- #1
me189
- 4
- 0
sand is pouring on a conveyor belt in a rate of rate of dm/dt kgs-1
if the velocity of the conveyor belt is v ms-1,
what is the power needed to run the conveyor belt.
My friend arrived at his answer in this way
F = dp/dt = d(mv)/dt = v (dm/dt) N
Power = Fv = (dm/dt)v2 Watts
(however the velocity of the accelerating sand is not constant, right?? it should be increasing linearly if acceleration is constant, the average velocity of the sand during acceleration should be v/2 ms-1, right?)
But i arrived at a slightly different answer in this way
Total change in the KE of a mass m of sand after falling on the belt = 0.5mv2
Power
= dW/dt
= d(0.5mv2)/dt
= 0.5 dm/dt v2 Watts
if the velocity of the conveyor belt is v ms-1,
what is the power needed to run the conveyor belt.
My friend arrived at his answer in this way
F = dp/dt = d(mv)/dt = v (dm/dt) N
Power = Fv = (dm/dt)v2 Watts
(however the velocity of the accelerating sand is not constant, right?? it should be increasing linearly if acceleration is constant, the average velocity of the sand during acceleration should be v/2 ms-1, right?)
But i arrived at a slightly different answer in this way
Total change in the KE of a mass m of sand after falling on the belt = 0.5mv2
Power
= dW/dt
= d(0.5mv2)/dt
= 0.5 dm/dt v2 Watts