Conveyor Belt's Power: Calculating Energy and Power in a Moving System

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SUMMARY

The discussion focuses on calculating energy and power in a moving conveyor belt system where sand falls at a rate of 0.5 kg/s onto the belt moving at 2.0 m/s. The power expended in driving the belt is calculated using the formula P = F.v, resulting in 2 W. The kinetic energy acquired by the sand is also derived, leading to a calculation of 1 W for the rate of change of kinetic energy. The difference between the power expended and the kinetic energy acquired is clarified through the application of relevant physics equations.

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Homework Statement


Sand falls at a rate of 0.5 kg/s onto a conveyor belt which is moving at a constant speed of 2.0 m/s.

Questions:

a. Find the power expended in driving the belt!
b. Find the rate at which the sand acquires kinetic energy!
c. Account for any difference between the answer in part (a) and (b)


Homework Equations


F = m.(dv/dt) + v (dm/dt)
P = W/t = F.v


The Attempt at a Solution


For question (a), i think:
P = F.v
= [m.(dv/dt) + v (dm/dt)].v ; dv/dt = 0
= (dm/dt) . v^2
= ... W

Is it right ?

For question (b) i have no idea about the question. The question asked about speed or Power ?

Thanks a lot.
 
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you know that power = change in energy per second?
Can you calculate the rate of change of KE per second of the sand?
 
do you mean like this ?
W = KE / t
= 0.5 m v^2 / t

m/t = 0.5 kg.
v = 2 m/s

so, W = 1 W ? is this right for question (b) ?

and

P = F.v
= [m.(dv/dt) + v (dm/dt)].v ; dv/dt = 0
= (dm/dt) . v^2
= 2 W for question (a) ?
 

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