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Energy question with constant velocity

  1. Mar 1, 2013 #1
    1. The problem statement, all variables and given/known data

    At what speed would a heavy object hit the ground if it was lifted vertically from the ground at a constant speed of 1.2 m/s for 2.5s and then dropped?

    2. Relevant equations

    Ep= mgh
    Ek= 1/2 mv^2

    3. The attempt at a solution

    Im not sure how to do this. Do i find Ek 1/2mv^2 going up and at the top Ek = Ep? then with Ep at the top drops down Ep=Ek=1/2m(v^2-vi^2)? But what do i do with the time?

    The answer is: 7.7m/s
     
  2. jcsd
  3. Mar 1, 2013 #2
    You need to find the potential energy just before the drop. You just learned how to do that is a similar problem :)

    Then you can apply your equations.
     
  4. Mar 1, 2013 #3

    CWatters

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    There are two ways to solve this problem:

    1) Use the equation that relates velocity and time to calculate the displacement (aka height). Then calculate the potential energy just before the drop. Then during the drop the PE will be converted to KE. Apply conservation of energy. An unknown will cancel. Solve for final velocity. Only then substitute the numbers.

    2) As above to work out the height. Then because gravity can be assumed constant over small distances you can apply one of the standard equation of motion to work out the final velocity from the height and acceleration (a=g)...
    http://en.wikipedia.org/wiki/Equations_of_motion#SUVAT_equations

    Edit: I suppose you should also state any assumptions made. For example about air resistance.
     
  5. Mar 1, 2013 #4

    rude man

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    Don't forget the cotton-picker is still moving when it reaches 2.5 s x 1.2 m/s. bSo at that point it has k.e. and p.e. But just before it reaches the ground it has only k.e.
     
  6. Mar 1, 2013 #5

    haruspex

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    I agree, but to get the given answer of 7.7 you have to ignore that :frown:.
     
  7. Mar 1, 2013 #6

    rude man

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    Not so! When you combine the k.e. at the point where the ball is released with the p.e. at that point and equate them to the k.e. at the bottom you get the advertised answer.

    EDIT; OK, you get 7.76 my way but if you assume the weight has stopped moving before being let go then it's 7.67.
     
    Last edited: Mar 1, 2013
  8. Mar 2, 2013 #7

    CWatters

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    Well spotted. I'd totally missed that.

    You could still solve it using either method (eg Energy or equations of motion) but they would need to be modified to take that into account.
     
  9. Mar 2, 2013 #8

    rude man

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    Don't feel bad - looks like the problem poser missed it too (his answer seems to assume at rest at 3m). :smile:
     
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