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Power, Work, friction

  1. Feb 16, 2017 #1
    1. The problem statement, all variables and given/known data
    A tractor moves, at constant velocity, on a flat surface with known friction coefficient μ (0.24), ten objects each of which has known mass m (12 kg). It does so for a known distance d (500 m), and it takes known time t (150s) to do so. Calculate the Power of the tractor.

    2. Relevant equations
    v = s/t
    P = W/t
    W = F*distance
    Kinetic energy = (0.5)*m*v2
    F = m*a
    Friction force = Normal force * μ

    3. The attempt at a solution
    I managed to get to the solution proposed by the workbook, but I had to do random guesswork, and I'm not convinced/still have some doubts.

    The way it's supposed to be calculated, apparently, is like this:

    Friction Force = m*g*μ
    Here m = 10*12, so:

    Friction Force = 10*12*9.81*0.24 = 282.5 N

    I'm kinda lost at this passage, because apparently the Friction Force equals the force of the tractor, at least in module...?

    If that were the case, then Force of the tractor is:
    F_tractor = 282.5 N

    Then W = F*distance
    W = 282.5 * 500 = 141 kJ

    P = 141K / 150 = 0.94 kW

    This is the correct result.

    However, before finding what seems to be the correct way to proceed, I previously attempted to proceed like this:

    Since velocity is constant, and it takes 150 seconds to do the 500 meters, then v = s/t, so
    v = 500/150 = 3.3 m/s

    Then I tried using the formula of kinetic energy, since Work = Kinetic energy (or at least it seemed to be that way in previous problems). Is this a mistake? Because I feel like it is (if anything because the numbers are waaay too small), but I can't understand why.

    K = (0.5) * 120 * (3.32) = 653.4 J

    Then I thought that the calculated energy was the result of all the forces that act on the mass(es), so I wrote down

    W = F*s
    W = K
    653.4 = F_all * 500 => F_all = 1.3 N

    Then I thought:
    F_all = Friction force (negative) + Tractor force (positive)
    => Tractor force = 1.3 + 282.5 = 283.8 N

    The numbers are so close that mere approximation can still make the 0.94 kW result come out in the end, but a net force of 1.3 N is nothing like what I'd expect from real-life experience, so...

    Anyway, I don't understand why in this context we can't say that Work = Kinetic energy = 0.5 * mass * velocity2. Is it because velocity is constant, so there is no acceleration? If not, then why doesn't that (Work = Kinetic energy) apply here (whereas I'm sure it applies in other situations; I've solved more than one problem by putting together K = 0.5*m*v2, W = F*distance, and W = K_end - K_start) ? There's definitely something wrong here, but I don't know what.

    Also, am I to understand that when there is no acceleration, the force needed to move a body (at constant velocity) is equal to the Friction force? I understand the concept that "when the sum of all the forces applied on a body is null, the body moves at constant velocity", which would confirm that the force of the tractor = -Friction force here...but what I don't understand is, how does the tractor move in the first place? Shouldn't there be a starting force that sets things in motion, gives that starting speed of 3.3 m/s, and THEN the tractor force, by equalizing the Friction force, makes sure that velocity remains constant?

    The questions in bold are the most important ones (the only ones really).

    Thanks in advance.
     
  2. jcsd
  3. Feb 16, 2017 #2
    Now consider another problem where the tractor had to pull those same 10 objects across a near frictionless surface. Would that change the kinetic energy of those objects? Would that change how much work the tractor had to do, or how much power it would have to provide?
     
  4. Feb 16, 2017 #3
    Yes, there has to be a starting force to set things in motion, but you don't have to consider that for this particular problem.
     
  5. Feb 17, 2017 #4
    Well, I'd say that the Kinetic energy would be the same, since it should still be:

    0.5 * m * v2

    Since velocity and mass are still the same 3.3 m/s and 120 Kg, nothing would change on that front.

    Would it change how much work the tractor has to do...well:

    Work = Force_all * distance
    Since Force_all in this case would be lower (Force_all = Force_tractor - Force_friction), then Work would be lower as well.
    Power is directly proportional to Work, so it would be lower as well.

    I see, thanks.

    Still, I don't quite understand why in some situations we can say Work = Kinetic energy, but here it's wrong (the first bolded question of the previous post I made). I'll try to re-phrase what I don't understand down below.

    Textbook definition of Work: Kinetic energy at the end - Kinetic energy at the start.
    How do you even apply this definition to this problem?

    If it were a situation, say, where you have a car of mass m that goes from zero to 100 m/s, then it would all work out just fine:
    Work by the car = 0.5*m*(1002)

    What about this situation? Starting and ending velocity are the same, and mass doesn't change either...so work = 0? It's blatantly wrong, but why? Why in this problem we have Work = Force*distance, but not Work = Kinetic_end - Kinetic_start ?
     
  6. Feb 17, 2017 #5
    Well, I guess you just have to look at the individual case to see if work is being done to alter that kinetic energy.

    If the driver disconnected that load from from the tractor and let it slide to a stop, then the work done by the friction would be equal to the change in kinetic energy. If the tractor was put into neutral (and there was no friction between the tractors tires and the ground), then the friction from the 10 objects would do work to stop the kinetic energy of the objects and the tractor. If it was a train (instead of a tractor) with frictionless wheels pulling the 10 objects, the friction would have to do a lot more work (and it would take a much greater distance) to dissipate the much greater kinetic energy of the train. I hope that helps?
     
  7. Feb 17, 2017 #6
    Mmh...I guess there isn't a clear "if this is here, don't use this formula" sign then. All right, thanks.
     
  8. Feb 17, 2017 #7
    If there is a clear sign, it may be whether or not the kinetic energy is being affected by the work. In this problem, the kinetic energy was not being affected because it remains constant the entire time.

    The problem could also have said that the tractor started from a stop and had a constant acceleration of 0.1 m/s^2 up to its 3.3 m/s cruising speed, and asked how much work was done in that segment of the trip. Work would be required to produce the kinetic energy. But work would also be required to overcome the friction during that portion also. So you would have to calculate both of those to get the overall work required.
     
  9. Feb 17, 2017 #8
    And since Kinetic energy depends from velocity, I believe it was actually sensible to state that this does depend on whether or not there is an acceleration, after all. I think that's how I'll consider it from now on: if velocity is said to be constant, then I'll avoid using that formula for kinetic energy (0.5mv2).
     
  10. Feb 19, 2017 #9
    Two days later, I got it.

    Sum of non-conservative forces Works = Mechanical Energy at the end - Mechanical energy at the start

    Non-conservative forces: friction; tractor.

    W_friction + W_tractor = 0.5*m*v_end2 - 0.5*m*v_start2

    With constant speed, the right side of the equation is zero, thus turning everything into:
    W_friction = -W_tractor

    Yay.
     
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