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The dreaded conveyor belt problem

  1. Apr 7, 2003 #1
    Not too hard I guess. Just wondered if I am correct.


    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?

    my answer:

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

    v = velocity of the conveyor belt.


    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?

    thanks in advance all.

  2. jcsd
  3. Apr 7, 2003 #2

    Tom Mattson

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    Hi, and welcome to PF.

    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,


    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?

    Where did the 4.5 J/s come from?

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

    No; make the corrections above and you will be OK.
  4. Apr 7, 2003 #3
    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.


    Last edited: Apr 7, 2003
  5. Apr 7, 2003 #4
    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.


    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?


  6. Apr 8, 2003 #5
    anybody have an opinion on my latest post?

  7. Apr 13, 2003 #6
    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
  8. Apr 13, 2003 #7


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    Looks fine to me.
  9. Apr 13, 2003 #8
    Yeah, I already figured out the answer. I got 1620 too using the same method.

    thanks for the reply, Peace out
    Last edited: Apr 13, 2003
  10. Apr 16, 2003 #9


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    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)

    Last edited: Apr 16, 2003
  11. Apr 16, 2003 #10
    GSRX720 was correct in his explination:

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

    peace out M2k
    Last edited: Apr 16, 2003
  12. Apr 16, 2003 #11


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    Hrm. What is the source of the answer?

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