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Finding Work Done on a Ball

  1. Apr 16, 2007 #1
    1. The problem statement, all variables and given/known data

    A 125g ball is thrown with a force of 85.0 N that acts through a distance of 78.0cm. The ball's velocity just before it is caught is 9.84m/s.

    Calculate the work done on the ball.


    2. Relevant equations
    W= F X d (delta displacement)


    3. The attempt at a solution

    W = 85N X 0.78m
    = 66.3 J

    But the answer in my physics textbook is 6.24 J. Can anyone please tell me what I am doing wrong?
     
  2. jcsd
  3. Apr 16, 2007 #2

    Doc Al

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    You found the work done by the applied "throwing" force. But perhaps they want the total work done on the ball due to all forces acting on the ball (like air resistance) up to just before it's caught. How would you calculate that?
     
  4. Apr 16, 2007 #3
    I would say that I would also have to find the force of gravity, because that is also acting on the object.

    Fg = mgh (delta height)
    = 0.125kg X 9.81 X 0.78m
    = 0.956475 J

    But even if I add this to the value of 66.3 J, I still wouldn't get the right answer.
     
  5. Apr 16, 2007 #4

    Doc Al

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    That's the hard way. (And what about air resistance?) Hint: Take full advantage of the information given.
     
  6. Apr 16, 2007 #5
    I am really not sure what to do. I am guessing that maybe I have to find the kinetic force? I am not sure, can you please elaborate more. Thanks.
     
  7. Apr 16, 2007 #6

    D H

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    What is the relationship between work and energy?
     
  8. Apr 16, 2007 #7
    Well, since the chapter is about Power, it says in my book that power is work divided by change in time, and power could also be energy divided by change in time.

    Also, I know about how when work is done, there is a change in energy.
     
  9. Apr 16, 2007 #8

    Doc Al

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    Now you're getting warm! :wink:

    Express that mathematically.
     
  10. Apr 16, 2007 #9
    W = E (delta energy)

    i remember reading in my book that work is equal to change in kinetic energy or gravitational potential energy. But I am not sure whether that's right.
     
  11. Apr 16, 2007 #10

    Doc Al

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    Look up the "Work-KE" theorem (or Work-Energy Principle) in your text. It states that the net work done by all the forces acting on a body equals the change in kinetic energy.

    Read this: Work-Energy Principle
     
  12. Apr 16, 2007 #11
    You are correct, but who is doing the work the system or is energy being transfered(work done by outside force) to the system, or is the system losing potential energy, and gaining negative work? So it is a matter of perspective, you can have positive work and negative work. If the system gains potential energy, there was positive work done, but the system is losing energy gained by work(negative work is being done) if it is transfer it via kinetic or other forms,friction etc...

    make sure you pay attention to what object or system they are referring to when talking about work, it is very important!! Get familiarized with what system means and how energy is transfered. Work has big part in it, its a term used most often.
     
    Last edited: Apr 16, 2007
  13. Apr 16, 2007 #12
    ya, but you see the work-energy theorem states that there must be a second kinetic energy and potential energy. the problem with this question is that the final velocity is 0, so that would mean the whole final energy would be 0.

    here are the calculations that i am trying to do:

    KEi + PEi + Wext = KEf + PEf

    (1/2)(0.125)(9.78^2) + (0.125)(9.81)(0.78) = (1/2)(0.125)(0^2) + (0.125)(9.81)(0.78)

    5.978025 + 0.956475 = 0 + 0.956475

    Now I am not sure what to do, because this formula doesn't lead me to anything.
     
  14. Apr 16, 2007 #13

    Doc Al

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    Find the total work done on the ball, by all forces, up until the moment just before it's caught. What's the change in KE of the ball?
     
  15. Apr 16, 2007 #14
    To find change in KE, I did:

    PEi

    (0.125)(9.81)(0.78)
    = 0.956475

    KEi

    (1/2)(0.125)(9.78^2)
    = 5.978025

    Total Change in KE : 5.978025 - 0.956475

    = 5.02155 is the total change in KE.

    I hope I am on the right track. Please tell me. thanks.
     
  16. Apr 16, 2007 #15
    ya, i am sorry i can't understand because the thing is that the formula for KE is:

    (1/2)mv^2

    Now, how can there be a change in kinetic energy while nothing is changing? because i am confused since velocity is not changing, so how can there be a change in KE.
     
  17. Apr 16, 2007 #16

    Doc Al

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    On what basis are you calculating PE? You are given no information about the height or change in height. (0.78 m is the distance over which force acts, not the change in height.)

    Forget PE. Find the change in KE.
    Use the correct "final" speed.
     
  18. Apr 16, 2007 #17

    Doc Al

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    What makes you think velocity is not changing? Someone just threw a ball! (The initial speed is zero.)
     
  19. Apr 16, 2007 #18
    oh my bad,

    i was using the wrong speed.
    ok, i recorrected my calculation:

    KEi
    (1/2)(0.125)(9.84^2)
    = 6.0516

    ok, so the initial speed is zero, that would just mean that the kinetic energy in the beginning is also zero, since the formula for kinetic energy is all multiplication and when we multiply anything by zero, it's zero.
    (1/2)(0.125)(0^2) = 0

    so, i am thinking that 6.05 is the change in kinetic energy.
    is that right? please help, thanks very much.
     
  20. Apr 16, 2007 #19

    Doc Al

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    That's what I would say. (6.05 Joules.)
     
  21. Apr 16, 2007 #20
    ya, but isn't that the kinetic energy? and the question is asking about the work being done on the ball. so that means the kinetic energy and the work being done are actually the same values?

    i guess maybe the textbook has the wrong answer. but it couldn't be wrong since it asks about how much energy was lost in the other part of the question.
     
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