Understanding the Discrepancy in Power Calculation for a Moving Conveyor Belt

In summary, the problem set asks for the average power required by a conveyor belt as it rapidly accelerates stationary packages of mass. However, the power calculation based on force and velocity does not match the power calculation based on kinetic energy. The discrepancy may be due to the fact that conservation of energy and momentum are both affected by outside forces.
  • #1
patricks1
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Hi, everyone, this is my first time posting.

I have a problem set that states "A machine at a packaging facility places stationary packages of mass m onto a horizontal conveyor belt that is moving packages steadily and horizontally at speed v. Once placed on the belt, the packages start moving at speed v; that is, they are rapidly accelerated."

It asks that I calculate the average mechanical power required by the belt by dividing the kinetic energy per package by the time interval between packages.

I have P = 1/2 m v^2 / T.

It then asks to compute the average force the belt is applying to packages.

I have F = mv/T

Then it says "power is force times velocity so the force answer can be turned into a power answer and compared to the power you computed above."

I get P = mv^2/T

Finally, the set asks "Do you have a discrepancy? If so, why? Which answer is more correct?"

I have wracked my brain for hours on this and cannot figure out why the two power formulae do not match. I've had one idea, but it's not really leading me to an answer; using the total time interval might be messing things up because power is only concerned with the time interval over which work is done. I thought if I knew the value for that time interval and plugged in the numbers the results of the formulae would match. However, I tried assigning values to the variable and plugging then in and the formula did not match. Can someone help please? Thank you!
 
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  • #2
patricks1 said:
Hi, everyone, this is my first time posting.

I have a problem set that states "A machine at a packaging facility places stationary packages of mass m onto a horizontal conveyor belt that is moving packages steadily and horizontally at speed v. Once placed on the belt, the packages start moving at speed v; that is, they are rapidly accelerated."

It asks that I calculate the average mechanical power required by the belt by dividing the kinetic energy per package by the time interval between packages.

I have P = 1/2 m v^2 / T.

It then asks to compute the average force the belt is applying to packages.

I have F = mv/T

Then it says "power is force times velocity so the force answer can be turned into a power answer and compared to the power you computed above."

I get P = mv^2/T

Finally, the set asks "Do you have a discrepancy? If so, why? Which answer is more correct?"

I have wracked my brain for hours on this and cannot figure out why the two power formulae do not match. I've had one idea, but it's not really leading me to an answer; using the total time interval might be messing things up because power is only concerned with the time interval over which work is done. I thought if I knew the value for that time interval and plugged in the numbers the results of the formulae would match. However, I tried assigning values to the variable and plugging then in and the formula did not match. Can someone help please? Thank you!
You have effectively used two conservation laws. Each has its limitations. Which might be in question here, and why?
 
  • #3
Hi, haruspex, thanks for the response. I think it's conservation of energy versus conservation of momentum. Perhaps it's accurate to say that power based on force and velocity is more accurate than that based on kinetic energy because there are likely factors that complicate the conservation of energy such as friction in the conveyor belt system. Does that sound right?
 
  • #4
patricks1 said:
conservation of energy versus conservation of momentum
Yes.
patricks1 said:
power based on force and velocity
That is still assuming conservation of work, effectively. What would render conservation of momentum invalid? Is it valid here?
patricks1 said:
factors that complicate the conservation of energy such as friction in the conveyor belt system.
Not necessarily friction. If the package drops onto a rubber belt it might be that there is no slippage, but losses occur as oscillations in the rubber, which dissipate. Any time there is a sudden acceleration you should suspect work is not conserved.
 
  • #5
Now that I think about it, an outside force can complicate both conservation of energy AND conservation of momentum, so neither power formula is totally valid here. I need to think about which one is more "messed up" if outside forces come into play.
 
  • #6
patricks1 said:
Now that I think about it, an outside force can complicate both conservation of energy AND conservation of momentum, so neither power formula is totally valid here. I need to think about which one is more "messed up" if outside forces come into play.
What outside force? The force of the belt, right? But that is the force we are interested in.
What else do you need in order to find the momentum change produced by a force?
 
  • #7
I was thinking the outside force would be something like the force of a box being dropped, causing oscillations, like you said, or friction.

To figure out momentum change, I need to know the time during which the force is applied. Ft = mΔv
That leads me back to my original thinking; we're not using the actual time during which the force is being applied; we're using a much longer time period finding the average.
 
  • #8
A classmate and I have come up with somewhat of an answer:

Power based on ΔKE is half the power based on force. This is because when based on force, the power formula uses the average velocity, while the ΔKE-based power uses only the final velocity. Given that the actual power required is likely higher than using conservation laws would suggest, because of forces such as friction opposing forward motion, it is probably more accurate to use force-based power given that its magnitude is higher.

I am not confident in this answer, but it's the best I can come up with. If anyone can let me know what they think of this, I'd appreciate it.
 
  • #9
patricks1 said:
To figure out momentum change, I need to know the time during which the force is applied. Ft = mΔv
You know the momentum change on each package from mΔv, and you want the average force over all time, not just the time while one package is changing speed.
patricks1 said:
Power based on ΔKE is half the power based on force. This is because when based on force, the power formula uses the average velocity, while the ΔKE-based power uses only the final velocity.
That is right, but a bit hard to put up as a rigorous argument. Better to use the momentum approach.
 

1. What is the momentum formula disagreement?

The momentum formula disagreement refers to a discrepancy between two different equations used to calculate momentum, the classical mechanics formula (p=mv) and the relativistic formula (p=γmv). This disagreement arises when dealing with objects traveling at high speeds, where the relativistic formula becomes more accurate.

2. Why is there a disagreement in the momentum formula?

The disagreement is due to the different assumptions and principles that the two formulas are based on. The classical mechanics formula assumes that objects are moving at non-relativistic speeds, while the relativistic formula takes into account the effects of special relativity at high speeds.

3. Which formula is more accurate?

The more accurate formula to use depends on the speed of the object being measured. At low speeds, the classical mechanics formula is more accurate, while at high speeds, the relativistic formula is more accurate. However, both formulas can be used interchangeably as the difference in accuracy is usually negligible for everyday objects.

4. How does the disagreement impact scientific calculations?

The disagreement may lead to slight discrepancies in calculated values for momentum. However, for most practical purposes, these discrepancies are insignificant and do not impact scientific calculations significantly. In cases where high precision is required, the relativistic formula should be used.

5. Are there any other factors that could affect the disagreement?

Yes, the disagreement can also be affected by other factors such as the direction of motion, the mass of the object, and the accuracy of the measuring instruments. These factors can further contribute to the discrepancies between the two formulas.

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