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Potential energy problem

  1. Dec 23, 2015 #1
    • Member warned to use the formatting template for homework posts.
    I present the following problem to my AP and have received a variety of answers, being a new physics teacher I am not sure that my answer is correct. I would love to see the solution and justification.
    Thanks
    One attempt based on the inference that one block is gaining U (A) and the other (B) 1 is loosing it
    1/2 mv2 b - 1/2 mv2a = 4.5 J

    In the figure below, the pulley is massless, and both it and the inclined plane are frictionless. The masses are released from rest with the connecting cord taut. Some time later the 2.0 kg block has a speed of 3 m/s. What was the change in potential energy of both blocks?
    upload_2015-12-23_9-47-49.png
     
  2. jcsd
  3. Dec 23, 2015 #2
    The string restricts the blocks to move at the same speed though so wouldn't 1/2 m(vb)^2 - 1/2 m(va)^2 = 0?
     
  4. Dec 23, 2015 #3
    The OP probably left out the subscripts in the mass terms. The two blocks have different masses.

    I don't see anything wrong with the energy approach - it should give the correct answer. Is there any particular reason why you're doubtful about it?
     
  5. Dec 23, 2015 #4
    If
    wouldn't that equate to them having the same kinetic energy? but since they are both moving at the same speed with different masses it can't equal zero?
     
  6. Dec 23, 2015 #5
    Yes indeed I overlooked the unequal masses
     
  7. Dec 23, 2015 #6
    One students argument in the change of U should be negative since in the end there is less potential energy...
    Another approach was to calculate the total energy 1/2 ma+mbv2 which ends up being 13.5 and since

    Emech =U + K U = -13.5
     
    Last edited: Dec 23, 2015
  8. Dec 23, 2015 #7
    I think the problem here is the "approaches" you present are not exactly complete enough for us to comment on their correctness. The energy approach is the right way to go about it, although whether the students execute it correctly is another question altogether.
    You just have to use
    [tex]\Delta \mathrm{KE} = - \Delta U[/tex]
    and make use of the fact that the cord remains taut throughout.
     
  9. Dec 23, 2015 #8
    The problem with the student solution is that he is ignoring the fact that the weight force of the block "A" is doing negative work. So the relationship ΔKE=−ΔU is not complete, in this case. What we need is to decompose the Weight of block "A" in the x-direction:
    WeightA = 9,81 N → Weight Ax = 9,81 × (sin 30°) = 4,905 N.
    Okay, now we need to figure out what is the work made by that force, I did using the block "B":
    WB = ΔKE = ½×2×(32) = 9J
    By definition: Work = F×Δx, therefore:
    ΔxB = 9J/19,62N ≅ 0,46 m
    If we know how much "B" has moved, and we know the angle of the plane, we can discover the variation of space made by the block "A":
    (Sin 30°) = ΔxB/ΔxA → ΔxA ≅ 0,91 m
    Now the easy part, just put the values in the relationship:
    ΔKE + Wother forces = -ΔU → ΔKE + Wweight of "A" = -ΔU → [½ × (1Kg +2Kg) × 32m/s] - (4,905N × 0,91m) = -ΔU = 13,5 - 4,5 = 9J
    If the system is composed of two objects, the potential energy of the variation is 4,5J for each block , which is the response obtained by Physicscookie. I think that's the way that we can solve the problem. I hope I've helped, and sorry for bad English.
     
    Last edited: Dec 23, 2015
  10. Dec 24, 2015 #9
    I would be very disturbed if [itex]\Delta KE + \Delta U \neq 0[/itex] for a closed system.
    No, the magnitude of the change in gravitational potential energy should be different for each block...

    If you want to do a force analysis, you shouldn't mix it up with the energy approach, and you need to consider the tension in the rope as well.
     
  11. Dec 24, 2015 #10
    Well, if there is a non-conservative force acting on a system (like friction force, or, in this case, the weight in x-direction), the relationship ΔKE = -ΔU is incomplete, because you need to add the work made by the non-conservative forces. The relationship becomes ΔKE + Workadditional = -ΔU.
    I saw that type of solution (using the work of the weight in the x-direction) made by a book called "Physics I, 12ª Ed." by Young & Freedman, in the chapter 7, exercise number 73, it's a little different, cause involves elastic potencial energy, but the ideia is there.
    Anyway, sorry if I made a mistake.
     
  12. Dec 24, 2015 #11
    "Weight", otherwise known as gravitational force, is conservative.
     
  13. Dec 24, 2015 #12
    Yes, I tried to solve the problem that way because of that exercise on the book, I was confused when I saw the solution, but I tried to see that force similarly to a force of friction made by the block. It seems that the solution is wrong, sorry.
     
  14. Dec 25, 2015 #13
    If you calculate potential energy as work of acting forces you cannot calculate again as potential energy. There is double calculation.
    $$ T = \frac{1}{2}(m_1+m_2)v^2 \,\,\,\, U = g(m_1h_1+m_2h_2) $$
    Take the pulley as reference point we have by geometry:
    $$ h_2 \to h \,\, and \,\, h_1 \to (l-h)\sin\phi \Rightarrow U(h) = gm_1(l-h)\sin\phi + gm_2h $$
    Now we have:
    $$ T = \frac{1}{2}(m_1+m_2)\left(\frac{dh}{dt}\right)^2 \,\,and\,\, U = m_1gl + (m_2-m_1\sin\phi)gh $$
     
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