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Lab based conservation of energy quesiton.

  1. Dec 1, 2005 #1

    404

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    Ok, there is a cart on a table(very little friction) with a string attached to it. The cart and the string goes over a pulley at the edge of the table, and has a weight hanging down the side of the table. So when you let go of the cart, it will accelerate until the cart hits the floor, then go in constant velocity until it runs into the pulley.

    So when you graph it's velocity vs time, you get a sloped line and a horizontal line, and we picked the max point (v at end of the acceleration) as the final velocity to use in the KE formula.

    The point of the lab is to prove (ignore friction)
    mgh(PE) = .5mv^2 (KE)
    We had a relatively high % error some of our trials, so I was wondering, what are you suppose to use for the mass from the PE side? (just mass of the weight or both that and the cart?). Also, same question with the m from the KE side... I thought I knew what I was doing, but I started questioning my calculations when I saw some rather high % errors...
     
  2. jcsd
  3. Dec 1, 2005 #2
    well, what all has potential energy? what kinds of potential energy are there?
    and what has kinetic energy?

    looking at your equations for the PE and the KE this is very easy to decide if you think about it a little bit.
    /s
     
  4. Dec 1, 2005 #3

    404

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    Well for the PE side I just used the mass of the weight
    for the KE side, I got both masses.

    I already did all the calculatiosn etc, just looking for some quick confirmation or where I went wrong.
     
  5. Dec 1, 2005 #4
    why would you use just the mass of the weight for PE?
    why would you use both the masses for the KE?

    if you can answer those you'll know where you went wrong.
    here's a hint: at the point on the graph where the slope becomes constant is where you got your v value from. what all is moving?
     
  6. Dec 1, 2005 #5

    404

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    Ok so only the cart is moving. So PE both and KE just cart mass?
    But I know that's wrong because then PE ends up around about 90 times bigger than KE.

    What is it then? will you please kindly just cut to the chase?
     
  7. Dec 1, 2005 #6
    why is it both on PE?
    what is the formula for PE?
    are they both off the ground?
     
  8. Dec 1, 2005 #7

    404

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    Ok post #4 you implied PE wasn't just the weight like I originally said, so the only other option is both(sure as well can't just be the cart).
    Lol...sorry but you're not helping much.... it's been a long day, tired, was justing looking for a quick answer. nvm, I'll just ask my teacher tomorrow.
     
    Last edited: Dec 1, 2005
  9. Dec 1, 2005 #8
    my the post #4 didn't imply anything.
    i just asked if you knew why you only used tha mass.
    look at the formulas you have:

    PE = mgh
    KE = .5mv²

    the only thing suspended above a surface is the weight, and the only thing moving at the point you took v (when slope is straight - the weight hit the ground like you said in the beginning) the only thing moving is the car.
     
  10. Dec 2, 2005 #9

    daniel_i_l

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    The potential energy of the hanging mass is conserved to the kinetic energy of both of the masses. After the hanging mass hits the floor there is no more PE so the horizontal mass continues at a constant maximum speed. But when you equate the PE and KE you have to use both of the masses for the KE.
     
  11. Dec 3, 2005 #10

    andrevdh

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    Your initial approach to the problem is correct. Did you include the mass of the cart? Did you measure the height of the hanging mass to the bottom of the mass? Is the mass of the pulley small? Can the friction of the pulley be ignored - especially when it is under load?
     
  12. Dec 4, 2005 #11
    daniel, you need to use both of the masses for both of the energies.
    PE = KE
    [tex] m_\textrm{mass} g h_\textrm{mass} + m_\textrm{cart} g h_\textrm{cart} = \frac{1} {2} m_\textrm{mass} {v_\textrm{mass}}^2 + \frac{1} {2} m_\textrm{cart} {v_\textrm{cart}}^2 [/tex]
    at t = 0 the mass is suspended in the air and the cart is on the table.
    he took the value of V where the slope was a constant (when the mass was lying on the floor) and the cart was rolling with constant v.
    so clearly some of the above terms vanish.
    in this level of lab it is safe to assume that pulleys are frictionless, and [itex] m_\textrm{pulley} << m_\textrm{cart} , m_\textrm{mass} [/itex] so the rotational energy is negligible.
     
    Last edited: Dec 4, 2005
  13. Dec 5, 2005 #12

    andrevdh

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    Another possibility is that the cart was slightly rotated with respect to the line of motion. In such a case energy will be lost due to friction as it is pulled along. One should therefore repeat the runs and find the "true" maximum speed of the cart.
     
    Last edited: Nov 29, 2006
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