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Homework Help: Really simple problem - can't get right answer!

  1. Sep 25, 2004 #1
    Suppose a 50kg block slides along a
    horizontal surface where the coefficient of kinetic friction
    between the block and the surface is uk = 0.60. A force F =
    400 N is now applied, where the
    angle of the force above horizontal is 20°.

    What is the magnitude of the acceleration of the block?

    0.54 m/s2
    2.31 m/s2
    3.27 m/s2
    6.78 m/s2
    8.11 m/s2

    Here's what i did:

    F(x-net) = F(applied) - F(friction) = ma
    a = F(applied) - F(friction) / m
    = (400)(cos 20) - (50 * 9.8 * 0.6) / 50
    = 375 N - 294 N / 50 kg
    = 1.67 m/s2

    I checked my numbers quite a few times and keep getting
    1.67, which is not one of the answer choices. The only thing
    I saw was that 3.72 m/s2 is exactly double what I got, but I
    can't find an explanation for that.

  2. jcsd
  3. Sep 25, 2004 #2
    The answer is 3.27.
    You gave the right anwser for the acceleration in the x-direction. Now when you look vertically in the y direction you still have the y-component of the 400 N-force that you forgot (gravity and normal force annihilate each other)

    This y-component = 400*sin(20)

    Then calculate a_y and you already have a_x =1.67 which is correct.

    The magnitude is sqrt((a_x)² + (a_y)²)

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