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Work Energy Theorem and speed

  1. Oct 31, 2007 #1
    1. The problem statement, all variables and given/known data

    You throw a rock of weight 20.4 N vertically into the air from ground level. You observe that when it is a height 14.9 m above the ground, it is traveling at a speed of 26.0 m/s upward.

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


    2. Relevant equations
    Wtotal = delta K

    K = mv^2

    v^2 = vi^2 + 2a(x-xi)

    3. The attempt at a solution


    I believe this can also be done using kinematic equation, since we are given vf,x,a, and we want vi.

    So if this is true, I will attempt both ways :D


    26^2 = vi^2 + 2(-9.8)(14.9-0)
    vi = 31.11 <--------> which is right o:)

    now, for using the work energy theorem.

    we first need the mass.

    using
    w=mg

    20.4 = m ( 9.8)

    m = 2.08

    K1 = 1.2(2.08)(v)^2
    K2 = 1.2(2.08)(26)^2

    K2 = 843.65
    K1 = 0

    ....how do I proceed?
     
  2. jcsd
  3. Oct 31, 2007 #2
    since there is no external force (such as friction or air resistance) there is no external work done on theb all except for gravity. So the total initial energy of the ball is equal to the total final energy of the ball

    what is the initial energy of the ball and the final energy of the ball?
     
  4. Oct 31, 2007 #3
    ummm...

    so Winitial = Wfinal.

    W=Fs

    W = (F)(14.9)...?
     
  5. Oct 31, 2007 #4

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No, it's not "work initial" = "work final", which in itself would not make sense.

    you may either use

    change of kinetic energy = total work done by gravity

    OR

    total final energy = total initial energy

    where total energy = kinetic energy plus gravitational potential energy
     
  6. Oct 31, 2007 #5
    delta k = wg ?

    if so, how do I apply this to my problem, as I have this:

    K1 = 1.2(2.08)(v)^2
    K2 = 1.2(2.08)(26)^2
     
  7. Oct 31, 2007 #6
    ok so u have the final and initial kinetic energies
    now how do you calculate the work done by gravity??
     
  8. Oct 31, 2007 #7
    w done by gravity is:

    w = mgsin(theta)

    w = 2.08 (9.8)(sin 90)
    w = 20.38
     
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