What is the power required for the belt to maintain its velocity

[Rasmus10]

Homework Statement

Corn is falling vertically down on a conveyor belt at a constant rate of 1 kg/s. The corn instantanoulsy gets the forward speed of the belt of v = 5m/s. What is the power required for the belt to maintain its velocity.

Homework Equations

p=mv
dp/dt=F
F*v=P
Ekin=1/2*m*v^2

The Attempt at a Solution

Well, i thought that the simple way to solve this was to say that the belt needs to apply kinetic energy to the corns. How much? 1/2*1kg*(5m/s)^2, and that every second so the effect would be 25/2 W = 12.5W.

However, that is not the correct answer according to my lecturers soloutions. He says it is 25. Well, i can think of; from a momentum kind of view. dp/dt=v*dm/dt=F
F*v=P so 5m/s*1kg/s*5m/s = 25W. What am i missing here?

Another approach. Correct me if I am wrong, but the energy needed to accelerate an object from f.x. velocity 0m/s to 5m/s is independent of the actual acceleration (even though its a non conservative force friction that is doing the acceleration?). To say, it deosn't matter if a large force is acting in a short time or a weak force in a long time? So
Faverage*Δt=Δp. From kinematics constant acceleration: x-x0=(v+v0)/2*t <=> if x0 and v0 is 0: x=v/2*t. Ok, so we multiply by this in the first equation to get the work: Fav*Δt*v/2=Δp. So if the acceleration time of the corn was 1 second, then i get the same result as i would get with energy considerations = 12.5W. Is this totally wrong?

Thanks for your help!

 

[Doc Al]

However, that is not the correct answer according to my lecturers soloutions. He says it is 25. Well, i can think of; from a momentum kind of view. dp/dt=v*dm/dt=F
F*v=P so 5m/s*1kg/s*5m/s = 25W. What am i missing here?

You are missing the fact that not all of the work done by the belt goes into increasing the translational KE of the corn. (The corn makes an inelastic collision with the moving belt.)