Say we are given a conveyor belt with sand falling onto it at rate $\Omega$. I am trying to find the power it takes for the belt to operate if it goes forward with constant velocity $v$, but using two different approaches I get two different answers.

The first way, I say that in one second, $\Omega$ amount of sand falls onto the belt, thus every second $\frac 1 2 \Omega v^2$ is required to continue moving the belt forward (simply by looking at kinetic energy difference) and $P=\frac 1 2 \Omega v^2$

The second way, I saw that $F=\frac {d(mv)} {dt}$, thus $F=\Omega v$. Finally, $P=Fv$ so $P=\Omega v^2$.

So which one is it: $P=\Omega v^2$ or $P=\frac {1} {2} \Omega v^2$, and where did I make an incorrect assumption in my two methods?


Your first answer gives you the rate at which the sand gains kinetic energy whereas your second answer gives you the rate at which work has to be done on the conveyor belt to keep it moving at constant speed.

When the sand falls on the belt in order to accelerate the sand to the same velocity as the belt there must be frictional forces between the sand and the belt.
There must also be relative movement between the belt and the sand during this acceleration phase as the sand cannot instantaneously to the velocity of the belt.
During the acceleration phase when there is slippage between the belt and the sand heat is generated and the rate of heat generation is the difference between your two answers.

So the motor which drives the belt increases the kinetic energy of the sand whilst heat is being generated due to the relative movement between the sand and the belt during the acceleration phase of the sand.

  • $\begingroup$ This argument about heat doesn't seem valid to me. The way OP derives the equations does NOT depend on anything related to heat. OP just considers the frictional force between the belt and the conveyor and calculates the work down by the frictional force on the sands. OP does not calculate anything about the power of motor. $\endgroup$ – Ma Joad May 1 '19 at 13:45
  • $\begingroup$ @MaJoad The mention of heat is there because of the relative movement between the sand and the belt with a frictional force producing the acceleration of the sand. $\endgroup$ – Farcher May 1 '19 at 13:48
  • $\begingroup$ Sorry if I have misunderstood you. Both expressions of OP deals with the work done by the frictional force on the sand. The two expressions deal with the same work done. Why do you think they are two different work done? How can you know it is because of the heat? $\endgroup$ – Ma Joad May 1 '19 at 13:55
  • $\begingroup$ @MaJoad It is because there is a discrepancy between the two methods. $\endgroup$ – Farcher May 1 '19 at 14:01

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