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Work Done by gravity(Conceptual help needed)

  1. Oct 14, 2011 #1
    1. The problem statement, all variables and given/known data
    System is initially at rest, if m2 falls a distance h. What is the work done by gravity on the system.
    Given: Theta, m1, m2, h
    The image looks fairly similar to this.
    prob31fricpullplaneyi0.gif

    2. Relevant equations
    I'm going to go ahead and assume we are using Work(gravity)=The change in gravitational potential energy(ΔU) for this problem


    3. The attempt at a solution
    So I posted this question somewhere else, and got the reply "Here loss in Gravitational potential Energy of M2 = M2gh

    Gain in Gravitational Poetntial Energy of M1 = M1gh/sinθ

    Therefore
    work done by gravity on the system
    = M2gh - M1gh/sinθ

    = gh(M2 - M1/sinθ)"

    M2 made sense to me, but the equation for m1, did not, so I questioned him on his method of arriving at that and he gave me the image he made...
    91357044-0b48-40e7-a234-7d0f52719d6d.png
    Ok, so I see what he did. Sin(θ)=h/x, fine, but in doing that, he is assuming the total height is h, and the problem states "m2 falls a distance h," not "height of the triangle is h," or "Both m1 and m2 start from h" Also, when plugging this into his m1 equation 'm1gh' for h, he used x, not h, and h would equal h=xsin(θ), but we don't know x. I think he justified it by stating x='m1 move up' which is not the case, as x is the hypotenuse, which would be the distance the block moves up the ramp, not the final height. If this were the case, then gravity would be also dependent on a displacement, as well as height, which would make it a non-conservative force, right? But we know gravity is conservative.

    Just confused tbh, I feel dumb for not being able to answer such a seemingly simple question :/
     
  2. jcsd
  3. Oct 14, 2011 #2
    When M1 moves a distance h along its incline, it rises a distance A
    where

    A = h * sin (theta)
     
  4. Oct 14, 2011 #3
    So what your saying is, the distance m1 moves up the slope is the same as the height m2 fell.
    Ok, but we still arn't given that they both start at the same height.... If we work out the math, having our potential energy reference at the bottom of the incline (h below m2).


    W(g)=ΔUg
    W(g)=(Ugf-Ugi)m1+(Ugf-Ugi)m2
    W(g)=(m1ghsinθ-m1gy)+(0-m2gh) Where y is the initial height of m1
    W(g)=g(m1hsinθ-m1y-m2h)
    Except we still don't know what y is....
     
  5. Oct 14, 2011 #4

    HallsofIvy

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    It doesn't matter if "they both start at the same height". All that is important is one rose and the other fell the same distance.
     
  6. Oct 14, 2011 #5
    Ok, but it does matter whether or not they start from the same height, at least how I view it. Because work done is CHANGE in Gravtational potential energy (Ugf-Ugi) for both m1 and m2 (system). You can only have one GPE reference point where U=0, and if they are initially at different heights, well, you have to many unknowns compared to your known values? Idk, my thinking is probably off.
     
  7. Oct 14, 2011 #6

    phinds

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    I'm just glancing at this but I don't see how different starting heights makes any difference since the two heights are closely fixed in relationship to each other. Basically, what matters is how far off the ground is the falling weight at the start. If you know that, you know how far it falls, and from the other parameters you know how far the slanted weight rises vertically.
     
  8. Oct 14, 2011 #7
    Lol, maybe if I draw a picture, you guys will understand what I mean. I think I'll do that.
     
  9. Oct 15, 2011 #8
    I'll try to explain what I mean to the people who tell me that m1 and m2 move up and down the same height.

    Imagine an extremely long incline plane, with a very small degree angle, like 15-25 degrees. Two masses are connected by an extremely long, rope that doesn't not stretch. m1's bottom is at the very end of this extremely long incline, while m2's top is at the very top of the incline, hanging down. If m2's top travels to the very bottom of the incline, m2 has to travel all the way up the incline to be displaced the same height vertically??? Right?? m2 is barely moving down, m2 is being displaced horizontally ALOT, just to reach the same height, which doesn't make sense when the rope isn't shrinking!

    But It doesn't even matter I guess, because if I'm not even using the right formula for this problem (W(g)=(Ugf-Ugi)m1+(Ugf-Ugi)m2) this might help me lol. Which formula do I use? If it is the right formula, then having different initial heights does mess it up because you can only have one potential energy reference frame! :/ :/
     
  10. Oct 15, 2011 #9

    phinds

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    Perhaps I'm misreading what you just said, but here's the thing. No one has been saying that the two vertical-motion componsents are identical. Quite the contrary. If a mass moves from the top of an incline down vertially to the ground 10 feet then another mass that is connected to it via a rope, and is on a long 10 degree incline it will move along the incline for 10 feet and its vertical motion will be 1.7 feet.
     
  11. Oct 15, 2011 #10
    Thank you! I seriously thought I was crazy for a moment in thinking this after HallsofIvy said "All that is important is one rose and the other fell the same distance." To me that sounds like the hieghts they each rise and fall are the same. On to the next thing, am I using the correct equation for this situtation like I stated in my last post?
     
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