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Work done = Fx

  1. Jan 13, 2016 #1
    1. The problem statement, all variables and given/known data
    A 4.0kg ball is thrown vertically up into the air with an initial velocity of 8.5ms−1 . By the time it is height h metres above the starting point, it has a velocity of 3.0ms−1 and has done 4.0J of work against air resistance. Find h.

    2. Relevant equations
    Work done = force x distance

    3. The attempt at a solution
    Work/force = distance. I have the work done, I need the force.
    Force = 4kg * [ (3 - 8.5) / t ]
    How am I supposed to find t without the distance?
     
  2. jcsd
  3. Jan 13, 2016 #2

    Doc Al

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    Use the work-energy theorem.
     
  4. Jan 13, 2016 #3

    TSny

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    If you use energy concepts, you will not need to know any forces or the time.

    How much mechanical energy does the ball have immediately after it is thrown?
     
  5. Jan 13, 2016 #4
    KE = 0.5 * 4 * 8.52 = 144.5 J
    I can't convert this to GPE without assuming that h is the max height.
     
  6. Jan 13, 2016 #5
    Take as the datum of GPE the release point of the ball.
     
  7. Jan 13, 2016 #6

    TSny

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    The initial mechanical energy does not get converted entirely to GPE. Note that the ball still has a speed of 3 m/s at the final position. Also, some of the mechanical energy is used to do work against air resistance.
     
    Last edited: Jan 14, 2016
  8. Jan 13, 2016 #7

    Ray Vickson

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    You should realize that your equation "Work done = force x distance" is FALSE if the force varies over the interval you are looking at. So, the equation fails if the force ##F## is either a function of time ##t## or position ##x##. In general, you need to do an integral. If the force varies, the work done between times ##t_1## and ##t_2## or positions ##x_1 = x(t_1)## and ##x_2 = x(t_2)## is
    [tex] \text{Work done} = \int_{t_1}^{t_2} F(t) x'(t) \, dt \; \Longleftarrow \: F = F(t)[/tex]
    or
    [tex] \text{Work done} = \int_{x_1}^{x_2} F(x) \, dx \; \Longleftarrow \: F = F(x)[/tex]
     
  9. Jan 13, 2016 #8
    4kg is not a force !!
     
  10. Jan 13, 2016 #9

    Doc Al

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    Luckily, the integration of the air resistance force has already been done.
     
  11. Jan 13, 2016 #10

    Ray Vickson

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    Right. But the OP wrote Force × distance, as though he believes that is always true. I was trying to get him to realize that is not the case in general, but of course it is SOMETIMES true.
     
  12. Jan 13, 2016 #11
    Didn't say it was.

    I'm too novice, too young and too stupid to understand that, but thanks. I'll do the Q in my own time.
     
  13. Jan 13, 2016 #12

    Doc Al

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    You don't need to 'convert' this to anything. It's the KE at the point of release.

    What's the total mechanical energy at the point of release? At the height h? What's the change in total energy?
     
  14. Jan 14, 2016 #13
    KE = 0.5 * 4 * 32 = 18 J
    18 - 144.5 + 4 = -122.5 J
    GPE = -122.5 = 4 * -9.8 * h
    ...
    h = 3.1m, correct answer. Thanks.
     
  15. Jan 14, 2016 #14

    TSny

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    You need to rethink the signs. In the expression GPE = mgh, g is a positive quantity. The GPE should be positive at the height h.
     
  16. Jan 14, 2016 #15
    So delta KE isn't negative either, why not?
     
  17. Jan 14, 2016 #16

    Doc Al

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    ΔKE is certainly negative since it slows down.
     
  18. Jan 14, 2016 #17

    Doc Al

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    Set it up carefully, and the signs will work out:

    Initial Energy = ?
    Final Energy = ?
    Δ Energy = Final Energy - Initial Energy = ?
     
  19. Jan 14, 2016 #18
    144.5 J
    18 J
    18 - 144.5 = -126.5 J
    GPE = -126.5 = 4 * 9.8 * h
    h must be negative?
     
  20. Jan 14, 2016 #19

    TSny

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    You are still writing the GPE at height h as a negative number. As an object gains height, it gains GPE. Since you are taking GPE = 0 at the starting point, then GPE must be positive at height h.

    The total mechanical energy at any point is E = KE + GPE. If there were no air resistance, the total mechanical energy, E, would not change. However, work had to be done against the air resistance and this uses up some of the mechanical energy (by converting some of the mechanical energy to heat energy, etc.).

    You know the value of E at the start. What is the value of E at height h? (Answer this based on the value of E at the start and the amount of work done against air resistance.)
     
  21. Jan 14, 2016 #20

    Doc Al

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    That's the initial KE. Since we can measure GPE from that starting point, the initial GPE = 0. So yes, that's the initial energy.

    That's the final KE, but what about GPE? (Symbolically, what's the GPE at height h?)

    Please read over TSny's posts; he's got you covered. :smile:
     
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