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Homework Help: Gravitational Potential Energy of a bag

  1. Aug 26, 2007 #1
    1. The problem statement, all variables and given/known data
    A mail bag with a mass 120kg is suspended by a vertical rope 8.0m long.
    a.) What horizontal force is necessary to hold the bag in a position displaced sideways 4.0m from its initial position?
    b.) How much work is done by the worker in moving the bag to this position?

    2. Relevant equations
    K1 + U1 + W_{other} = K2 + U2

    3. The attempt at a solution
    i dont understand the problem but here's what i drew so far.... can u tell me if this correct

    1/2(mvi^2) + mgy + F x d = 1/2(mvf^2) + mgyf
    stuck here
    Last edited: Aug 26, 2007
  2. jcsd
  3. Aug 26, 2007 #2


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    For part a) use a free body diagram... use sum of forces = 0 at equilibrium, to calculate this force...

    For part b) use conservation of energy... but remember that the bag is at rest at the beginning and at the end of the process of moving, so no kinetic energy. work done is just final energy - initial energy
  4. Aug 26, 2007 #3
    a.) Equilibrium:
    Sum Fy = 0 = Fgrav - N
    Fgrav = N
    Sum Fx = None

    Fgrav = 120kg(9.8m/s^2)
    = 1176N

    i think Fgrav is the only force to be moved

    b.) W = mgy2 - mgy1
    = (120kg)(9.8m/s^2)(-4.0m) - (120kg)(9.8m/s^2)(-8.0m)
    = 4704J
  5. Aug 26, 2007 #4


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    Part b is right. But part a isn't. You need to consider 3 forces... the horizontal force F, gravity and tension in the rope. The tension has a vertical and horizontal component.

    If you do sum Fy = 0, and sum Fx = 0, you should be able to solve for F. You have two equations with two unknowns (tension and F).
  6. Aug 26, 2007 #5
    Fy = 0 = Fw - T
    Fw = T
    (120kg)(9.8m/s^2) = T
    T = 1176N

    Fx = 0 = F

    F = 0N?

    is it the part when the person already lifted it 4.0m or from the 8.0m?
  7. Aug 26, 2007 #6


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    You should calculate these for when it is lifted up... not at the bottom.
  8. Aug 27, 2007 #7
    tension has a vertical and horizontal component:

    using pythagoras
    L = sqrt (8^2 - 4^2)
    L = 6.92820m

    F x L = W
    since W = 4709J

    F = (4709J) / (6.92820m)
    F = 678.9639N < --- horizontal component

    yeah i know that because this is part a when you dont know the work...
    but this is all i got
    what do you think?
    Last edited: Aug 27, 2007
  9. Aug 28, 2007 #8
    Hi can u check my work above (part a) if its correct?
  10. Aug 28, 2007 #9
    You've done the 4m wrong. It says displacement SIDEWAYS.

    That means that the Triangle (relative to the top angle) is:
    Hyp = 8m
    Opp = 4m
    Adj = sqrt( 8^2 - 4^2 )
    = sqrt (64 - 16)
    = sqrt (48)
    ~ 7

    Also, if you had looked, 4^2 + 4^2 =/= 8^2
  11. Aug 28, 2007 #10
    so my part b
    (b.) W = mgy2 - mgy1
    = (120kg)(9.8m/s^2)(-4.0m) - (120kg)(9.8m/s^2)(-8.0m)
    = 4704J)

    is wrong? in which i assumed the initial y to be downward from the bag? which was presumed correct by learningphysics?
    Last edited: Aug 28, 2007
  12. Aug 28, 2007 #11
    position displaced sideways 4.0m from its initial position?

    The word "displacement" implies that the absolute distance between the final position and initial position is 4m. But the addition of the word "sideways" means that we're only looking at the horizontal translation. If you had used an absolute 4m displacement, that could be argued correct.
    But using a vertical displacement of 4m?
    No, that can't be argued correct at all.

    So, it is wrong. But, take a lesson from this. Read the question very carefully, and sometimes you'll see they're asking for something completely different from what you thought they wanted, or you'll see that there's extra information given to confuse you, etc.
  13. Aug 28, 2007 #12


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    I apologize Eduardo, BlackWyvern is right. The question says 4m displaced sideways...

    So using pyth. theorem... sqrt(8^2 - 4^2) = sqrt(48) = 6.928

    so y2 = -6.928 m. y1=-8.0m. So that's what you need to use for part b. so i'm getting mgy2 - mgy1 = 1260.4J

    Part a), is still wrong method wise (I'm getting the same number as you, but this is just coincidence I think)... don't use work for part a)... use sum of forces...

    for part a),
    [tex]Tsin\theta - mg = 0[/tex] (that's the y direction)
    [tex]Tcos\theta - F = 0[/tex] (that's the x direction).

    Solve for F. Coincidentally I'm getting F = 678. 9639N...
    Last edited: Aug 28, 2007
  14. Aug 28, 2007 #13
    a.) Its in equilibrium:
    forces sums to zero --->

    F = Tcos(thet)
    W = Tsin(thet)

    now need to find thet;
    cos(thet) = 4/8
    (thet) = cos^-1(0.5)
    (thet) = 60

    mg = Tsin(thet)
    ((120kg)(9.8m/s^2))/(sin(60) = T
    T = 1357.927833 N

    F = 1357.927833 N cos(60)
    F = 678.9639N

    b.) W = ?

    W = U2 - U1
    W = mgy2 - mgy1
    W = (120)(9.8)(-sqrt(48)) - (120)(9.8)(-(8.0)))
    W = 1260.433001

    Thanks for all the help
    Thanks learning physics
  15. Aug 28, 2007 #14


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    everything looks good. you're welcome eduardo.
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