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Work Problem

  1. Mar 3, 2016 #1
    1. The problem statement, all variables and given/known data
    A man (75kg) is sledding down a 20m high hill that is at an angle of 30 degrees. He is given a push that has him initially going 7m/s. He experiences a force of friction of 22N on his way down the hill.

    a) How much work is done by friction on his way down the hill?

    b) How fast is he going when he gets to the bottom?

    2. Relevant equations
    W=Fdcos(Θ), MEi + W = MEf

    3. The attempt at a solution
    For (part a) I thought you had to use 1/2mvi2 + W = mgh but I feel like I am messing something up with the potential and kinetic energy.

    For (part b) I know you solve for final velocity but I'm not sure if you use 1/2mvi2+ mgh + W = 1/2mvf2 after getting the answer for work.
     
  2. jcsd
  3. Mar 3, 2016 #2
    a) What's the basic definition of work?

    b) Have in mind that, when the man is pushed, he is at the top of the hill so he has both kinetic and potential energy. When he gets to the bottom, he has only kinetic energy. It's just conservation of mechanical energy.
     
  4. Mar 3, 2016 #3

    haruspex

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    You don't have enough information to tackle it from that end.
    What other equations can you think of for work done by a force?
     
  5. Mar 3, 2016 #4
    a) Using W=Fdcos(theta) I don't know what the distance is, do I solve for the hypotenuse?

    b) If he only has kinetic energy then wouldn't it only be 1/2mv2i + W = 1/2mvf
     
  6. Mar 3, 2016 #5

    haruspex

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    Yes, why wouldn't you?
     
  7. Mar 3, 2016 #6
    Did that and used W=Fdcos(theta) and got 190.525J, now I'm just confused on what I am missing for part b.
     
  8. Mar 3, 2016 #7

    haruspex

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    Looks wrong to me. Please post your working.

    For part b), the equation you originally posted was correct, but we need to see how you apply it.
     
  9. Mar 3, 2016 #8
    a) I did it over and first I did cos(30) * 20 to solve for the hypotenuse and got 17 m. Then I did 22 * 17 * cos30 and got 324J.
     
  10. Mar 3, 2016 #9

    haruspex

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    That is not going to give you the hypotenuse. Did you draw a diagram?
     
  11. Mar 3, 2016 #10
    Yeah I did, the first time I did it I just did the sin(30) = 20/x and got 10 for x.
     
  12. Mar 3, 2016 #11
    Are you sure that sin(30) = x/20?
     
  13. Mar 3, 2016 #12
    Sine is opposite over hypotenuse and we are giving the opposite which is 20m unless I am putting my angle in the wrong spot.

    Sin(30) is equal to 0.5 which means that x would have to be 40 which I calculated wrong in the first place.
     
  14. Mar 3, 2016 #13
    Yes, x must be 40.
     
  15. Mar 3, 2016 #14
    Alright so knowing that using W=Fdcos(theta) I should get 762J.

    part b) If that is work then I can now solve for vf by using the original formula and get 427.32 m/s? Seems a bit large to me.
     
  16. Mar 3, 2016 #15

    haruspex

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    Still wrong. How do you get that? What is theta there?
     
  17. Mar 3, 2016 #16
    Did 22N*40m*cos(30) and got that.
     
  18. Mar 3, 2016 #17
    Regarding item (b), remember to take the square root!
     
  19. Mar 3, 2016 #18

    haruspex

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    In the standard equation W=Fd cos(theta), how is the angle theta defined?
     
  20. Mar 3, 2016 #19
    Oh I forgot the angle is defined by the direction of the force and distance, in this case the force and distance are opposite to each other so would the angle be 180 degrees?
     
  21. Mar 3, 2016 #20

    haruspex

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    Whether they are opposite or in the same direction depends on how you are defining the positive directions, but yes, that will give the right magnitude.
     
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