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Work-Energy Problem

  1. Jan 26, 2014 #1
    1. The problem statement, all variables and given/known data

    A child lifts a bucket full of water from the bottom of a 5 meter deep well by pulling on a rope. The 10.0 kg bucket accelerates at a rate of 0.250 m/s/s.

    Determine:

    a. the work done on the bucket by the child
    b. the work done on the bucket by gravity
    c. the total work


    2. The attempt at a solution

    Calculating the applied force by the child:

    Fnet=ma
    Fapp - mg = ma
    Fapp = ma + mg
    Fapp = (10 kg)(0.25 m/s/s) + (10 kg)(9.81 m/s/s)
    Fapp = 100.6 N

    Wchild = F x d
    = (100.6 N)(5 m) = 503 J

    Wgravity = F x d
    = (9.81 m/s/s x 10 kg)(5 m) = 490.5 J

    Wtotal = Wchild + Wgravity
    = 503 J - 490.5 J = 12.5 J
     
    Last edited: Jan 26, 2014
  2. jcsd
  3. Jan 26, 2014 #2

    BvU

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    How come I think 503 + 490.5 = 993.5 ?
     
  4. Jan 26, 2014 #3

    BvU

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    What I mean is: you (correctly) introduce a minus sign, because ... ?
     
  5. Jan 26, 2014 #4
    The gravitational work is resistive, and in the opposite direction.
     
  6. Jan 26, 2014 #5

    BvU

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    Excellent. Opposite for work is a bit strange, after all it's a number, not a vector like F.
    What you want to do is distinguish between work where force and d point opposite (such as this gravity work) and work where force and d are in the same direction (student pulling the rope).

    Your choice for the positive direction is upwards; that way you get + 503 and -490.5.

    Now for the hard part: what can be considered to be the total work ?

    To begin with: why is there a difference and where did it go ?
     
  7. Jan 26, 2014 #6

    BvU

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    Excellent is too much, because I read again and find: resistive. That is not correct. Why not?
     
  8. Jan 26, 2014 #7
    My textbook states that if work is being done in an opposite direction from the direction of motion, then it is called resistive work. It's not aiding the movement of the object, rather it is "holding it back".
     
  9. Jan 26, 2014 #8
    The total work can be considered the "net force" x distance. Because this is the actual work that brings the bucket up the well.

    The difference is because of the two different directions; however, the child pulled with enough force to counteract the gravitational force and thus, pulled the bucket up.
     
  10. Jan 26, 2014 #9

    BvU

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    Back to excellent! (Textbooks are right most of the time). I was triggered the wrong way by the wording. Resistance can convert work into heat (like when you drag something over a rough floor). It can also oppose another force and still be conservative as it is called. So I learn too! Conservative is the opposite of dissipative; it means you can get it back: in the bucket case: the student can drop the bucket back in the well and get back a lot of kinetic energy that, at landing, gets dissipated into a huge splash! What fun.

    Now about total work. Suppose the student pulls a little less hard. Or harder. What would be the difference ?
     
    Last edited: Jan 26, 2014
  11. Jan 26, 2014 #10
    If the student pulled harder, there would be more force applied, over the same distance, thus the there would be more work done by the student.

    If he pulled a little less hard, less work would be done because less force is applied.
     
  12. Jan 26, 2014 #11

    BvU

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    Yes. Gravity work would remain the same, right? Where does the excess, or if you want: difference, go ?
     
  13. Jan 26, 2014 #12
    The excess work aids the student in pulling the bucket up since the gravitational work would remain identical.
     
  14. Jan 26, 2014 #13

    BvU

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    Check your 503 J expression. Which term causes the difference? What is its effect by the time the bucket is above?
     
  15. Jan 26, 2014 #14
    The force causes the difference, right?
     
    Last edited: Jan 26, 2014
  16. Jan 26, 2014 #15
    Is part (c) wrong?
     
  17. Jan 26, 2014 #16

    BvU

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    Wrong is a bit too severe. In my view (opinion!) the total work is what the student has done.

    I had 2 questions, you answered the first one: the force. What part of the force? the 503 and 490.5 have something in common and some difference.
     
  18. Jan 26, 2014 #17

    PhanthomJay

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    no, total work is the net work done on the object by all the forces acting on the object, which in this example is the sum of the work done by gravity and the work done by the child. Or alternatively, the work done by the net force.
     
  19. Jan 26, 2014 #18

    BvU

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    Well PV, now you have two experts haggling over your head ! My argument for the total work being what the student has done is simple: It takes energy to bring something above ground and that means work has to be done. If you use an electric motor, you will have to pay for the total, not for the net work....

    But I still miss your ideas on where the 12.5 J has gone ?
     
  20. Jan 26, 2014 #19
    So, according to you, part (c) has been correctly calculated?

    The 12.5 J is the net work that has caused the acceleration of the bucket.
     
  21. Jan 26, 2014 #20

    PhanthomJay

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    The food bill so to speak is for the energy used by the child. But nonetheless, by definition of total work and its use in the work - energy theorem, the total work is the net work or 12.5 J. Answer c is correct.
     
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