Post a New Question


posted by on .

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
    F = mu m g
    = .25 * 11.2 * 9.81

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

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


    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

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question