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MCAT question on energy

  1. Jun 13, 2007 #1
    1. The problem statement, all variables and given/known data

    OBjects A and B are placed on a vertical spring. A has twice the mass as B. If the spring is depressed and released, propelling the objects into the air, then A will:

    a) rise 1/4 as high as b
    b) rise 1/2 as high as b
    c) rise the same amount as b
    d) rise twice as much as b

    2. Relevant equations
    1/2kx², 1/2mv², mgh, projectile equations, etc.


    3. The attempt at a solution

    I really dont understand the logic in the solution (answer is C). It would make sense to me that the energy from the spring is transfered to kinetic, and then to gravitational potential energy for the masses. In this case, the mass of the objects would make a difference in determining the height (since its "mgh".)

    The only reason I can think is that there is more energy transfered from the spring to the heavier block.

    If it were just 1 block on the spring, its simply 1/2kx² = mgh, and you plug and chug. In this sense, if each object were sprung up seperately, the lighter one would definitly go higher.

    Basically my problem deals with the logic in having both objects on the same spring, and how that affects the calculations/result of energy transfer.

    Thanks
     
  2. jcsd
  3. Jun 13, 2007 #2

    Doc Al

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    You are over-complicating things. Think about this: Imagine you had these two objects in your hand and you tossed them into the air. What would happen?
     
  4. Jun 13, 2007 #3
    They would reach the same height.

    Id like to understand why though, in terms of energy transfer.
     
    Last edited: Jun 13, 2007
  5. Jun 13, 2007 #4
    you could think about it as an elastic explosion
     
  6. Jun 13, 2007 #5

    Dick

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    In terms of energy transfer the initial velocity of each mass is the same. Call it v. Then the initial kinetic energy (1/2)*m*v^2 is equal to the final potential energy m*g*h. If you solve for h you should notice that the mass cancels.
     
  7. Jun 14, 2007 #6
    why is V the same? because they can be treated as one body?
     
  8. Jun 14, 2007 #7

    Dick

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    It's because the objects will fly off of whatever platform is accelerating them upward as soon as the acceleration of the platform becomes greater than g in a downward direction. So the criterion for flying off is independent of mass.
     
  9. Jul 6, 2010 #8
    "It's because the objects will fly off of whatever platform is accelerating them upward as soon as the acceleration of the platform becomes greater than g in a downward direction. So the criterion for flying off is independent of mass. "

    I am having a similar problem as the OP. How do you prove that the objects leave the spring surface at the same velocity?

    The analogy with throwing two objects into the air only works if one assumes that the objects leave the person's hands at the same speed (which would mean a different throwing forces due to the different weights)
     
  10. Jul 6, 2010 #9

    Dick

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    At the point of flying off the platform the platform is no longer exerting a force on the objects, but the objects and the platform are travelling with the same velocity and acceleration. Since gravity is now the only force acting on the objects their acceleration is g downward. Hence the acceleration of the platform is also g downward. That's when they 'fly off'. Nothing to do with the mass.
     
  11. Jul 6, 2010 #10
    I Just re-read the question; I didn't realize they were on the spring on the same time

    On first read I got the impression that the objects were tested on the spring separately hence my confusion
     
  12. Jul 6, 2010 #11

    Dick

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    You actually raise a good point. I don't think the question is actually all that clear that they are on the spring at the same time. I'd rewrite it to make that explicit. If they aren't, then you can't really answer the question. That's why I read it as "on the spring at the same time".
     
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