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Work and the Frictional Force (Grade 12 Physics)

  1. Jun 16, 2016 #1
    1. The problem statement, all variables and given/known data
    A boy does 465 J of work pulling an empty wagon along level ground with a force of 111 N [31° below the horizontal]. A frictional force of 155 N opposes the motion and is actually slowing the wagon down from an initial high velocity. The distance the wagon travels is _____.

    2. Relevant equations
    W = Fdcosθ

    3. The attempt at a solution
    465 J = (111 N)d(cos31°)
    d = (465 J)/(111 N)(cos31°)
    d = 4.89 m

    Where does the frictional force come in here? Does it actually affect the distance travelled or does it just slow down the wagon?
     
  2. jcsd
  3. Jun 16, 2016 #2
    you are given the total work done in moving a distance say d on the incline-
    so write down the total work done as force times the distance - but this force should be net force....
    the wagon must be overcoming the frictional force as well as gravitational pull so correct your work done expression.
     
  4. Jun 16, 2016 #3
    I'm still confused...

    Do I need to look at force from both a vertical and horizontal component to solve the problem? Many thanks for the response!
     
  5. Jun 16, 2016 #4

    cnh1995

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    Is the wagon moving along level ground or along an incline? From the wording of the problem, I imagine the wagon is moving along level ground and the boy is pulling it at an angle 31° below the horizontal level. Is that right?
     
  6. Jun 16, 2016 #5
    I believe that that is the case.

    So if the wagon is moving along level ground, would I calculate the net force using the horizontal component of the applied force and the frictional force? And then input into the formula for work?
     
  7. Jun 16, 2016 #6

    cnh1995

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    Yes.
     
  8. Jun 16, 2016 #7
    Great! Thank you so much for your help! :)
     
  9. Jun 16, 2016 #8
    Upon calculating, I got a negative value for distance (-7.77 m). For some reason, this doesn't make sense to me. It seems with the wording of the question that the boy is moving forwards, but I may be mistaken.
     
  10. Jun 16, 2016 #9

    cnh1995

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    The net force is coming out to be negative, which means there is retardation of the body. This means the body is losing energy. This makes the work done on the body negative. So, negative work and negative force will give you a positive displacement.
     
  11. Jun 16, 2016 #10
    So, I would use the value of -465 J when calculating the distance due to the fact that the work is acting in the same direction as the net force?
     
  12. Jun 16, 2016 #11

    cnh1995

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    Work does not act on a body. It is a scalar quantity. Displacement and force are in the opposite directions, which gives a negative work.
     
  13. Jun 16, 2016 #12
    Okay, that makes sense.

    So...
    W = Fd
    -465 J = (-59.9 N)d
    d = (-465 J)/(-59.9 N)
    d = 7.61 m
     
  14. Jun 16, 2016 #13

    ehild

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    The force of friction does not matter when you calculate the displacement. Given the force the boy exerts, given the angle between force and displacement, and given the work of the boy. The boy's force and work are independent of the force and work of friction, the friction slows down the wagon.
     
  15. Jun 16, 2016 #14

    ehild

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    That would be the net work done on the wagon. Why is it relevant when you are given the force the boy exerts, the work the boy does and the question is the distance traveled during the work is done?
     
  16. Jun 16, 2016 #15
    Okay. So I should ignore the frictional force all together in this instance?
     
  17. Jun 16, 2016 #16

    ehild

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    Yes. Your first solution was correct.
     
  18. Jun 16, 2016 #17
    Okay! Thank you so much for your clarification, ehild! :smile:
     
  19. Jun 16, 2016 #18

    ehild

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    You are welcome :smile:
     
  20. Jun 16, 2016 #19

    ehild

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    You are wrong. The work of the boy is given, read the problem text more carefully. Also, the wagon moves on level ground, gravitation is irrelevant.
     
  21. Jun 16, 2016 #20
    thanks for pointing the error- perhaps hurriedly i missed the point of horizontal displacement.. certainly .i should be more careful.
     
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