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Homework Help: Inclined Plane Problem with change of COM

  1. Oct 14, 2012 #1
    1. The problem statement, all variables and given/known data

    A block of mass 10kg moves along a surface inclined 30° to the horizontal. The centre of
    gravity of the block is elevated by 3m and the kinetic energy of the block decreases by 50J.
    The block is acted upon by a constant force Rparallel to the incline and by the force of
    gravity. Assume frictionless surfaces and let g= 9.81 m/s**2.
    Determine the magnitude and direction of the constant force R.

    2. Relevant equations

    ΔK = m*(Vb^2 - Va^2)*(1/2) , ΔU=mgz

    3. The attempt at a solution
    Since i am unable to express mathëmatically the ""The centre of
    gravity of the block is elevated by 3m and the kinetic energy of the block decreases by 50J""
    i will try to ignore it for now.

    The force R can be either applied to the left or the right,we don't know yet.I will assume that it is towards the right,if not then the sign will be minus in my final calculations.

    -- ΣFy = 0 , N=Wy=mgcos30°
    ( the COM being the origin and x axis parallel to the inclined surface)

    -- ΣFx = Wx - R

    i) How do we even change the COM of a block ?
    ii)Why is the kinetic energy lost ( 50J)not equal to the potential energy gained ?
    In other words 50 <> m*g*( (h+3)-h) which makes sense mathematically but since energy is conserved where does it go ?

    I am really confused guys !
  2. jcsd
  3. Oct 14, 2012 #2

    Check this (Work-energy theorem).

    Also, at the beginning the task says that the force R is parallel to the incline.
  4. Oct 14, 2012 #3
    Thank you for your reply.

    Since it says that we are only interested in the change in PE then PE = 3mg right ?

    so the change in energy is 50 J therefore 50 - 3mg = constant ?

    What about kinetic energy ?
  5. Oct 14, 2012 #4
    Your kinetic energy decreases by 50 J. Your potential energy increases by almost 300 J (3*10*9,81).

    That means you have wasted only 50 J of kinetic energy to lift the block 3 meters up, but to do that you need ~300 J. That means some force is acting up the way.

    Check the formula here.

    What is the initial kinetic and potential energy?
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