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Derivations of specific values in Physics

  1. Feb 17, 2015 #1
    1. The problem statement, all variables and given/known data
    I am wondering if anyone knows of a document that shows derivations of specific values in Physics? For example, I need to find μs and am given the distance traveled and time (1.2 km in 17 sec).

    2. Relevant equations
    I know the equation for μs, Ffrictions⋅FN.

    3. The attempt at a solution
    I can use this to calculate velocity, etc., but I am at a loss of how to proceed and my text book doesn't have an example. Mastering Physics doesn't provide a "walk me through a similar problem" feature.
     
  2. jcsd
  3. Feb 17, 2015 #2

    Nathanael

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    You need to know more than that to solve the problem. I suspect the piece of information you left out is that after these 17 seconds (and 1.2 km) the object ends up at rest?
    In the future (and perhaps in your next post) you should copy the whole problem statement, preferably word for word.

    In physics, there aren't formulas for every kind of problem; it is up to you to put the pieces together.
    I'll help you get started: The equation you wrote for the force of friction says that the force is constant, which means that the acceleration is constant.
     
  4. Feb 17, 2015 #3

    mfb

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    Staff: Mentor

    Physics is the art of learning how to solve those problems if you are not given the formula to plug in values (that's the part a computer can do).

    There has to be more information to solve the problem. Does the object stop at the end?
     
  5. Feb 17, 2015 #4
    This is the question: Assuming a constant acceleration and no slipping of tires, estimate the coefficient of static friction needed for a drag racer to cover 1.2km in 17s , starting from rest.
     
  6. Feb 17, 2015 #5

    lightgrav

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    so, what "constant acceleration" would be needed?
     
  7. Feb 17, 2015 #6
    I understand that, I just don't have a strong grasp of the relationships yet. If there was a tool that showed the relationships, or derivations, then it would be easier to understand.
     
  8. Feb 17, 2015 #7
    4.15 m/s2
     
  9. Feb 17, 2015 #8

    Nathanael

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    How do you get this?
     
  10. Feb 17, 2015 #9
    vavg = Δx/Δt = 1200m/17sec = 70.6 m/s

    v = v° + at => v/t = a => 70.59 m/s / 17 s = 4.15 m/s2
     
  11. Feb 17, 2015 #10

    Nathanael

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    You have the equation "Δv/Δt = v/t = a" ... but Δv ≠ vavg. (You used vavg for the Δv). To do it it your way you have to note that vfinal = 2vavg

    This is how I think about it:
    vavg = 0.5(vfinal+v0) = 0.5(vfinal) = 0.5(a)(Δt) ... [This only applies when v0=0]
    vavgΔt = Δx
    Put the first equation into the second equation and you get:
    0.5(a)(Δt)2 = Δx ... [Again, it only applies when v0=0]


    Once you get the correct acceleration, you have to figure out what is causing this acceleration. (What is the net force on the car?) Draw a free-body-diagram.
     
    Last edited: Feb 17, 2015
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