Question

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. $\frac{Mv}{2}$ newton
• B. Zero
• C. Mv newton
• D. 2 Mv newton

Answer 1

The correct option is C Mv newton

$F_{ext.}=\frac{dp}{dt}=\frac{d\left(mv\right)}{dt}$;
m = mass of system as conveyor belt with sand
drops at time t.
$F_{ext.}=m\frac{dv}{dt}+v\frac{dm}{dt}$
but $v\to$constant, so,$\frac{dv}{dt}=0$
$F_{ext.}=v\frac{dm}{dt}=Mv$
$F_{ext.}$ is in direction of $\overrightarrow{v}$of belt.
OR
Since $m.\frac{dv}{dt}=F_{ext.}+F_{reaction}$ ........................(i)
Consider belt as a system with variable mass
$\frac{dv}{dt}=0$, m = mass of system at time t
$F_{reaction}=v_{rec.}\frac{dm}{dt}=-v\frac{dm}{dt}$ .................(ii)
$\Rightarrow F_{ext.}=v.\frac{dm}{dt}=Mv$.
$\overrightarrow{F_{ext.}}$is in direction of $\overrightarrow{v}$ to keep belt moving with constant velocity.