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Two masses on two inclined planes -- What is magnitude of velocity

  1. Apr 14, 2015 #1
    1. The problem statement, all variables and given/known data
    Two boxes of equal mass m are connected by a light string over a massless pulley, and rest on surfaces of inclination θ and φ, θ>φ. The boxes are released from rest. The coefficient of kinetic friction between the boxes and the surfaces is μ. Determine the magnitude of the velocity of the boxes when they have moved distance s.

    2. Relevant equations
    W=(1/2)mv^2

    3. The attempt at a solution
    Well, this is rather simple question when using the work energy equations, setting the total force on each of the masses in x direction times the distance equal to 1/2 mv^2. I have gotten the right answer this way.
    But I've realized that I could probably solve it using another method, which was to use integration by changing the acceleration I get into v*dv/ds.

    IMG_0265.JPG

    Sorry for the bad quality picture and I don't know how to turn it around.
    But basically, I get an equation v=√[((g)(sinθ-sinφ)-μ(cosθ+cosφ))s] but when I plug it in a sample question, the velocity comes out to be a bit off.
    Any help or criticism will be appreciated.
     
  2. jcsd
  3. Apr 14, 2015 #2

    RaulTheUCSCSlug

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    What you can do is besides looking at it through a conservation of energy point of view, you could look at it using your kinematics equations. You would just need to adjust the acceleration accordingly. But your best bet seems to be using conservation of energy which is what you are using. Your picture is to blurry for me to read, but once you calculate the work done by the friction over the total span of the inclined plane, then you can subtract that from the total energy, and use the remainder to figure out the kinetic energy at the bottom since the total potential energy will be equal to the kinetic energy and work done by the friction. (P_total= K_final+W_friction)
     
  4. Apr 14, 2015 #3

    RaulTheUCSCSlug

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    I'm not sure if this is what you did, but it could be one of the approaches you could use.
     
  5. Apr 14, 2015 #4
    Hi! I am very sorry for the picture quality. Yes. To me, that was the intuitive way. But the work I have shown in the picture was finding the tension for each mass, equating them and solving for acceleration. Then, I converted that acceleration into v*(dv/ds ) and then use integration in order to find the equation for v in terms of s.
    Do you know if converting a into v*(dv/ds) is allowed? Because I think that is the only place where I could have possibly gone wrong.
     
  6. Apr 15, 2015 #5

    haruspex

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    I don't see how using energy balance would lead to a different equation from the one you obtained with a = v dv/ds.
     
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