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Velocity Down a Hill

  1. Sep 29, 2012 #1
    1. The problem statement, all variables and given/known data
    A San Francisco cable car loses its brakes and runs down a hill 30.4 m high. Calculate the velocity of the car at the bottom of the hill.


    2. Relevant equations
    Height = 30.4 m
    Time = ???
    Acceleration due to Gravity = 9.8 m/s2
    Initial Velocity = 0 m/s
    Final Velocity = ???


    3. The attempt at a solution
    3.1 m/s??

    I don't understand this problem. Does this answer look correct??
     
  2. jcsd
  3. Sep 29, 2012 #2

    PhanthomJay

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    I believe you are familiar with the conservation of energy principle ? Please show how you arrived at your answer. It is not correct.
     
  4. Sep 29, 2012 #3
    Yes, I understand that it is the principle that says: Energy cannot be created or destroyed, although it can be changed from one form to another.

    I don't understand how that law applies to velocity though. Can you give me a hint? I've ben working on this problem for an hour straight and it's driving me nuts! :yuck:
     
  5. Sep 30, 2012 #4
    The earth is doing work by pulling the car down to the bottom of the hill.
    Where all the energy goes at he bottom of the hill?
     
  6. Sep 30, 2012 #5
    I guessed 25 m/s as the answer, and it turned out to be correct. :rofl: Go figure!
    I still want to know how to go about solving this problem though.

    In response to your question: I believe the energy is transferred from potential energy to kinetic energy as it goes down the hill. I also believe that the car reaches its maximum velocity at the bottom of the hill.

    I need someone to explain to me how to convert potential energy to kinetic energy to velocity at the bottom of the hill, when all we know is that the hill is 30.4 m high.

    PE = mgh ... We do not know the mass of the vehicle, so potential energy cannot be determined.
    KE = ½mv2 ... We do not know the mass of the vehicle OR the velocity, so kinetic energy cannot be determined.
    Velocity = Distance/Time ... We do not know the time it took for the car to reach the bottom of the hill, so velocity cannot be determined.

    Am I missing something???
     
  7. Sep 30, 2012 #6

    PhanthomJay

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    You are missing the fact that all the PE at the start is converted to KE at the finish. The energy at the start and finish must be the same, only the form has changed from potential to kinetic. Do you need to know the mass?
     
  8. Oct 1, 2012 #7
    We don't need to know the mass. I just thought I needed to know it in order to solve for the potential energy.

    So here is what I understand:

    PE = mgh ... Potential Energy equals mass times gravity times height.
    KE = ½mv2 ... Kinetic Energy equals one-half times mass times velocity squared.
    PE = KE ... Potential Energy equals Kinetic Energy because PE is converted to KE when it reaches the bottom of the hill.
    mgh = ½mv2 ... Therefore, the formulas for PE and KE equal each other (according to the principle of conservation of energy).

    How do I solve for PE when I only know gravity = 9.8 m/s2 and height = 30.4 m, but I do not know the mass of the vehicle???
     
  9. Oct 1, 2012 #8

    CAF123

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    If all the potential energy is converted to kinetic energy then, by the conservation of (mechanical) energy, you have [itex] mgh = \frac{mv^2}{2}. [/itex] Solve for [itex] v[/itex].
     
  10. Oct 1, 2012 #9
    Wait a sec... :bugeye: I think I just figured it out.

    mgh = ½mv2
    [STRIKE]m[/STRIKE]gh = ½[STRIKE]m[/STRIKE]v2
    gh = ½v2
    (9.8)(30.4) = ½v2
    595.84 = v2
    24.4098340838 = v

    The velocity of the car at the bottom of the hill is 24.41 m/s!

    Did I do it right???
     
  11. Oct 1, 2012 #10

    PhanthomJay

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    Yes, it's velocity is independent of its mass, as you have discovered.
     
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