Solving Conveyor Belt Problems: Force, Magnitude, and Power Calculations

[DivD]

i have a test tomorow and our teacher gave us an example of a problem that might be coming tomorow. the question is
Sand falls on a conveyor belt at 60 kg per second and the belt moves at 2 meters per second.
a) what is the force moving the sand?
b) what is the magnitude of the force?
c) force exerted on belt by the engine.
d) the engine has 40% efficiency. how much is the input power?

well i am guessing that the first three parts will be the same based on some of the other questions i looked at https://www.physicsforums.com/archive.../t-150868.html
do we multiply the rate at which the mass is changing times the velocity to find the force on the belt? so would the force on the belt be 120 N since 60 x 2?? and the magnitude would be the same I am guessing. and the force exerted on the belt by the engine would also be 120 N right? but I am a bit lost on how to convert it to power. can anyone help me please? i have the test tomorow and i need to know if i am correct or not. thank you in advance

 

[rl.bhat]

It is the problem of varying mass.
At any instant let the moment of the system be
P = m*v
External force = dP/dt = v*dm/dt, because dv/dt = 0
Power supplied by the external force = F*v.

 

[DivD]

It is the problem of varying mass.
At any instant let the moment of the system be
P = m*v
External force = dP/dt = v*dm/dt, because dv/dt = 0
Power supplied by the external force = F*v.

yea but these are just the formulas. i don't get the answer. like are my answers right? thank

 

[rl.bhat]

Your answer for the force is correct.

 

[DivD]

Your answer for the force is correct.

for the power required for the belt it would be 240 watts since power=F*v=240*2
since it is 40% efficient would that mean that the total power is 600 watts since 40% of 600 is 240?? and is the answer 120 N for the first three parts? or only for one of the three parts?

 

[yeteendra]

# force required to move the sand is frictional force between the sand and belt
# the magnitude is 120N
# since efficiency is 40%
then 40% of P = dW/dt = Fv
0.4 P= 120 X 2 = 240
P = 600W