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Conservation of energy stone problem

  1. Oct 2, 2006 #1
    I have this problem for class and I'm kind of stuck on all of them. Any help would be appreciated.

    a stone of weight w is thrown vertically with initial speed v0. air drag force dissipates at a rate of f*y of mechanical energy as the stone travels distance y.
    a) show that the maximum height of the stone is

    h = (v0)^2/(2g(1+f/w))

    b) show that the speed of the stone upon impact is

    v = v0*((w-f)/(w+f))^(1/2)

    thanks in advance
     
    Last edited: Oct 2, 2006
  2. jcsd
  3. Oct 3, 2006 #2

    HallsofIvy

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    That's poorly written. Force does not "dissapate". I think what is meant is that the air drag force causes the energy to be lost at that rate. Since Energy = force* distance, saying the energy dissapates at rate f*y (y is distance) means that the drag force is the constant f.
    The total force on the stone, as it is going up, is F= ma= -mg- f and so the acceleration is a= -g- f/m= -g-fg/w= -g(1+1/w) (w= mg so m= w/g). Integrate that to find the velocity function, integrate the velocity function to find the height function.
    At its highest point, the velocity of the stone is 0. Solve that equation for t and calculate the height of the rock at that t.

    Be careful with this one. On the way down, the resistance force is upward: F= ma= -mg+ f so a= -g(1- f/w). Integrate that to find the velocity function (with v(0)= 0 at the top of the throw) and integrate again to find the height function (with h(0)= (v0)^2/(2g(1+f/w))). Set the height= 0 (hitting the ground) and solve for t. Use that t to find the velocity at impact.

    Or, perhaps simpler, since the resistance force is constant, find the total energy dissapated and subtract that from the initial kinetic energy. Use that lower kinetic energy to find the velocity on impact.
     
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