# Homework Help: Force and Power

1. Jan 26, 2009

### ritwik06

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Total work done on a body= Change in kinetic energy
Force=dP/dt

3. 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
but the answer at the back of my book says 50 watt.
I am confused. Where am I wrong?

2. Jan 26, 2009

### 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.

3. Jan 26, 2009

### ritwik06

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.

4. Jan 26, 2009

### 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.

5. Jan 27, 2009

### ritwik06

Physics is full of such things, isnt it? I learned it after I joined PF
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 isnt 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?

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

6. Jan 28, 2009

### ritwik06

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

7. Jan 30, 2009

### ritwik06

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 W

I 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 isnt anyone trying to help me?????

Last edited: Jan 31, 2009