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Comparing two energy systems - two boxes being pushed

  1. Oct 17, 2015 #1
    1. The problem statement, all variables and given/known data
    Starting from rest, two identical boxes are pushed through the same distance. Box A experiences a force F, while box B experience a force 2F. What is true about their final speeds?


    - The final speed of box A is twice that of box B.
    - The final speed of box A is equal to that of box B.
    - Something else.
    - The final speed of box A is half that of box B.

    I also got a hint: "Choose the box as your system, what does the energy principle tell you about the change in kinetic energy? What can you then conclude about the final speeds?"

    2. Relevant equations
    The chapter this is relevant to is about energy, but I'm not entirely sure if I need an energy equation because I think this is more concept-focused. The equation I tried to use was F=mv.

    3. The attempt at a solution
    I tried thinking about F=mvA/t and then 2F=mvB/t. I solved for v in both of these and got vA=Ft/m and vB=2Ft/m. This seems to indicate that one velocity is twice the other. However, I tried that answer and it was wrong. I only have one more try so I want to be fully sure that I understand before I just start guessing.
    Right now, I'm torn between saying that the velocities are equal, and that the answer is something else entirely.
    Based on the hint above, I'm inclined to think that the two systems have equal total energy, but I'm not sure how this relates to their final velocity.

    Thanks in advance for any help!
     
  2. jcsd
  3. Oct 17, 2015 #2

    gneill

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    Staff: Mentor

    Logically, if the forces applied are different then you should expect different results.

    Looking at the hint, consider the work being done in each case. What equation applies to work done here?
     
  4. Oct 17, 2015 #3
    Well, W=F*d. Since the distance is equal for both boxes, but the force is doubled for the second box, the work must also be doubled. But I guess I'm not understanding what this has to do with their velocities. Can you point me in the right direction?
     
  5. Oct 17, 2015 #4

    gneill

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    Staff: Mentor

    What form does the work that is done take? Hint: The unit of work is energy.
     
  6. Oct 17, 2015 #5
    Work is done in joules. Joules can be written as kg * m2 /s2. So I could separate that into N*s * m/s or momentum times velocity, right? That gets me closer to the velocity I think.
     
  7. Oct 17, 2015 #6

    gneill

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    Staff: Mentor

    Okay, it's good to think about looking at the units. That can often help when you're looking for relationships between values in a problem. It doesn't give you the value of any proportionality constants that might be involved though.

    If you take the component units of the Joule you could also split them up as: ##kg \frac{m^2}{s^2}## which looks a lot like mass times velocity squared, or ##M~V^2## :wink:

    In this case the problem author wants you to practice with the work-energy theorem. Work done on a system ends up as energy in a certain form. What form is that?
     
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