Dropping sand onto a conveyor belt

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Homework Help Overview

The problem involves a conveyor belt system where sand is continuously dropped onto the belt. The discussion focuses on calculating the power supplied to maintain the belt's velocity and the rate of change of kinetic energy (KE) of the sand. Participants explore the relationship between the power input and the energy associated with the sand and the belt.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculations for power and kinetic energy, questioning the apparent discrepancy between the total power input and the energy gained by the sand. They explore the need for energy to maintain the belt's speed despite the added mass.

Discussion Status

The discussion is active, with participants providing insights into the energy dynamics of the system. Some have suggested that the belt requires additional energy to maintain its speed with the added mass of sand, while others are questioning the assumptions about the kinetic energy of the belt and the sand.

Contextual Notes

Participants are considering the implications of a frictionless scenario and how it affects the movement of sand on the belt. There is an ongoing examination of the energy distribution in the system, particularly regarding the power input from the motor and its effects on both the belt and the sand.

serverxeon
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Homework Statement



Sand is placed onto a horizontal conveyor belt.
rate of sand placed = 0.5 kg s-1
velocity of conveyor belt = 2 m s-1

a) what is the power supplied by the system to keep the velocity constant?
b) what is the rate of change of KE by the sand
c) why is the answer in a and b different?

--------------------------

I know the answer to part (a) comes from
P = Fv = dp/dt * v = dm/dt * v2 = 2W

and the answer to part (b) is
just 1/2* (m/t) * v2 = 1W

my question is, where does the other 1W goes to?
2W is inputted, but only 1W goes into the KE of the sand!
 
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the first is the total work for belt+sand, the second is just for the sand.
 
can u elaborate? why does the belt needs energy?
the before and after KE of the belt remains constant. no?
 
Can you think of anywhere else the energy can go?

without the extra energy, belt must go slower with extra mass on it, therefore it needs more energy to maintain it's speed. Try it with a moving plate instead of the belt, and you just drop one lump mass on it.
 
I understand where you are coming from.

But,

looking at the scenario of in one second, in the case where the motor to the belt is running,
Comparing the KE of the belt before and after, -> no change as the belt will remain at velocity v!
comparing KE of sand before and after -> gained 1J

Still does not account for the 2J put in by the motor per second!
 
Last edited:
serverxeon said:
I understand where you are coming from.

But,

looking at the scenario of in one second, in the case where the motor to the belt is running,
Comparing the KE of the belt before and after, -> no change as the belt will remain at velocity v!
comparing KE of sand before and after -> gained 1J

Still does not account for the 2J put in by the motor per second!

If the belt were frictionless the sand wouldn't move along with the belt. It would just sit where it was dumped while the belt slides underneath it. Does that give you a hint?
 

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