Find Extra Force & Power to Keep Conveyor Belt Moving

[ritwik06]

Homework Statement

Sand drops are falling gently at the rate of 2 kg/sec on a conveyor belt moving horizontally with a velocity of 5 m/s. Find the extra force and power required to keep the conveyor belt moving at constant speed.

Homework Equations

Total work done on a body= Change in kinetic energy
Force=dP/dt

The Attempt at a Solution

For finding the extra force I use:
Force=dP/dt
Change in momentum of sand=2*5=10 Kg m/s per second
therefore F=10 Newton
here the answer matches with the one given in my book.
for power
Power=WorkDone/Time
Total work done on a body= Change in kinetic energy
=0.5*2*5*5=25 J
Therefore the answer =25 watt
but the answer at the back of my book says 50 watt.
I am confused. Where am I wrong?

 

[Dr.D]

Every second, 2 kg of sand has to be brought up to a speed of 5 m/s, so the change in energy is
deltaT = (1/2)*M*(V2^2-V1^2) = (1/2)*2*(5^2 - 0^2) = 25 J
This is the work to be done every second, so that the power is
P = deltaT/deltat = (25)/1 = 25 W
I think you are wrong only in relying on the back of the book.

 

[ritwik06]

Every second, 2 kg of sand has to be brought up to a speed of 5 m/s, so the change in energy is
deltaT = (1/2)*M*(V2^2-V1^2) = (1/2)*2*(5^2 - 0^2) = 25 J
This is the work to be done every second, so that the power is
P = deltaT/deltat = (25)/1 = 25 W
I think you are wrong only in relying on the back of the book.

Thanks a lot sir. But if I see it this way, the scenario is different.
the extra force as needed was 10 N. The point of application moves 5 m in a second. SO the net work done in one second could possibly be =5*10=50 J. and power=50/1= 50 watt

i am confused. Please explain the fallacy in this method.

 

[Dr.D]

You bring up a very interesting point, something that I had overlooked. Thank you.

If we start with Newton's second law applied to a single sand particle in the horizontal direction, the only force acting is the friction force of the belt and the equation reads
f = m * d^2x/d^2 = m * dx/dt * d (dx/dt)/dx
which can be re-arranged to express the differeential work done on the particle by friction
dW = f*dx = m * dx/dt * d(dx/dt) = d(m*v^2/2)
Now express the friction power, Pf as
Pf = dW/dt = f *dx/dt = d(m*v^2/2)/dt = 10*5 = 50 W

Now this is friction power, and as we have both observed, only half of it winds up in the eventual load kinetic energy. The remaining part is dissipated as heat - it is lost from the mechanical system. Thus dribbling material onto the conveyor is a relatively inefficient way of putting it onto the conveyor compared to putting it on from a chute where it has a velocity that almost matches the conveyor velocity when it lands.

There is a minor fallacy in your statement. The point of application of the friction force most likely does not move 5 m in one second, that is to say, this force is not active through a 5 m distance. It is active through whatever distance the grain of sand slips on the conveyor belt. Once slipping stops, the friction force goes to zero.

 

[ritwik06]

You bring up a very interesting point, something that I had overlooked. Thank you.

Physics is full of such things, isn't it? I learned it after I joined PF

If we start with Newton's second law applied to a single sand particle in the horizontal direction, the only force acting is the friction force of the belt and the equation reads
f = m * d^2x/d^2 = m * dx/dt * d (dx/dt)/dx
which can be re-arranged to express the differential work done on the particle by friction
dW = f*dx = m * dx/dt * d(dx/dt) = d(m*v^2/2)
Now express the friction power, Pf as
Pf = dW/dt = f *dx/dt = d(m*v^2/2)/dt = 10*5 = 50 W

I understood the whole thing until the last line;
Pf = dW/dt = d(m*v^2/2)/dt = 10*5 = 50 W
d(m*v^2/2)/dt=m*v*(dv/dt)
m*dv/dt= friction=10 N isn't it?

but "v" varies over time. Why should I put it 2? Isn't power a function of time? If we take the force constant over time, the acceleration is also a non zero constant. So velocity is a function of time. Right?

Now this is friction power, and as we have both observed, only half of it winds up in the eventual load kinetic energy. The remaining part is dissipated as heat - it is lost from the mechanical system. Thus dribbling material onto the conveyor is a relatively inefficient way of putting it onto the conveyor compared to putting it on from a chute where it has a velocity that almost matches the conveyor velocity when it lands.

There is a minor fallacy in your statement. The point of application of the friction force most likely does not move 5 m in one second, that is to say, this force is not active through a 5 m distance. It is active through whatever distance the grain of sand slips on the conveyor belt. Once slipping stops, the friction force goes to zero.

Thanks for the explanation. I have understood this part well. Thanks.

 

[ritwik06]

I understood the whole thing until the last line;
Pf = dW/dt = d(m*v^2/2)/dt = 10*5 = 50 W
d(m*v^2/2)/dt=m*v*(dv/dt)
m*dv/dt= friction=10 N isn't it?
but "v" varies over time. Why should I put it 2? Isn't power a function of time? If we take the force constant over time, the acceleration is also a non zero constant. So velocity is a function of time. Right?

This post is just to bring the topic back to light. I still have this confusion. I urge you to please help me.

 

[ritwik06]

If we start with Newton's second law applied to a single sand particle in the horizontal direction, the only force acting is the friction force of the belt and the equation reads
f = m * d^2x/d^2 = m * dx/dt * d (dx/dt)/dx
which can be re-arranged to express the differeential work done on the particle by friction
dW = f*dx = m * dx/dt * d(dx/dt) = d(m*v^2/2)
Now express the friction power, Pf as
Pf = dW/dt = f *dx/dt = d(m*v^2/2)/dt = 10*5 = 50 W

I still have a doubt with this question. When the frictional force is constant, till the relative slipping is there. The velocity changes with time. Therefore \vec{F}.\vec{v} also changes, so the power generated should be variable over time.

Now consider one more situation:
Say. My second method was absolutely wrong I guess. But you were right, there was only friction force between the conveyor and the sand. It attains a velocity of 5m/s in one second. It does not travel 5 m in that second.
Average force=10 N
Acceleration=10/2=5 m/s^2
displacement=5/2
Average power=Work/Time
=(10*5/2) /1
=25 WI am thoroughly confused over this. Force on a particle is dP/dt
How is power defined?
dW/dt, right??
If I want to calculate the average power= delta (W)/delta (t)
delta W= F(avg)*delta (x)
where x is displacement.But, when I reread the question, I found out that the question asks for the power generated by conveyor. I think the time period of action of friction is very small ->0
So in that case, I can say that average power is \vec{F}.\vec{v}here v=2 and f=10.Why isn't anyone trying to help me?