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Elastic potential energy completely lost to projectile?

  1. Nov 27, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose you have a mass/spring system (with spring constant k) buried underground such that its relaxed (equilibrium) position is level with the ground. Then suppose someone takes a small ball of mass m and compresses the spring a distance xo below the mass/spring's equilibrium position. If someone releases the spring, what is the maximum height of the ball relative to the ground? Assume there is no energy lost to resistive forces and that the ground is level.

    2. Relevant equations
    The elastic potential energy of a spring, kinetic energy equation, gravitational potential energy equation, and the conservation of mechanical energy.


    3. The attempt at a solution
    I solved this by defining the gravitational potential energy of the ball to be zero where the spring is at rest, that is at a distance xo below ground level. This means the total energy is .5kxo2 + mgh = .5kxo2. When the spring is released, this energy is converted into kinetic energy then converted back into potential energy. The ball will have gravitational potential energy at this point too. This means the ball's maximum height will be reached when the sum of the elastic potential energy of the spring and the gravitational potential energy of the ball is equal to .5xo2.

    When this happens, I'll say the ball is at a distance hmax above the point of zero gravitational potential energy and the spring is at a distance xf above its equilibrium position. Thus, .5kxo2 = .5kxf2 + mghmax. Since hmax is equal to the ball's height above the ground plus the ball's initial depth below the ground, hmax = xo + xf.

    This means .5k(xf2 - xo2 ) + mg(xo + xf) = 5k(xf2 - xo2) + mgxo + mgxf = 5kxf2 + mgxf + (mgd - .5kd) = 0 where d = xo since it is a given quanitity (just to make the equation clearer). Solving the quadratic equation then gives the maximum height of the ball.


    However, the correct answer goes like this. It starts off like mine, but it states that the spring gives all of its original elastic potential energy to the ball. Thus, .5xo2 = mghmax --> hmax = xo2/(2mg). Then, because h is defined as distance above the point xo below the ground, the maximum height of the ball relative to the ground is hmax - xo.

    I'm having trouble swallowing this because doesn't the conservation of energy refer to systems, and isn't the system in this case the mass/spring-ball-earth system? So if the correct answer is that the spring gives all of its stored energy to the ball, then what happens to the spring? Does it convert all of its potential energy to kinetic energy, transfer it to the ball, then come to a sudden stop all at its equilibrium position (I do realize the conversion of potential to kinetic energy is gradual)? If so, will someone please explain why?
     
  2. jcsd
  3. Nov 27, 2009 #2

    mgb_phys

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    Homework Helper

    You ignore the speed of the spring when you assume all the energy goes into the ball - it's just a simplification to make the question solvable by the level of maths you are using.
    You also ignore bigger effects like air resistance, so details like the heating of the spring don't matter.

    Note you are also free to pick potential energy as zero at any point you want, it's probably easier to measure it from the surface. So at the surface - with the full extent of the spring, all the spring energy has been transferred into kinetic energy of the ball, which will become potential energy of the ball at the top of it's flight
     
  4. Nov 27, 2009 #3
    Okay, I should be gald of the simplification then, even though I got the problem wrong >_<. I also did originally solve this by defining potential energy relative to the surface, but I wrote it out the way I did to save on writing. Thanks for explaining!
     
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