Dropping sand onto a conveyor belt

[serverxeon]

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?

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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!

 

[Simon Bridge]

the first is the total work for belt+sand, the second is just for the sand.

 

[serverxeon]

can u elaborate? why does the belt needs energy?
the before and after KE of the belt remains constant. no?

 

[Simon Bridge]

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.

 

[serverxeon]

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!

 

[Dick]

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?