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Conservation of energy and change in energy

  1. Jul 1, 2012 #1
    Suppose i have a ball of mass m and if i left the ball at a particular height 'h', then i have done some work on it or by the law of conservation of energy i have transferred some energy to the ball. Now suppose i release the ball isn't the energy that i have transferred changing because it moves with a particular acceleration and hence a changing velocity and also the potential energy keeps on decreasing as the height decreases, then doesn't the energy necessarily change?

    So how is energy constant. The ball falls down and is then at rest and hence has no potential energy or kinetic energy (mgh=0; 1/2mv^2=0), hence the potential energy given to the ball mgh now becomes 0 (mgh-->0). Thus energy has been destroyed isn't it??
     
  2. jcsd
  3. Jul 1, 2012 #2
    The velocity increases thus decrease in potential energy is balanced by increase in kinetic energy
     
  4. Jul 1, 2012 #3
    But what about the bottom most position? The velocity is zero and also the height is zero (object at rest). Where did the mgh + 1/2mv^2 go?
     
  5. Jul 1, 2012 #4
    There must be a force to "brake" the object to a halt. Have you seen an object just stop when it reaches the ground. Try dropping a glass.
     
  6. Jul 1, 2012 #5

    jtbell

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    Staff: Mentor

    If the ball is made of putty or something like that, it smushes to a stop when it hits the ground. The collision with the ground is completely inelastic, and the kinetic energy of the ball is converted to thermal energy ("heat") and a bit of sound energy.

    If the ball and the ground are both, say, steel, the collision is elastic and the ball bounces back up with almost as much KE just after the bounce as it had just before the bounce. In practice no collision is completely elastic, so you get a little bit of energy loss to "heat".
     
  7. Jul 1, 2012 #6
    Thanks...^
     
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