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Work and Energy Problem. Why not include PE?

  1. Dec 15, 2014 #1
    1. The problem statement, all variables and given/known data
    Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loose gravel are constructed to stop runaway trucks that have lost their brakes. The combination of a slight upward slope and a large coefficient of rolling friction as the truck tires sink into the gravel brings the truck safely to a halt. Suppose a gravel ramp slopes upward at 6.0 degrees and the coefficient of rolling friction is 0.40.

    Use work and energy to find the length of a ramp that will stop a 15,000 kg truck that enters the ramp at 35 m/s.

    2. Relevant equations
    W= ΔKE + ΔU

    3. The attempt at a solution

    W = KEf-KEi +Uf-Ui

    Given Ui = 0 and KEf = 0,
    W = -KEi + Uf
    F⋅d = 1/2mVi^2 + mghf
    F= mgsin(Θ) + μmgcos(Θ)
    (mgsin(Θ) + μmgcos(Θ))d = 1/2mVi^2 + mghf

    This is where I get stuck. In the end, I am left with two variables hf and d. I can't actually solve for one variable without the other. Where did I go wrong?
     
  2. jcsd
  3. Dec 15, 2014 #2
    I would help, but i need to review my concepts on this topic first ?:) Gimme a day, i'll get back to you.
     
  4. Dec 15, 2014 #3
    Cool, Thanks! In the mean time I'll keep trying to work on it. I think the problem lies in the fact that i'm using potential energy, Uf. I don't think I am supposed to incorporate it, based on similar examples in my textbook. However, I also don't understand why I wouldn't incorporate it either.
     
  5. Dec 15, 2014 #4
    Update: I solved the problem without incorporating the potential energy and the solution was correct. My answer came out to be .12km. However, I still don't fully understand why potential energy is not incorporated.
     
  6. Dec 15, 2014 #5

    haruspex

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    Your mg sin(theta) is the component of gravity acting down the slope. Working against that is what gaining gravitational PE is all about, so you have included PE change. The mistake would be to include it again.
     
  7. Dec 15, 2014 #6
    I"m still not sure that I understand that. To clarify, I'll try to break down my method a bit more. Perhaps that will help show more thoroughly where exactly my understanding is flawed.
    To start, we know that WTOT = ΔKE + ΔU
    WTOT = KEf - KEi + Uf - Ui
    Because the truck goes from moving to stationary KEf = 0 and Ui = 0
    WTOT = - KEi + Uf
    WTOT = -.5mV2 + mghf

    From drawing a free body diagram of the mass, it is known that:
    N-mgcos(Θ)=0
    and
    mgsin(Θ) = 0

    Thus,
    WGravity = mgsin(Θ)d
    and,
    Given F = μkN
    WFriction = μkmgcos(θ)d

    Combining that:
    WTOT = WGravity + WFriction = mgsin(Θ)d + μkmgcos(θ)d

    Plugging this back into the original equation:
    WTOT = -.5mV2 + mghf
    mgsin(Θ)d + μkmgcos(θ)d = .5mV2 + mghf

    Given all of that, I still struggle to understand how WGravity in my equation accounts for PE rather than simply having a close relationship with PE. I understand that we are dealing with the force downward, which is counteracting movement and thus causing the truck to stop going upwards along the hill. I understand that that this is what PE is all about, given PE is equal to mgh. However, I can't figure out why we wouldn't still need to include mghf separately. I feel like I am missing an important piece of understanding this relationship. My textbook goes over this concept briefly but does not provide an in depth explanation. I would really appreciate more clarification!

    Perhaps a better question to ask in order to help me understand the problem is this:
    How do you use Work and energy to find the height of the ramp, given all of the factors and values in the original question would remain the same.
     
  8. Dec 15, 2014 #7

    haruspex

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    When you throw a ball in the air, the work done against gravity is the work that is shifted from KE to PE. If you are going to count the total energy of the ball as KE+PE, then it does not also lose energy in working against gravity.
     
  9. Dec 16, 2014 #8

    vela

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    The work-energy theorem says ##\Delta KE = W##, where ##W## is the work done by the net force acting on an object. We can separate this work into contributions from conservative and non-conservative forces, giving ##\Delta KE = W_\text{cons} + W_\text{nc}##. With the concept of potential energy, we move the work done by conservative forces to the other side of the equation, and identify ##-W_\text{cons}## as the change in potential energy: ##\Delta KE-W_\text{cons} = \Delta KE + \Delta PE = W_\text{nc}##. If you have a conservative force and use the corresponding potential energy, you should not also calculate the work done by the force, otherwise you're double-counting.
     
  10. Dec 16, 2014 #9

    haruspex

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    If you want me to identify the exact spot where your reasoning goes wrong then you will need to define all these variables precisely.
    But it still seems to me that your mistake is in not understanding that work done against gravity means the work that goes into gaining gravitational PE. If you count that PE as part of the total energy of the truck then it still has that work and has not 'spent' it fighting gravity.
    You are counting it as work lost from the KE twice, once as mgh (why do you write mghf?) and once as mgsin(Θ)d. Note that sin(Θ)d = h.
     
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