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Need Help Tonight

  1. Oct 29, 2007 #1
    1. The problem statement, all variables and given/known data
    Traveling at a speed of 58.0 km/h, the driver of a car locks the wheels by slamming on the brakes. The coefficient of kinetic friction is 0.720. How far does the car skid before coming to a rest


    2. Relevant equations
    f=ma?
    uk=Fk/Fn
    ??


    3. The attempt at a solution

    i dont know where to start because no mass is given
     
  2. jcsd
  3. Oct 29, 2007 #2
    Let the mass equal an arbitrary value m. Start working with what you know and see if anything happens to the mass. You might not need it after all.
     
  4. Oct 29, 2007 #3
    i think i should be using an equation that would make the masses cancel, but i dont know what equation that is
     
  5. Oct 29, 2007 #4
    What do you know about the frictional force?

    You probably (or you probably should) know that it is equal to uN where u = coefficient of friction and N = mg.

    Being that this is the only force acting in the horizontal direction, what can you say about the net force and, thus, the acceleration of the car?

    Take it one step further and you've got the solution.
     
  6. Oct 29, 2007 #5
    lol im still drawing a blank because of that mass
    Fk=(.720)(m)(9.8m/s^2)
    you cant put zero for mass and i dont see how can go away or how to solve for it.....??
     
  7. Oct 29, 2007 #6

    Kurdt

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    If you know friction is the only force acting and that is [itex] F = \mu_k mg [/itex] and that force is also [itex] F= ma[/itex] I can easily see how the mass will cancel and you can solve for acceleration.
     
  8. Oct 29, 2007 #7
    ok
    so ukmg=ma
    uk(g)=a
    a=(.720)(9.8m/s^2), a=7.056m/s^2
    the question is find the distance it takes to stop...i know final velocity, acceleration, and am looking for distance which means i need initial velocity....?
     
  9. Oct 29, 2007 #8
    o lol
    nvm
    forgot to read back through to find initial velocity
     
  10. Oct 29, 2007 #9
    Which is given, can you solve it now?
     
  11. Oct 29, 2007 #10
    yep
    thanks....there may be more questions coming soon.
    have a huge test tomorrow and i dont really understand this stuff
     
  12. Oct 29, 2007 #11
    well this is what i did

    [tex]v^{2}=v_{0}^{2}+2a_{x}(x-x_{0})[/tex]

    solve for x

    x-initial & v-final = 0

    [tex]\Sigma{F_{x}}=ma_{x}[/tex]
    [tex]-\mu_{k}N=ma_{x}[/tex]

    [tex]\Sigma{F_{y}}=0[/tex]
    [tex]N=mg[/tex]

    plug in normal in the X solve for acceleration, and plug that into the first equation, and don't forget to convert your initial velocity.
     
  13. Oct 29, 2007 #12
    what about
    "a 250lb block rests on a horizontal table. The coefficient of kinetic friction between the block and table is .30. What force is required to pull the block at a constant speed if (a) a horizontal rope is attached, or if (b) the rope makes an angle of 25 above the horizontal.

    for part (a) used (.30)(250)=75N whichh is the frictional force..is that the final answer?

    For part (b) i used Uk=(Fk)/(Fn) where Uk=.30, and Fn=Wsin(theta).....which now that i look at it doesnt seem right
     
  14. Oct 29, 2007 #13
    you must convert your weight ... lb to kg

    N = (kg*m)/s^2
     
  15. Oct 29, 2007 #14
    i did, its 25.5kg
     
  16. Oct 29, 2007 #15
    that's not what i got
     
  17. Oct 29, 2007 #16
    how about 113.40kg.
    but for the inlined rope the x-comp. is Wsin(25) and the y-comp. is Wcos(25) so x comp=105.65 and y-comp=229.58 and pythag gets you back to 250.0008278.....? im lost
     
  18. Oct 29, 2007 #17
    you don't need the Pythagorean theorem. solve it the same way you solve for F in part a.

    look at your forces in the X and Y, and then solve 1 equation and plug into the other.
     
  19. Oct 29, 2007 #18
    Uk=(Fk)/(Fn)
    Fn would be Wcos(25) so (.30)(250cos(25))=Fk=67.97 which is less than 75 (from part a) and my teacher said it would be SLIGHTLY less, so maybe...? assuming my part a is right
     
  20. Oct 29, 2007 #19
    anyone?
     
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