction of the velocity of the belt and equals $Mv$ newton.