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Conservation of energy theorem problem

  1. Nov 10, 2008 #1
    1. The problem statement, all variables and given/known data
    2 boxes are connected by a massless rope. the system is frictionless
    m1 = 3kg
    m2 = 5kg
    [​IMG]
    Find m1's speed when it is 1.8 m above the ground.
    2. Relevant equations
    K = 0.5mv^2
    U = mg delta y

    3. The attempt at a solution
    v1=v2 obviously
    also, delta y of m1 = 1.8, whereas delta y of m2=-1.8
    At the beginning:
    Etot = K1 + K2 + U1 + U2
    since K1, K2 and U1 are = 0, Etot = U2 = m2*g*5 = 25g

    When m1 is 1.8 meters above the ground:
    Etot = K1+K2+U1+U2
    25g = 0.5m1v2+0.5m2v2+1.8m1g-1.8m2g

    solving for v gives 8.37,
    but the answer is 2.97
    can anyone help me find what is wrong in my reasoning?
     
  2. jcsd
  3. Nov 10, 2008 #2
    m2 ends up not at -1.8m, but at (5-1.8)m.

    Otherwise, you did the right thing.
     
  4. Nov 11, 2008 #3
    Also, be careful about your velocities. Though it doesn't matter in this problem since velocities are squared, it should be:

    [tex]
    v_1 = -v_2
    [/tex]
     
  5. Nov 11, 2008 #4
    I thought that what is important is the difference of height, and not the height itself. since the formula is U = mg delta y.. then delta y = [(5-1.8] - 5] = -1.8
    humm can you please tell me where I am mistaken?

    EDIT: I also see in the book U=mgh. If I had used this formula I woulnd't have had problems. What is the difference between both? When to use one instead of the other?

    thank you
     
  6. Nov 11, 2008 #5
    You did NOT use the change in height: you wrote down the initial potential energy as mgh_2, with h_2=5m, then in the final potential energy you used h_2=-1.8m. That's inconsistent. Either you use the change in kinetic energy, or you use the correct initial and final potential energies.
     
  7. Nov 11, 2008 #6
    You have to set an absolute coordinate system and work from there. i.e. the ground is y=0.

    The [tex] \Delta y [/tex] is most likely referring to the fact that potential energy is always relative with respect to some reference point. If we set the ground to be y=0 then [tex] \Delta y [/tex] is just y ( [tex] \Delta y = y - 0 = y [/tex].

    On the other hand, if we set sea level to be 0 but then do experiments in a city not at sea level, well [tex] \Delta y = y - y_0 [/tex] where y_0 is the height above sea level, and y is height with respect to sea level. (Though at this point, it would make more sense to set y_0 to be zero, and take y to be height above the ground you're using).

    With this in mind, you can see that [tex] \Delta y [/tex]
    is almost always just y.
     
    Last edited: Nov 11, 2008
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