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Kinetic and Potential Gravitational Energy Problem

  1. Jul 18, 2011 #1
    1. The problem statement, all variables and given/known data

    a) A watermelon with a mass of 2.0kg falls out of a tree house that is 5.4m above the ground. What is the speed of the watermelon just before it hits the ground?

    First I found Eg to be 105.9 J based on the formula Eg = mg (h2 - h1).
    Since the watermelon hits the ground, Eg = Ek. Then I found the velocity to be 10.29m/s from the formula v = sq.rt (2 Ek / m).

    I (hopefully) did this part properly.

    b) A cantaloupe with a mass of 0.45kg falls out of the other side of the tree house. It hits a tree branch at a speed of 6.3 m/s. How high is the tree branch from the ground?

    This is the part I'm having a bit of trouble with.

    The only thing I can think of using is the formula for height.

    h = ((v1)^2 - (vf)^2) / 2g
    h = (0 - (6.3m/s)^2) / (2 * 9.81m/s^2)
    h = -2.02m

    Then 5.4m - 2.02m = 3.38m from the ground to the branch.

    When I do it this way, I'm not using the information given in the question, which was the mass, 0.45kg. So, I think I'm doing it incorrectly. Can anybody help me out? Thanks!
     
  2. jcsd
  3. Jul 18, 2011 #2
    When you equate the kinetic and potential energies, the mass gets cancelled.

    [itex]mgh=\frac{1}{2} m v^{2}[/itex]

    Also the equation you used reads as:

    [itex]v_{f}^{2}-v_{i}^{2}=2gh[/itex]

    You switched the position of the final and initial velocities, giving a negative height; how can a height be negative? If you switch those two, you get 2.02 m covered out of 5.4 m. So the distance of the branch from the ground is 5.4-2.02 = 3.38 m.
     
  4. Jul 18, 2011 #3
    Thanks for the response!

    Hmmm....my book says h = ((v1)^2 - (vf)^2) / 2g. I just triple checked.

    Oh well, I just assumed that the negative height meant that it was being measured from the top down instead of the bottom up.

    But is 3.38m the correct answer then? I would have thought it would have something to do with the cantaloupe being 0.45kg, otherwise why would they give you that information? Unless they're trying to trick me and doing a good job of it haha
     
  5. Jul 18, 2011 #4
    v1 should be the final velocity in that case (6.3 m/s) and vf should be the initial velocity (0 m/s). To see why the mass does not affect the result, you can use energy conservation:

    Suppose the branch is h meters above the ground. So the height covered by the cantaloupe is (5.4-h) meters. Hence the amount of potential energy converted to kinetic is mg(5.4-h). Using energy conservation,

    [itex]\frac{1}{2} m v^{2} = mg(5.4-h)[/itex]

    So the m will cancel. Solving through for h will give the same answer you got from kinematics.
     
  6. Jul 18, 2011 #5
    Ok, thanks a lot, much appreciated!

    Just one more quick question if you don't mind.

    What are the basic energy conversions that take place when someone at the top of a snowy hill toboggans down?

    Is it just gravitational potential energy - kinetic energy

    Or, is it gravitational potential energy - kinetic AND thermal energy (due to friction on the way down)?
     
  7. Jul 18, 2011 #6
    In real life, the potential energy at the top would be converted into kinetic energy & thermal energy as you mentioned, but also into sound. This means you can't simply equate kinetic and potential energies to find velocity at the bottom of the hill!
     
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