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Work-Energy Theorem

  1. Sep 24, 2006 #1
    Please check to see if my calculations are correct. I am unsure if I used the right values and setup.

    You throw a rock of weight 21.0 N vertically into the air from ground level. You observe that when it is a height 14.8 m above the ground, it is traveling at a speed of 25.7 m/s upward.

    a. Use the work-energy theorem to find its speed just as it left the ground;Take the free fall acceleration to be g = 9.80 m/s^2 .

    m_rock = 21.0 N/ 9.8 m/s^2 = 2.14285 kg

    F*d = 0.5*m*(v_2)^2 - 0.5*m*(v_1)^2
    14.8 m*(-21 N) - 0.5*m*(v_2)^2 = -0.5*m*(v_1)^2

    v_1 = sqrt[(F*d - 0.5*m_rock*(v_2)^2)/(-0.5*m_rock)]
    = sqrt[((14.8m*-21.0N) - 0.5*2.14 kg*(25.7 m/s)^2)/(-0.5*2.14 kg)]
    = 30.8 m/s ??

    b. Use the work-energy theorem to find its maximum height.
    Take the free fall acceleration to be g = 9.80 m/s^2 .

    F*d = 0 - 0.5*m*(v_1)^2
    d = [- 0.5*m*(v_1)^2]/[F] = 0.5*(v_initial)^2/g

    For v_initial ,am I supposed to use the value that I found in Part A?

    d = 0.5*(30.8 m/s)^2/g = 48.4 m ??

    Thanks.
     
  2. jcsd
  3. Sep 24, 2006 #2

    radou

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    Homework Helper

    Yes, looks perfectly OK.
     
  4. Sep 25, 2006 #3
    So it is correct that I used -21.0 N for the force and 30.8 m/s for the initial velocity?

    Thanks again.
     
  5. Sep 25, 2006 #4

    radou

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    Yes, it is correct, since the work done by the weight has the opposite direction to the velocity.
     
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