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Two objects connencted - finding KE before one reaches the floor

  1. Jun 3, 2008 #1
    1. The problem statement, all variables and given/known data
    Two boxes are connected to each other as shown. The system is released from rest and the 1.00 kg box falls through a distance of 1.00 m. The surface of the table is frictionless. What is the kinetic energy of box B just before it reaches the floor?

    2. Relevant equations
    I believe you should be using an equation such as Ei=Ef
    kinetic energy - (1/2)mv^2
    and potential energy - mgy


    3. The attempt at a solution
    so Ei = (1/2)mvi^2 + (1/2)mv^2 + mgy +mgy
    = (1/2)(3)(0)^2 + (1/2)(5)(0^2) +(3)(9.8)(0) + (1)(9.8)(1)

    Ef = (1/2)mvi^2 + (1/2)mv^2 + mgy +mgy
    = (1/2)(3)(vi)^2 + (1/2)(5)(vi^2) +(3)(9.8)(0) + (1)(9.8)(.01)

    the velocity would be the same for each since they are connected

    or would I do something like
    = (1/2)(3+1)v^2 + (1/2)(9.8)(1.0)
    solve for velocity...? and then use this velocity to find the KE


    thanks for the help!
     

    Attached Files:

  2. jcsd
  3. Jun 3, 2008 #2
    next time use imageshack for image posting.

    Is there any reference to the height, in the problem? if so - you should consider the potential energy mgh.
     
    Last edited: Jun 3, 2008
  4. Jun 3, 2008 #3
    Well there is a hint that says: The gravitational energy will appear as kinetic energy of both boxes. (I'm not sure what this means?) It also says the answer is suppose to be 2.45 J.
    I'm not getting anywhere near that.
     
  5. Jun 3, 2008 #4
    Please label your masses m1 and m2. You are given that the fallen object's mass = 1kg. It does not match with the equations below.
    See, you put m=3, m=5, then switch to m=3, m=1. It makes no sense. Label your variables.

    By the Work-Energy Theorem, [tex]\frac{1}{2}(m_{1}+m_{2})v^2=m_{2}gy[/tex]
    where [tex]m_{1}[/tex] is the mass of the box on the table, and the [tex]m_{2}[/tex] is what I'm presuming to be a box hanging from a pulley( I don't see a diagram)
    Let y = 1m (the distance it falls)

    Yes.
     
  6. Jun 3, 2008 #5
    Yeah I realized that after because I was trying to do the problem while looking at an example that had (3 and 5 kg boxes) sorry!!

    I solved for velocity and did v= 1.56 m/s

    so K= 1/2mv^2 + mgy
    K = (1/2)(1)(1.56)^2 + (1)(9.8)(0)

    but this does not equal 2.45 J ...what am i missing?
     
  7. Jun 4, 2008 #6

    alphysicist

    User Avatar
    Homework Helper

    Hi crazyog,

    I don't think the velocity you solved for is correct; it looks like you might have used m=3kg and 5kg for the kinetic energy terms, instead of m=1kg and 3 kg.

    If that's not what happened, please post the numbers you used that gave a speed of 1.56 m/s.

    (At this point, your attachment is not approved, so I may be not understanding the problem. However, I did get the answer you said was correct.)
     
  8. Jun 4, 2008 #7
    Yeah I noticed that I wrote the wrong numbers because I was working off of an example and got mixed up, but I don't know how to change that. Sorry!

    I used
    1/2(m1+m2)v^2=(m2gy)
    1/2(1+3)v^2 = (1)(9.8)(1m)
    solved for v^2 and got 1.56 m/s
     
  9. Jun 4, 2008 #8
    where do I go from here?
     
  10. Jun 4, 2008 #9

    alphysicist

    User Avatar
    Homework Helper

    What I was saying in my last post, is that this equation does not give 1.56 m/s. If you use the wrong masses on the left, (with m=3 and 5), then you get 1.56 m/s, or if you don't multiply by the 1/2 you get 1.56. So I think you either used the wrong masses or had a math error.

    Try solving the above equation again, and you'll get a larger answer for the velocity; what do you get?



    Also, you can check your answer. For example, if we try plugging in v=1.56 m/s in the above we get for the left side:

    (1/2)(1+3) 1.56^2 = 4.8672

    which does not equal the right side, so 1.56 is not a solution of the above equation.
     
    Last edited: Jun 4, 2008
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