# The dreaded conveyor belt problem

• mighty2000
In summary: Well, it's true that the power is 1620 Watts at any instant in time, but the question was asking for the power required to operate the conveyor belt. This implies that you want the total power needed to keep the conveyor belt moving at a steady speed of 4.5 m/s while continuously pouring gravel onto it at a rate of 80 kg/s. So, you need to find the average power required. And since the power is constantly changing, you need to integrate over time to get the average power.That is where the 810 Watts comes in. It is the average power required to operate the conveyor belt at a steady speed of 4.5 m/s while continuously pouring gravel onto it at a rate of
mighty2000
Not too hard I guess. Just wondered if I am correct.

question:

Gravel is continiously being poured at the rate dm/dt = 80kg/s on a conveyor belt moving at 4.5 m/s. What is the power required to operate the conveyor belt?

power required = work done on the gravel per second=ma(v)

v = velocity of the conveyor belt.

so

P=80.0 kg/s * 4.5 m/s * 4.5 J/s =1620 Watts

Now I have a problem with my formula.
will this give me watts or some other unit of measurement?

also, is the problem even close to being correct?

M2k

Hi, and welcome to PF.

Originally posted by mighty2000
Gravel is continiously being poured at the rate dm/dt = 80kg/s on a conveyor belt moving at 4.5 m/s. What is the power required to operate the conveyor belt?

power required = work done on the gravel per second=ma(v)

Not quite. The power is a force times a velocity. The velocity is the velocity of the belt (because that is what you are trying to move. What is the force? It comes from the stream of matter hitting it. In this case,

F=(dm/dt)*u

where u=the velocity of the gravel as it hits the belt. You have not provided information on this, and it is needed to solve the problem.

Maybe you left something out?

P=80.0 kg/s * 4.5 m/s * 4.5 J/s =1620 Watts

Where did the 4.5 J/s come from?

Now I have a problem with my formula.
will this give me watts or some other unit of measurement?

No, the units are not right. J/s=Watts.

also, is the problem even close to being correct?

No; make the corrections above and you will be OK.

Thanks for the reply Tom. You say that u is needed to solve the problem but that is the problem. Is there another approach that I need to take to solve this then? And I did leave something out. What is the power required to operate the conveyor in watts.

Thanks

M2k

Last edited:
Ok I think I may have it.

if watts are in units of J/s then the change in KE vs time would equal the power in watts.

thus:

dKE/dt = 1/2 dm/dt Vcb^2

=(.5)(80)(4.5)^2 = 810 J/s or 810 Watts.

which is half of the gravel's rate of change of momentum in the horizontal direction.

Any Ideas?

Thanks

M2k

anybody have an opinion on my latest post?

Thanks

I did it like this, but I'm not sure if it's correct.

Like Tom said: P=F*v

and F=dp/dt with p=m*v
but here m is time-dependant and v is constant.
So: F=dp/dt=dm/dt*v
and P=80*4.5^2=1620 Watts

Looks fine to me.

Yeah, I already figured out the answer. I got 1620 too using the same method.

thanks for the reply, Peace out

Last edited:
I think 810 is the right answer, via two approaches:

(a) I think mighty2000's derivation yielding 810 Watts is nearly correct:

The final kinetic energy of a mass m of gravel is (1/2) m v2. If we presume it started at rest, then the net work done at time t is (letting t = 0 be when the first particle of gravel has reached speed v):

W(t) = (1/2) m(t) v2 + C

Where C is a constant that equals the work done on the gravel that hasn't quite been accelerated to v. (C is a constant because dm/dt is a constant, so the only difference between the picture at two different times is the amount of mass that has already reached speed v)

Differentiating WRT time gives:

P = (1/2) dm/dt v2 = 810 Watts

(b) Instantaneous power is the product of force and instantaneous velocity. Average power is the product of force and average velocity. If we presume a constant force is applied to bring individual gravel particles up to speed from rest, then the average velocity is half the final velocity, so the average power is P = F (1/2 v_max) = 1/2 * 1620 Watts = 810 Watts. (If we presume nonconstant force, we get a fun integral that will also result in 810 Watts)

Hurkyl

Last edited:
GSRX720 was correct in his explination:

F=dp/dt with p=m*v
but here m is time-dependant and v is constant.
So: F=dp/dt=dm/dt*v
and P=80*4.5^2=1620 Watts

so 1620 is the correct answer. Thanks for the vote of confidence though

peace out M2k

Last edited:
Hrm. What is the source of the answer?

Hurkyl

## 1. What is the "dreaded conveyor belt problem"?

The "dreaded conveyor belt problem" is a hypothetical scenario that is used to test decision-making and problem-solving skills. It involves a conveyor belt with items moving towards a destination, and the participant is tasked with stopping or redirecting the items before they reach the destination.

## 2. Why is the conveyor belt problem considered difficult?

The conveyor belt problem is considered difficult because it requires quick thinking and the ability to make decisions under pressure. The items on the conveyor belt are constantly moving, and the participant must make split-second decisions to stop or redirect them. It also involves multiple variables and potential consequences for each decision made.

## 3. What skills are necessary to solve the conveyor belt problem?

To successfully solve the conveyor belt problem, one must have strong critical thinking skills, the ability to analyze and evaluate different options, and the capacity to make efficient decisions under pressure. It also requires creativity and the ability to think outside of the box.

## 4. Are there any strategies for solving the conveyor belt problem?

Yes, there are various strategies that can be used to solve the conveyor belt problem. Some common approaches include breaking down the problem into smaller parts, considering all possible options and their consequences, and using trial and error to test different solutions. It is also helpful to stay calm and focused while making decisions.

## 5. How can practicing the conveyor belt problem be beneficial?

Practicing the conveyor belt problem can help improve decision-making and problem-solving skills, which are essential in many fields of science. It can also help develop quick thinking and the ability to handle stressful situations. Additionally, it can be a fun and challenging way to exercise the brain and improve cognitive abilities.

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