Calculating Driving Force on a Conveyor Belt with Baggage Load?

[gaobo9109]

A conveyor belt consists of a horizontal endless belt running over driving rollers, moving at a constant speed of 1.5m/s. To keep the belt moving requires a greater driving force than for an empty belt. On average, the rate at which baggage is placed on one end of the belt and lifted off at the other end is 20 kg per second.

For this question, the answer is simply 1.5 x 20 = 30N, but I am not sure of the exact explanation for such working.

My own attempt at the solution is at follow:

Rate of change of mass dm/dt = 20
Change of momentum of one baggage = m(1.5 - 0)
Force needed to this change of momentum = dp/dt = dm/dt x v = 1.5 x 20 = 30N

I am not sure whether my reasoning is correct

 

[thebigstar25]

I agree with the answer ..

Force = dp/dt , p = mv , then, F= d(mv)/dt = dm/dt *v + dv/dt *m .. since it said that the velocity is constant, then dv/dt = 0 .. that yields to F = dm/dt *v , which is the answer you got .. :)