Physics

A 2000 {\rm kg} car traveling at a speed of 13 {\rm m/s} skids to a halt on wet concrete where mu_k = 0.50. How long are the skid marks?

I tried doing this many times and each answer is wrong. Can someone help?

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  1. Fc = mg = 2kg * 9.8N/kg = 19.6N = Force
    of car.

    Fc = 19.8N @ 0 deg.(Hor. plane).

    Fp = mg*sin(0) = 19.8sin(0) = 0 = Force
    parallel with plane.

    Fv = 19.6cos(0) = 19.6N = Force perpendicular to plane.

    Ff = u*Fv = 0.5 * 19.6 = 9.8N.

    Fn = Fp - Ff = 0 - 9.8 = -9.8N = Net
    force.

    Fn = ma,
    a = Fn/m = -9.8/2 = -4.9m/s^2.

    Vf^2 = Vo^2 + 2ad,
    d = (Vf^2-Vo^2) / 2a,
    d = (0-(13)^2 / -9.8 = 17.2m. = Length
    of skid marks.

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    posted by Henry
  2. Correction:
    mass of the car = 2000kg: NOT 2kg.

    Fc = 2000kg * 9.8N/kg = 19,600N.

    Fp = 19,600sin(0) = 0.

    Fv = 19,600cos(0) = 19,600N.

    Ff = u*Fv = 0.5 * 19,600 = 9800N.

    Fn = Fp - Ff = o - 9800 = -9800N.

    a = Fn/m = -9800 / 2000 = -4.9m/s^2.

    The remaining calculations are correct
    as is.

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    posted by Henry
  3. Thank you very much!

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  4. Glad i could help!

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    posted by Henry
  5. work = change in kinetic energy
    friction force X distance = change in KE
    -mu_k n x = 0 - 1/2 m v^2
    [Remember, friction does negative work)
    mu_k m g x = 1/2 m v^2
    Simplifying:
    x = v^2/(2 mu_k g) = 17.24 m

    Notice that in this solution, the mass of the car is not needed.

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    posted by CANTIUS

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