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How much work can be done given force, mass, angle & displacement?

  1. Feb 10, 2014 #1
    1. The problem statement, all variables and given/known data
    a person uses a cord to pull a boat of 1000 kg along 50 meters. The cord makes a 45° angle. If the applied work to the boat is 40 N, how much work is done?

    Given:
    mass = 1000 kg
    displacement = 50 meters
    the angle = 45°
    applied work = 40 N



    2. Relevant equations

    Work = Fd
    Weight = mg
    Work = Fd cos θ

    3. The attempt at a solution

    First I converted the mass to weight:

    1000 kg/ 9.81 m/s = 101.9 N

    Then I added that to the other force so, 101.9 N + 40.0 N = 141.9 N. *I don't know if I should add or subtract*

    Then I used the W = Fd cos θ equation:

    W = (141.9 N * 50 m)(cos 45°) =

    5016.9 J


    Is this correct?
     
    Last edited: Feb 10, 2014
  2. jcsd
  3. Feb 10, 2014 #2
    You can't just add the weight and the applied force together. They're vectors, so they have to be handled separately, or by components. Calculate work done by each force separately, and by the two components of the applied force, and then add those together into the net work.
     
  4. Feb 10, 2014 #3
    Is it,

    (101.9 N (cos 45)) + (40 N (cos 45)) = 100.4 N

    100.4 N * 50 m = 5019.2 J

    Yes?

    :confused:
     
  5. Feb 10, 2014 #4
    No, you have it a bit mixed up. Just think about the problem for a second. The displacement is straight to the right. The weight points straight down. The applied force points up and to the right, at a 45 degree angle. Calculate work done by weight as:$$W_{weight} = mg \cdot d \cdot cos\alpha$$ and $$W_{app} = F_{app} \cdot d \cdot cos\beta$$ where [itex]\alpha[/itex] is the angle between the weight force and the displacement, and [itex]\beta[/itex] is the angle between the applied force and the displacement.
     
  6. Feb 10, 2014 #5

    So;

    Wweight = 1000 kg * 50 m * cos 45? 35355.3

    +

    Wapp = 40 N * 50 m * cos 45? 1414.2

    1414.2 + 35355.2 = 36769.4 J ???:confused:
     
  7. Feb 10, 2014 #6

    BvU

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    Funny calculation. is the boat really being pulled up? i.e. is the displacement in the vertical direction also 50 m? Or are we pulling a boat along a canal over a towpath (explains the 45 degrees -- keeps the feet dry) and is the water level at the start the same as at the destination ?
     
  8. Feb 10, 2014 #7
    the boat is being pulled along the canal by the person towards the board walk. The problem in my book does not say whether the water level is at the same as at the destination so I presume it is.
     
  9. Feb 10, 2014 #8

    BvU

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    And: what's this "applied work to the boat is 40 N" ?? Work is Newtons times meters. We interpret the 40 N as a force of 40 N, but is this correct ? 40 N is like pulling with one finger....
     
  10. Feb 10, 2014 #9

    BvU

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    So: no work done there. 35355.2 J saved.

    By the way: if your given data are 1 or 2 digits, don't present the final result with 6 digit precision. 35 kJ looks a lot better (but we just erased that...).
     
  11. Feb 10, 2014 #10
    Gah, I'm confused. I'm sorry if it doesn't make sense I'm translating it to English.

    I'll restate it, this time more carefully.

    a person uses a cord to pull a boat that has a mass of 100 kg along 50 m by 50 m. along the board walk. The cord makes a 45 degrees angle. If the applied force to the boat is 40 N, how much work is realized?

    What I did (with the help of jackarms) was to calculate each force individually using the equation W = (F) (d) (cos θ) and then add them both for the result.
     
  12. Feb 10, 2014 #11
    What do you mean saved? How do I solve it?

    IDK. The other one made sense to me.
     
  13. Feb 10, 2014 #12

    BvU

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    Yes, so the boat doesn't move up or down: Wweight path is 0 m. No Wweight. Only Wapp. Checked the number, same result. One finger can do the work ;-)
     
  14. Feb 10, 2014 #13

    BvU

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    IDK meaning ?
     
  15. Feb 10, 2014 #14

    gneill

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    So a picture of the problem would look something like this?

    attachment.php?attachmentid=66489&stc=1&d=1392079010.gif

    Presumably the only relevant motion is horizontal (50 m). The weight of the boat is supported by buoyancy, and no work is needed to float the boat :smile: What portion of the 40 N force is acting in the direction of motion?
     

    Attached Files:

  16. Feb 10, 2014 #15
    Then how do I do it?
     
  17. Feb 10, 2014 #16

    gneill

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    Text speak: IDK = I Don't Know. "Text speak" is not allowed here. fixedglare, please use proper English and grammar for posts in the forums.
     
  18. Feb 10, 2014 #17

    BvU

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    Dumb me no tekst speak. I also drew a picture. No bird pulling the boat, just a person -- with feet dry. Cheaper boat - same result.
     

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    • Boat.jpg
      Boat.jpg
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  19. Feb 10, 2014 #18
    I'm sorry, I'll be more careful.
     
  20. Feb 10, 2014 #19

    gneill

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    :smile:
     
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