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How much work must be done to stop a 980 kg car traveling at 108 km/h?

  1. Dec 3, 2004 #1
    How much work must be done to stop a 980 kg car traveling at 108 km/h?
    what i did was first convert the km/h to m/s:
    [tex]108 * 1000 / 3600 = 30[/tex]
    then plug the mass and velocity into the kinetic energy forumula [tex]K.E. = \frac{1}{2} mv^2[/tex]
    [tex] \frac{1}{2}980(30^2) = 441000 [/tex]
    I have checked and rechecked it and i keep coming up with that answer but its wrong. Where am I going wrong or is there more to do?

    Another problem i need help with is this:
    At an accident scene on a level road, investigators measure a car's skid mark to be 88 m long. It was a rainy day and the coefficient of friction was estimated to be 0.46. Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes.
    I don't even know where to get started on this one especially without a mass, but my teacher says the mass doesnt matter.

    Any help on these would be greatly appreciated.
  2. jcsd
  3. Dec 3, 2004 #2
    The first one seems right. Are you sure the answer is correct?
    For the second one, write out the equation that relates friction force with acceleration, and you should see that the mass cancels out.
  4. Dec 3, 2004 #3
    What's the correct answer for the first one?
    As far as I can see - you've done it right.
  5. Dec 4, 2004 #4
    yea - thats what i thought, but its not somehow, i think the teacher is wrong

    im still not sure what to do on the second one,
    Here's what i know:
    [tex]F_{F} = \mu * F_{N}[/tex] or [tex]F_{F} = \mu * 9.8m[/tex]
    and [tex]P.E. = .5mv^2[/tex]
    and [tex]W = Fd[/tex] or [tex]W = mad[/tex]
    those are the only formulas i know of that might help with this problem but I dont know what to do with them...
    i tried setting the frictional force formala equal to the potential energy forumula and came up with:
    [tex] \mu * 9.8 = .5v^2[/tex]
    [tex] .46 * 9.8 = .5v^2[/tex]
    [tex] 4.508 = .5v^2[/tex]
    [tex] 9.016 = v^2[/tex]
    [tex] 3.003 = v[/tex]
    that was wrong, then i noticed i didnt even use the distance provided, but I dont know what else to do?
    Last edited: Dec 4, 2004
  6. Dec 5, 2004 #5
    F_f = 4.5m

    so that means ...

    F_{net} = ma - F_f


    ma = 4.5m


    a = 4.5


    v^2 = u^2 + 2ax

    v = 0 - cuz it stops, and u - is what you are trying to find...
    take a = 4.5 and x = 88, so

    u^2 = -2ax

    and i found that

    u = 28.1

    I am not sure if this is right though
  7. Dec 5, 2004 #6
    now - that i think about it - i think MY force equation is wrong..
  8. Dec 5, 2004 #7
  9. Dec 5, 2004 #8
    thanks alot i understand now... and that site is awesome
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