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?
Answer
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.
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