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The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.20 with the floor. If the train is initially moving at a speed of 47 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?

  • physics -

    47 km/h = 13.056 m/s

    Compute the acceleration(a)when the stopping distance is X meters:

    V^2 = 170.44 m^2/s^2 = 2 a X
    a = V^2/(2X) = 85.22/X

    M g * us = M a when a is as high as possible without slipping

    M's cancel. You know the static fricion coefficient us. Substitute for a and solve for X

  • physics -


  • physics -

    coefficient of static friction = 0.20
    fs = static frictional force
    mass = m
    g = 9.8m/s^2
    as we know that fs = fnormal*coefficient of static friction
    so, fs= mg*0.20
    as fnormal= mg = 9.8m N
    now, acc to newtons law
    fnet = fs
    ma = 9.8*0.20*m
    m's get cancel
    a= 1.96m/s^2
    applying the formula
    v^2= u^2 + 2ax
    now substitute the value in this equation
    you get, x= 43.1m
    don't forget to change speed into m/s

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