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Work and force

  1. Apr 14, 2008 #1
    1. The problem statement, all variables and given/known data
    a 50 kg diver steps off a diving board and drops straight down into the water. The water provides an upward average net force of 1500 N. If the diver comes to rest 5.0m below the water's surface, what is the total distance between the diving board and the diver's stopping point under water?


    2. Relevant equations



    3. The attempt at a solution
    the total force would be Fn-Fg which is 1500-50*9.81=1009.5N
    W=1009.5*(x+5)

    i am completely lost on what to do next. please give me a clue or something.
     
  2. jcsd
  3. Apr 14, 2008 #2

    Doc Al

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    That's the total force on the diver when he's in the water. What about before he hits the water?
     
  4. Apr 14, 2008 #3
    oo it wud b fg which is 50*9.81 which is 490.5 . but that doesnt help.
     
    Last edited: Apr 14, 2008
  5. Apr 14, 2008 #4

    Doc Al

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    Sure it helps. There are several ways to solve this kind of problem. One way would be to compare the work done on the diver before he hits the water to the work done on him after he hits the water. After all is said and done, what must the total work (by all forces) be on the diver?
     
  6. Apr 14, 2008 #5
    well you dont know the distance so you cannot find the total work done.
     
  7. Apr 14, 2008 #6

    Nabeshin

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    Think what total work means in reference to his original state (position and velocity) and final state.
     
  8. Apr 14, 2008 #7

    Doc Al

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    The distance is what you are asked to find. Try calling the distance from board to water surface D (or whatever). Now set it up and see if you can solve for D. (Then use it to get the total distance.)
     
  9. Apr 14, 2008 #8
    you cannot because you would have two unsolved, the work net and the distance.
     
  10. Apr 14, 2008 #9

    Doc Al

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    Try it and see. The only unknown is the distance.
     
  11. Apr 14, 2008 #10
    W=1009.5*(x+5) and W=490.5x
    are those two equations correct?
     
  12. Apr 14, 2008 #11

    Doc Al

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    No.

    The diver travels a distance x before hitting the water. What net force acts? What's the net work? Is it positive or negative?

    The diver travels 5 m under the water. What net force acts? What's the net work? Is it positive or negative?

    What must those two work contributions add to? (Hint: What's the change in KE?)
     
  13. Apr 14, 2008 #12
    well when he is in water the work he does would be 1500N*5m because thats the distance he z traveling and the net force is 1500. so the work done would be 7500J.
    The work he does under water would not be the same as above water would it?
     
  14. Apr 14, 2008 #13

    Doc Al

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    1500 N is the force of the water, not the net force. You found the net force in post #1.

    Something like that. What are the signs of the two work contributions?
     
  15. Apr 14, 2008 #14
    the work done above water would be negative while the word done below water would be positive. but how can both of them have the same magnitude? that's the part i don't understand.
     
  16. Apr 15, 2008 #15

    Doc Al

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    Just the opposite. Above the water, the work done on the diver is positive: the force (gravity) acts in the same direction as the displacement. Below the water, the net force acts up while the displacement is still down; so the work on the diver is negative.
    Above the water, the diver's KE increases as he falls (since work is being done on him); below the water, it decreases (since negative work is being done on him). It all has to balance out, since he starts with 0 KE and ends up with 0 KE.
     
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