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A mass on an incline

  1. Jan 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Attached

    2. Relevant equations
    kinetic energy = (1/2) m v^2
    potential energy = mgh

    3. The attempt at a solution
    Did I do this correctly,

    At the top, kinetic energy is 0 since it starts at rest. At the bottom we choose the potential to be zero
    So using conservation,
    [tex] mgz= \frac{1}{2}m\vec{v_f}^2[/tex]
    Then, substitute [tex] z=s(t)sin\theta[/tex],
    And [tex] \vec{v_f} = \vec{a}t [/tex]
    where [tex] \vec{a} = \frac{\vec{F}}{m} = \frac{mgsin\theta}{m} = gsin\theta [/tex]

    To get [tex] mgs(t)sin\theta = \frac{1}{2}m(gtsin\theta)^2 [/tex]
    solve for [tex] s(t) = \frac{gt^2sin\theta}{2} [/tex]
     

    Attached Files:

  2. jcsd
  3. Jan 25, 2016 #2

    cnh1995

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    Homework Helper

    Looks correct to me! You can verify it by putting θ=90° and θ=0°. When θ=90°, it's a free fall, so, s(t)=½gt2, the 2nd kinematical equation with u=0. With θ=0, the body will not move at all, so, s=0.
     
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