Solving Frictional Force of Sand on Conveyor Belt

[abcd8989]

Homework Statement

Sand is poured onto a converyor belt at a constant rate of 25kg/s. The belt keeps moving horizontally at a constant speed of 4m/s.

Homework Equations

What is the fricional force acting on the sand?

The Attempt at a Solution

According to the solution, the answer can be calucated using "impulse":
Taking right hand side be positive.
friction =[ 25 X (+4) - 0 ] /1
= 100 N
I don't understand why such approach works. The teacher said that the sand experience kinetic friction at the instant when it touches the belt, and that some time(though very small) is required for the sand to accelerate from 0m/s to 4m/s. However, I don't know why the above equation can solve the problem. Why the time of impulse can be regarded as 1? It is not given that the sand accelerates from 0m/s to 4m/s in 1 second.

 

[Mindscrape]

You're right, this is a bogus solution. The problem really ought to be done with the true definition of F.

\Sum F_{ext} = \frac{dp}{dt} = m \frac{dv}{dt} + v \frac{dm}{dt}

where p is momentum.

So there is no acceleration from the conveyer belt, which makes dv/dt=0. There is however mass constantly being added to the conveyer belt at 25kg/s.

F_f=v_{conveyer}\frac{dm}{dt}

so F_f = 4m/s*25kg/s