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Waterslide and Energy Conservation

  1. Nov 5, 2014 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    Conservation of energy, F=ma

    3. The attempt at a solution

    (a) By the conservation of energy, we use points A and C as the initial and final, ## \frac{1}{2} m v^2 - \frac{1}{2} m 2.5^2= (m)(g)(9.76) \Rightarrow v = 14.1 m/s ##

    (b) Since ## W_{nc} = \Delta KE - \Delta PE, W = -\frac{1}{2} \cdot 80 \cdot 13.8^2 = --7952 J ##

    (c) F = W / d = 146 N


    I do not know what to do for part (D). Can somebody check if parts A-C are correct and give maybe a start at part D?
    Last edited: Nov 5, 2014
  2. jcsd
  3. Nov 5, 2014 #2


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    Part a): I think you forgot about the rider/sled's initial kinetic energy. Recall that the rider/sled leaves point A after being pushed with a speed of 2.5 m/s.

    Parts b) and c): Conceptually, they look correct to me, but you'll have to redo them since they rely on the answer from part a).

    Part d): I think it's asking you to calculate the normal force.
  4. Nov 5, 2014 #3
    I have fixed A-C. Is the answer to part D 771?
  5. Nov 5, 2014 #4


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    I suppose it depends on how many significant figures you are using to represent g. But if you are using 9.8 m/s2 for g, then yes, I think that's right.

    (if you're using 9.81 m/s2 for g, there may be a very minor rounding error)
  6. Nov 5, 2014 #5


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    By the way, for part c), the water also provides a vertical force that counteracts gravity. So there's the horizontal force that you calculated above, and also this vertical force.

    I'm not sure if your textbook wants you to consider that too, but technically it is a part of the total force that the water exerts on the sled. So you might want to find the magnitude of the vector sum and use that as the answer for part c).
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