Post a New Question

physics

posted by .

A stock room worker pushes a box with mass 11.2 kg on a horizontal surface with a constant speed of 4.50 m/s. The coefficient of kinetic friction between the box and the surface is 0.25. a) What horizontal force must be applied by the worker to maintain the motion? b) If the force calculated in parts (a) is removed, how far does the box slide before coming to rest

  • physics -

    a = 0
    so the sum of forces is 0
    F is applied
    - mu m g is friction force
    so
    F = mu m g
    = .25 * 11.2 * 9.81

    then
    m a = mu m g
    a = mu g
    solve for a

    time to stop first
    0 = 4.50 - a t
    solve for t

    then

    x = Xo + Vi t + (1/2) a t^2
    x = 0 + 4.5 t - .5 a t^2

  • physics -

    Vfinal = vi + at not negative it should be + sign i don't know pla reply

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question