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Homework Help: How fast is the wagon going after moving 82.1 m up the hill?

  1. Nov 2, 2007 #1
    1. The problem statement, all variables and given/known data

    A 40.5 kg wagon is towed up a hill inclined at 16.5 degrees with respect to the horizontal. The tow rope is parallel to the incline and exerts a force of 125 N on the wagon. Assume that the wagon starts from rest at the bottom of the hill, and disregard friction and significant figures.

    The acceleration of gravity is 9.81 m/s^2

    How fast is the wagon going after moving 82.1 m up the hill? Answer in m/s

    2. Relevant equations

    Force = mass * acceleration
    Fg = ma
    Vf^2= 2ad + Vi^2

    3. The attempt at a solution

    40.5 kg * 9.81 m/s^s = 397.305 kg/m/s^2 or 397.305 N

    Tilting the time of reference, so that you just need to find the Fg of the diagram.
    Fg in the x direction = 397.305 cos 16.5 = 380.8438748
    Fg in the y direction = 397.305 sin 16.5 = 112.8407165

    Just adding the forces in the x direction
    380.8438749 - 125 = 255.9438748
    255.9438748 / 40.5 kg = 6.319601847 = acceleration in the x direction

    Acceleration in the y direction = 0 because the wagon isn't moving up and down

    Vf^2 = 2ad + Vi^2
    Vf^2 = 2 (6.3196017847)(82.1) + 0 ^2
    Vf^2 = square root of 1037.678623
    Vf = 32.21301947 m/s

    I got 32.21301947 m/s as my answer

    Did I do anything wrong?
  2. jcsd
  3. Nov 2, 2007 #2


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    Homework Helper

    You have it backwards here. Fg in the x direction should be 397.305 sin 16.5 = 112.84. Check your trig. Other than that, I think you've got the right idea.
  4. Nov 4, 2007 #3


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    Homework Helper

    You have the force acting down the slope as 380.8 N, with 120 N acting up the slope, from the tow rope.

    That leaves you with 260.8 N down the slope moving the wagon - which means the wagon is going to move backwards.

    But the wagon is already at the bottom of the slope!! It can't move backwards anymore.

    This should have pointed out the error in your work.

    The actual error is a trig one.

    The component of the wagon's weight down the slope is mg.sin(alpha) - where here alpha = 16.5 degrees

    Apart from that, your working is fine.
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