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Work and energy problem

  1. Feb 10, 2008 #1
    [SOLVED] work and energy problem

    1. The problem statement, all variables and given/known data
    The first part asks
    A car (m = 890.0 kg) travelling on a level road at 27.0 m/s (60.5 mph) can stop, locking its wheels, in a distance of 60.0 m (196.9 ft). Find the size of the horizontal force which the car applies on the road while stopping on the road. First solve this problem using work/energy concepts and then check your answer using kinematics/force law concepts.

    I solved this by finding the acceleration and plugging into formula F=ma=5402N

    The second part asks
    Find the stopping distance of that same car when it is traveling up a 18.9deg slope, and it locks its wheels while traveling at 27.0 m/s (60.5 mph). Assume that muk does not depend on the speed.


    2. Relevant equations



    3. The attempt at a solution[/b

    I drew a free body diagram and came up with the following relationships
    F-fk-mgsin(18.9)=ma where F=5402N
    N-mgcos(18.9)=0

    How do I solve for the horizontal distance? Is it just the vertical component of the Force?
     
  2. jcsd
  3. Feb 10, 2008 #2

    Doc Al

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    Staff: Mentor

    That F=5402N doesn't belong here--that was the friction force from the first part.
    Redo this without that extra force.
    Good.
    Solve for the acceleration. (Or the net force parallel to the road.)
     
  4. Feb 10, 2008 #3
    OK, so i have F-ukmgcos(18.9)-mgsin(18.9)=ma. Can I solve for uk from the previous part of the problem and use it in this equation? If so, I still have one equation and 2 unknowns (F and a).

    To solve for x i think I can use the equation Vf^2=V0^2+2ax, where Vf=0 and V0 is given.
     
  5. Feb 10, 2008 #4

    Doc Al

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    Staff: Mentor

    What does the F stand for? You already have friction and weight accounted for.
    Absolutely.
    Lose the F!
     
  6. Feb 10, 2008 #5
    F is the horizontal force (same as what the first part of problem asked for)

    Well, I tried losing the F, which doesn't make much sense to me, and have -ukmgcos(18.9)-mgsin(18.9)=ma --> a=-14.57 m/s^2

    use Vf^2=V0^2+2ax --> (27)^2=-2ax --> x=25.01m which STILL isn't right! :(
     
  7. Feb 10, 2008 #6
    and i found uk with this equation from the first part

    5402-ukmg=ma where a=-6.075
    uk=1.23
     
  8. Feb 10, 2008 #7

    Doc Al

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    Staff: Mentor

    In the first part, F is the net force on the car, which is just the friction. In the second part, you will use the [itex]\mu_k[/itex] from the first part, but not the F. You are already including friction as well as gravity.
    What did you use for [itex]\mu_k[/itex]?
     
  9. Feb 10, 2008 #8

    Doc Al

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    Staff: Mentor

    Nope. Your value of force (5402N) is the friction!
     
  10. Feb 10, 2008 #9
    ah ok so uk=.619
     
  11. Feb 10, 2008 #10
    and equation becomes -(.619)(890)(9.8)cos(18.9) - (890)(9.8)sin(18.9) = ma , solve for a, plug in (27)^2=-2ax
     
  12. Feb 10, 2008 #11
    and x=40.89, great thank you!!!!!
     
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