How Is Power Calculated in a Conveyor Belt System?

[ReMa]

I have two questions regarding this, any help would be GREAT!

1st Problem:

Homework Statement

A conveyor belt lifts 1000kg of rocks per minute a vertical distance of 10m. The rocks are at rest at the bottom of the belt and are ejected at 5 m/s. The power supplied to this machine is:

A) 1000W
B) 1260W
C) 1630W
D) 1840W
E) 2100 W

Homework Equations

P=Fv

p=mv
[tex]\sum[/tex]F[tex]\Delta[/tex]T=m(vf-vo)

The Attempt at a Solution

vo = 0m/s
vf = 5m/s

1000kg/min = 16.67kg/s

sumF[tex]\Delta[/tex]T=m(vf-vo)
sumF=m(vf-vo)/[tex]\Delta[/tex]T
sumF=m(vf)-m(vo)/[tex]\Delta[/tex]T

since vo = 0/ms

sumF=m(vf)/[tex]\Delta[/tex]T
sumF=m/[tex]\Delta[/tex]T (vf)

subbing in values...

m/dT = 16.67kg/s

F = (16.67)(5) = 83.35N

and now for Power...
P = Fv
P = 83.35(5) = 416.75W

Now, I know I've done something wrong, especially as I haven't accounted for height. I feel like I am EXTREMELY off with attempting to solve this so i'd appreciate any help.2nd Problem:

Homework Statement

Water runs out of a horizontal pipe at the rate of 120kg/min. It falls 3.20m to the ground. Assuming the water doesn't splash up, what average vertical force does it exert on the ground?

Homework Equations

P=Fv

p=mv
sumF[tex]\Delta[/tex]T=m(vf-vo)

The Attempt at a Solution

SOLVED THIS ONE - if anyone else needs the solution though let me know and i'll post it.
Still need help on the first though... workin' on it however.

 

[RoyalCat]

For the first problem, ignore the forces within the machine. Ask yourself what happens to the rocks every second (Think energy rather than momentum)

 

[kerimek]

Hi there! It looks like you're on the right track with the first problem, but there are a few things that need to be clarified. First, the power supplied to the machine is not the same as the force exerted on the rocks. Power is the rate at which work is done, which is equal to force multiplied by velocity (P=Fv). In this case, the power supplied to the machine is the product of the force exerted on the rocks (which you calculated correctly as 83.35N) and the velocity at which the rocks are ejected (5m/s). So the correct answer for the power supplied to the machine would be 83.35N x 5m/s = 416.75W.

Now, in order to solve for the force exerted on the rocks, you need to consider the change in momentum of the rocks. The equation for momentum is p=mv, where p is momentum, m is mass, and v is velocity. In this case, the rocks are initially at rest (vo=0m/s) and are ejected at a velocity of 5m/s (vf=5m/s). So the change in momentum, or impulse, is given by the equation \Delta p = m(vf-vo). Since the mass of the rocks is 1000kg, the impulse is equal to 1000kg x (5m/s - 0m/s) = 5000kgm/s. This is equal to the force exerted on the rocks (83.35N) multiplied by the time it takes for the rocks to be lifted (1 minute or 60 seconds). So the equation is \Delta p = F \Delta t, or 5000kgm/s = 83.35N x 60s. Solving for the force, we get F = 5000kgm/s / 60s = 83.33N, which is the same answer you calculated earlier.

So in summary, the power supplied to the machine is 416.75W and the force exerted on the rocks is 83.33N. I hope this helps! Let me know if you have any further questions.