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2 work questions that got me stumped

  1. Nov 19, 2009 #1
    Two Work problems I can't figure out. I don't really know where to start?
    Question1) A delivery person carries a 215N box up stairs 4.20m vertically and 6.80m horizontally.
    a)How much work does the delivery person do?(Answer: 903 J)
    b)How much work does the delivery person do in carrying the box down the stairs?(-903J)

    Question2) A laborer pushes a wheel barrow weighing 200N at 25.0 degrees above the horizontal. If he pushes it a distance of 25.0m, how much work is done?(Answer: 10 700J)

    Can someone explain in full detail how to reach their answers? I tried them and get nowhere close.

    Relevant Equations
    W=Fs
    W=Fscostheda
    W=fs=mgs

    I know that in Q1) a) if you go W=(215N)(4.20m) = 903J However, what do we do about the 6.80m? Also.. Going down the stairs becomes negative 903J? Is that becomes they go downwards with the box? I really dont know how to properly display my answers for this.
     
    Last edited: Nov 19, 2009
  2. jcsd
  3. Nov 19, 2009 #2
    Let's start with problem #1:

    Would you not agree that because the box has no velocity at the top of the stairs the total work done on the box is zero? Also, which forces do work on the box?

    When can work be negative? Only when the force is opposite the displacement.
     
    Last edited: Nov 19, 2009
  4. Nov 19, 2009 #3
    I wouldn't agree with you. If the displacement is 0 than the work would be 0, however there is a displacement of 4.20m, plus the final answer is also 903 Joules or Newton Metres in the back of the text. So the answer cannot be 0.
     
    Last edited: Nov 19, 2009
  5. Nov 19, 2009 #4
    I'm not saying the answer is zero. We are being asked for the work done by the man; I'm saying the total work done is zero.

    [tex]\Sigma W=fd=\Delta Ke=0[/tex] -(at the top of the stairs)

    Now which forces do work on the box?
     
    Last edited: Nov 19, 2009
  6. Nov 19, 2009 #5

    berkeman

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    What force opposes the vertical movement of the box? So how much work is done in that part? What force opposes the horizontal movement of the box? How much work is done in that part?
     
  7. Nov 19, 2009 #6

    berkeman

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    Is the delta-KE meant to indicate some kinetic energy component in this question?
     
  8. Nov 19, 2009 #7
    I'm saying that:

    [tex]Ke_f - Ke_i = 0 = \Sigma W[/tex]

    because the box starts at rest and finished at rest the total work done on the box better be zero.
     
  9. Nov 19, 2009 #8

    berkeman

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    Not so. The OP could use PE changes to calculate the work done, or multiply the force by the distance through which the force was exerted. KE has nothing to do with this problem, and the work done on the box is definitely not zero.
     
  10. Nov 19, 2009 #9
    Total work done = 0

    the man does positive work and gravity does negative work of equal magnitude. Thus resulting in zero net work on the box.
     
  11. Nov 19, 2009 #10

    berkeman

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    No, srmeier, that is just plain incorrect. I don't want to give too much away here in this thread, because we will end up answering the OP's question for him. We should probably carry this on via PM. So PM me what the delta-PE is for the box in the movement described above, and tell me where that change in potential energy came from. Thanks.
     
  12. Nov 19, 2009 #11
    Ok I am getting two different answers and it is just confusing me. This is what I did for 1 a)
    W=Fs
    W=(215N)(4.20m)
    W=903Nm or J

    1 b)
    W=Fs
    W=(-215N)(4.20m)
    W=-903Nm or J

    These are both the final answers in the book, however I am still confused about what we do with the horizontal 6.80m.
     
  13. Nov 19, 2009 #12
    Question 2
    W=Fs(costheda)
    W=(200N)(25.0m)(cos25.0degrees)
    W=4531 N

    This is incorrect though.. I don't know.. This is just frustrating me now. They don't give examples for this in the book.
     
  14. Nov 19, 2009 #13

    berkeman

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    Good. There is no force opposing the horizontal motion, so there is no work done horizontally in this problem.
     
  15. Nov 19, 2009 #14

    berkeman

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    Hint -- if he pushes 25m total at an angle of 25 degrees, what is the total vertical displacement?
     
  16. Nov 19, 2009 #15
    I don't know how to get the vertical.. There is no example.

    Would I use trig to find the vertical? hypotaneuse being 25, inside angle being 25, and figure out the opposite side(vertical)?
     
  17. Nov 19, 2009 #16

    berkeman

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    Yep.
     
  18. Nov 19, 2009 #17
    I got 10.25m for the vertical..

    I popped that in the W=Fs and W=FsCostheda equations but I still don't get 10 700J.
     
  19. Nov 19, 2009 #18

    berkeman

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    Yeah, not sure why that isn't giving the answer they provide. Does the problem say anything else? Is it copied into your post correctly?
     
  20. Nov 19, 2009 #19
    I must dispute berkeman's method.

    "A laborer pushes a wheel barrow weighing 200N at 25.0 degrees above the horizontal. If he pushes it a distance of 25.0m, how much work is done?"

    He is pushing the wheel barrow in the horizontal direction (not up an incline). the vertical displacement is zero. Although, the man does do work.

    (note: even with my changes the answer isn't 10,700J. Also, if the wheel barrow weighted twice as much and had a vertical displacement of 25m the work done by gravity still wouldn't reach 10,700J. Please check your book.)
     
    Last edited: Nov 19, 2009
  21. Nov 19, 2009 #20
    I got 4300J. It does say 10700J in book. It is mistake.
     
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