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1.0kg physics book on slope, and Newton's Third Law?

  1. Jan 12, 2010 #1
    1. The problem statement, all variables and given/known data

    A 1.0 kg physics book is on a 20 degree slope. It is connected by a string to a 500 g coffee cup dangling at the bottom side of the incline. The book is given a push up the slope and released with a speed of 3.0 m/s. The coefficients of friction are us = .50 and uk = .20.

    a. How far does the book slide?
    b. At the highest point, does the book stick to the slope, or does it slide back down?

    2. Relevant equations

    vf^2 = vi^2 + 2as

    3. The attempt at a solution

    So I found four forces: Tension of the rope pulled by the weight, fk (friction force), force of gravity at a 20 degree angle, and normal force.
    n = mgcosx = (9.8)(1)cos20 N.........fk = uk*n = (.2)n = 1.842 N
    T = mass of cup * 9.8 = 4.9 N
    Fg = mgsinx = (1)(9.8)sin20 = 3.35 N

    ...so I get (Fnet)x = ma = -4.9N - 1.842N - 3.35N = -10.095
    ...so... a = (-10.095)/(1) = -10.095 m/s^2

    But when I put this value of acceleration into vf^2 = vi^2 + 2as, or 0=9 + 2(-10.095)s, I get .446m.
    The correct answer is .67m. What did I do wrong? I'm pretty sure it's around the (Fnet)x part.

    Thanks for any help.
  2. jcsd
  3. Jan 13, 2010 #2


    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi Jthunder! Welcome to PF! :smile:
    No, the cup is accelerating, so T will be different.

    Either do two F = ma equations (one for the book, one for the cup), and eliminate T, or just treat the book and the cup as a single body, as if it was one-dimensional, and calculate the external forces on it, along that dimension. :wink:
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