Question 243: Laws of Motion

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 $\upsilon $ m/s will be :

Accepted Answer

$F = {d \over {dt}}\left( {Mv} ight) = v{{dM} \over {dt}} + M{{dv} \over {dt}}$
As v is a constant,
$F = v{{dM} \over {dt}}$
But ${{dM} \over {dt}} = M\,kg/s$
$ 	herefore $ To keep the conveyer belt moving at v m/s,
force needed = vM newton.

Answer

${{Mv} \over 2}$ newton

Answer

zero

Answer

M$v$ newton

Answer

2 M$v$ newton

Laws of Motion

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 \upsilon m/s will be :

Choose the correct answer: