- #1
Bruce_Pipi121
- 9
- 4
- Homework Statement
- A conveyor belt is driven at velocity ##v## by a motor. Sand drops vertically on to the belt at a rate of ##m~kg~s^{-1}##.
What is the additional power needed to keep the conveyor belt moving at a steady speed when the sand starts to fall on it?
- Relevant Equations
- ##E_K=\frac{1}{2} m v^2##, ##p=mv##, ##F=\frac{\Delta p}{\Delta t}##
I can understand how it can be explained using the change of momentum to find the average force on the belt, then us $$P=Fv$$ to get the power.
What I didn't understand is why the energy solution won't work. If in 1 second, sand with mass m has fallen on the belt and accelerated to speed v, then the sand has gained kinetic energy $$E_K=\frac{1}{2} m v^2$$. As the mark scheme does confirm this is the amount of energy gained by sand, it states that this cannot be the only energy involved here, otherwise the acceleration of a grain of sand will be infinitely large.
I couldn't understand what does it have to do with the acceleration of sand, after all, if we assume the belt is way more massive and sticky, then the acceleration of sand in this case would indeed be very very large.
What I didn't understand is why the energy solution won't work. If in 1 second, sand with mass m has fallen on the belt and accelerated to speed v, then the sand has gained kinetic energy $$E_K=\frac{1}{2} m v^2$$. As the mark scheme does confirm this is the amount of energy gained by sand, it states that this cannot be the only energy involved here, otherwise the acceleration of a grain of sand will be infinitely large.
I couldn't understand what does it have to do with the acceleration of sand, after all, if we assume the belt is way more massive and sticky, then the acceleration of sand in this case would indeed be very very large.