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

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The discussion focuses on calculating the power and kinetic energy associated with a conveyor belt system where sand falls at a rate of 0.5 kg/s onto a moving belt at 2.0 m/s. For part (a), the power expended in driving the belt is derived using the formula P = F.v, leading to a calculation of 2 W. In part (b), the kinetic energy acquired by the sand is questioned, with an attempt to calculate it resulting in 1 W. The difference between the power calculated in part (a) and the kinetic energy in part (b) is acknowledged but not fully resolved. Overall, the thread highlights the application of physics equations to solve energy and power problems in a dynamic system.
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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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