Sand is being dropped on a conveyer 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

Question

Sand is being dropped on a conveyer 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) (Mv/2)newton (B) Zero (C) Mv newton (D) 2 Mv newton

Tardigrade Admin · Accepted Answer

F=(d/d t)(M v)=v (d M/d t)+M (d v/d t) As v is a constant, F=v (d M/d t) . But (d M/d t)=M kg / s ∴ To keep the conveyer belt moving at v m / s, force needed =v M newton.

Q. Sand is being dropped on a conveyer belt at the rate of $M k g / s$ . The force necessary to keep the belt moving with a constant velocity of v m/s will be

$\frac{M v}{2}$ newton

Zero

Mv newton

2 Mv newton

$F = \frac{d}{d t} (M v) = v \frac{d M}{d t} + M \frac{d v}{d t}$
As $v$ is a constant,
$F = v \frac{d M}{d t} .$
But $\frac{d M}{d t} = M k g / s$
$∴$ To keep the conveyer belt moving at $v m / s$ , force needed $= v M$ newton.

Q. Sand is being dropped on a conveyer belt at the rate of Mkg/s. The force necessary to keep the belt moving with a constant velocity of v m/s will be

2Mv​newton