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Another work-energy theorem problem

  1. Nov 23, 2007 #1
    [SOLVED] Another work-energy theorem problem...

    1. The problem statement, all variables and given/known data

    An automobile traveling at 45 km/h is brought to a stop in 60 m. Assuming that the same conditions (same braking force, neglect reaction time) hold for all cases, (a) what would be the stopping distance for an initial speed of 90 km/h? (b) What would be the initial speed for a stopping distance of 100 m?

    2. Relevant equations


    K= [tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex]


    3. The attempt at a solution

    For (a):
    Using the equation w=[tex]\Delta[/tex]K, I was substituting Fd in for w and [tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex] , making the equation Fd= [tex]\Delta[/tex][tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex]. I was going to plug in the given information and then solve for d...but I came across a problem because I don't know what the mass is and I don't know how to use [tex]\Delta[/tex] while solving either.

    For (b):
    I was going to use the same method to solve for v in the equation Fd= [tex]\Delta[/tex][tex]\frac{1}{2}[/tex]m[tex]v^{2}[/tex] but I encountered the same problems as before.

    Thank you for your help.
    Last edited: Nov 23, 2007
  2. jcsd
  3. Nov 23, 2007 #2

    Doc Al

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    Staff: Mentor

    Make use of the fact that the friction force (and the mass) is the same for all cases.
  4. Nov 23, 2007 #3
    Oh ok, so I could cancel out the mass because it's constant?
  5. Nov 23, 2007 #4

    Doc Al

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    Staff: Mentor

    Yes, you end up not needing to know the actual mass.
  6. Nov 23, 2007 #5
    ok, cool, I'll try it out...hopefully it'll work!! Thanks so much!
  7. Nov 24, 2007 #6
    Doc Al, I tried what you told me, and this is what I got...


    At first, I didn't have the acceleration but I used a kinematic equation for figure it out:


    60= 0+90+[tex]\frac{1}{2}[/tex]a
    -60 m/[tex]s^{2}[/tex]=a

    The negative made sense to me since the car was slowing down and then I used this to find the distance.



    *cancel all "m"s*

    a(d)= [tex]\frac{1}{2}[/tex][tex]v_{1}^{2}[/tex]-[tex]\frac{1}{2}[/tex][tex]v_{0}^{2}[/tex]

    d= [tex]\frac{1}{2}[/tex][tex](0)^{2}[/tex]-[tex]\frac{1}{2}[/tex][tex](60)^{2}[/tex] / -60

    d=[tex]\frac{1}{2}[/tex][tex](60)^{2}[/tex] / -60


    Unfortunately, my answer is wrong...the correct answer is supposed to be 240m.




    a(d)= [tex]\frac{1}{2}[/tex][tex]v_{1}^{2}[/tex]-[tex]\frac{1}{2}[/tex][tex]v_{0}^{2}[/tex]

    a(100)= [tex]\frac{1}{2}0^{2}[/tex]-[tex]\frac{1}{2}V_{0}^{2}[/tex]

    For this question, I applied the same method as part a but I (again) didn't have the acceleration so I was trying to use a kinematic equation to solve it but all of the equations have to have the value for initial speed, which is what I am solving for. The correct answer for this part is 58km/h.

    Please help! I'm lost!:cry:
    Last edited: Nov 24, 2007
  8. Nov 25, 2007 #7

    Doc Al

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    Staff: Mentor

    You can't use that particular kinematic equation to find the acceleration, since you don't have the time. But you don't need to find the acceleration.

    This is all you need. Now make use of the fact that the friction force (F) and the mass (m) are constant. Hint: That tells you that stopping distance (d) is proportional to [itex]\Delta v^2[/itex].

    Putting all the constants to one side:

    [tex]\frac{2F}{m} = \frac{\Delta v^2}{d}[/tex]

    You can use that constant ratio to solve all parts of this problem.
  9. Nov 26, 2007 #8
    sorry for being so annoying... what i understood from your last post was that since F and m are constant, they don't have to included in the equation so after solving the ratio...

    [tex]\frac{2F}{m}[/tex] = [tex]\frac{[tex]\Delta[/tex][tex]v^{2}[/tex]}{d}[/tex]

    [tex]\frac{2}[/tex] = [tex]\frac{-90^{2}}{d}[/tex]

    2= [tex]\frac{-8100}{d}[/tex]

    -4050 km= d

    -4.05 m= d

    am i doing something wrong?? (i surely am...but could you explain some more... thank you SO much! )
  10. Nov 26, 2007 #9

    Doc Al

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    Staff: Mentor

    Since this is constant:

    [tex]\frac{2F}{m} = \frac{\Delta v^2}{d}[/tex]

    You can set up the ratio like this:

    [tex]\frac{(\Delta v^2)_1}{d_1} = \frac{(\Delta v^2)_2}{d_2}[/tex]

    Using the first set of data gives you:

    [tex]\frac{(\Delta v^2)_1}{d_1} = \frac{(45 km/h)^2}{(60 m)}[/tex]

    The second set of data gives you:

    [tex]\frac{(\Delta v^2)_2}{d_2} = \frac{(90 km/h)^2}{(d_2)}[/tex]

    Set those ratios equal and solve for [itex]d_2[/itex].
    Last edited: Nov 26, 2007
  11. Nov 27, 2007 #10
    alright...thanks a lot for all your help!! :)
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