Physics

A block of mass m1 = 28 kg rests on a wedge of angle θ = 53∘ which is itself attached to a table (the wedge does not move in this problem). An inextensible string is attached to m1, passes over a frictionless pulley at the top of the wedge, and is then attached to another block of mass m2 = 4 kg. The coefficient of kinetic friction between block 1 and the plane is μ = 0.9. The string and wedge are long enough to ensure neither block hits the pulley or the table in this problem, and you may assume that block 1 never reaches the table. take g to be 9.81 m/s2.


The system is released from rest as shown above, at t = 0.

(a) Find the magnitude of the acceleration of block 1 when it is released (in m/s2).

(b) How many cm down the plane will block 1 have traveled when 0.50 s has elapsed?

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asked by sel
  1. I will assume the inclined block wins and slips down the hill
    Force up on inclined block = T
    force down slope on inclined block = 28 (9.81 )sin 53 - 28 (9.81)(.9)cos53
    a = acceleration down slope
    28 a = 28 (9.81 )sin 53 - 28 (9.81)(.9)cos53 - T


    Force down on hang mass = m g = 4*9.81
    force up on hang mass = T
    a = acceleration up

    4 a = T- 4*9.81
    or
    28 a = 7 T - 28*9.81
    so
    28 (9.81 )sin 53 - 28 (9.81)(.9)cos53 - T = 7 T - 28*9.81
    or
    sin 53 - (.9)cos53 = 8 T/(28*9.81) - 1
    or
    8 T/(28*9.81) = 1 + sin 53 - (.9)cos53
    solve for T
    then go back and get a
    then get d from
    d = (1/2) a t^2
    remember to multiply by 100 because it asks for cm, not meters

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    posted by Damon

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