Calculate Power Needed to Run Conveyor Belt with dm/dt kgs-1

[me189]

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

 

[Doc Al]

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

Your solution assumes energy conservation, but realize that the collision of sand and belt is inelastic.

Your friend's solution is correct.

 

[me189]

But is applying P = Fv right?
for P = Fv, the v basically implys displacement over time, which is the average velocity of the sand during acceleration , isn't it?
but at the instant that the sand drops on the conveyor, the velocity must be smaller than the velocity of the conveyor. Then it seems to be strange to apply P = Fv by using the velocity of the conveyor as the average of the sand during acceleration.

Also, if i "put" the sand on the conveyor instead of "dropping" it onto the conveyor, it should yield the same result, isn't it??

 

[Doc Al]

But is applying P = Fv right?
for P = Fv, the v basically implys displacement over time, which is the average velocity of the sand during acceleration , isn't it?
but at the instant that the sand drops on the conveyor, the velocity must be smaller than the velocity of the conveyor. Then it seems to be strange to apply P = Fv by using the velocity of the conveyor as the average of the sand during acceleration.

Not sure what you're saying here. What matters is the rate of change of momentum of the sand. In every unit of time the change in momentum is (Δm)v. (All bits of sand have the same change in velocity.) So the rate of change of that momentum is (Δm/Δt)v = (dm/dt)v. That's the force required.

Also, if i "put" the sand on the conveyor instead of "dropping" it onto the conveyor, it should yield the same result, isn't it??

No. The height from which the sand is dropped--if that's what you're getting at--is not relevant. (It's not even mentioned in the problem.)

 

[me189]

i understand that the force required is (dm/dt)v
but in the formula P = Fv, v means the velocity of sand when the force is applied, isn't it? However, the velocity of the sand bit is changing when the force is applied.
The initial velocity of the sand bit (when it is drop on the conveyor) is zero.
The final velocity of the sand bit is v ms^-1, the same as the conveyor.
Shouldnt the v in P = Fv be [ (0+v)/2 ] ms^-1??

If the sand bit is "put" on the conveyor "gently", will the collision become more elastic in this case?? (and ultimately become completely elastic??)

 

[Doc Al]

i understand that the force required is (dm/dt)v
but in the formula P = Fv, v means the velocity of sand when the force is applied, isn't it?

No, it's the velocity of what's applying the force, which is the belt.

If the sand bit is "put" on the conveyor "gently", will the collision become more elastic in this case?? (and ultimately become completely elastic??)

No. What's inelastic about it is the fact that the sand 'collides' with the moving belt. Its the horizontal speed that counts.

 

[me189]

thx a lot!
i understand the case now^^