Question about gravitational potential energy

  1. I am wondering how this all works and am hoping to get some guidance. If a 1000 lb weight freefalls vertically from 660 ft the energy is 894,289 joules. If the fall is down a 30 degree slope that is 1300 ft long, then it takes 17 seconds and the potential energy can be said to be 53 KW. If, and I really mean if, I did that math right then my question has to do with converting that 53KW into kinetic energy. If the 1000 lb. weight is brought to a stop by a mechanical device, rather than impacting the earth, does the 53KW transfer to the mechanical device? If so does all or most of it transfer? Secondly, if the mechanical device takes longer than the 17 seconds of freefall to bring the weight to a stop is the 53KW of potential energy still available? My guess is that no matter how long the mechanical device takes to bring the weight to a stop the potential energy, of 53KW, would remain the same.
     
  2. jcsd
  3. Be careful with terminology: "potential energy can be said to be 53 KW" makes no sense. You mean, the average power is 53 kW. Energy is measured in joules. Energy is conserved and power is not. If you take the same energy and spread it over a longer time, the power will decrease. So your last sentence is totally wrong.

    Depending on how efficient the mechanism is, almost all of the 894289 joules can be recovered by the machine. Of course there's always some inefficiency, but there's no hard limit on efficiency like there is in a heat engine.
     
  4. You are talking about a potential energy of 53 KW, energy is measured in Joules, not in Joules per second, which is W.
    And if I understand your question correct, yes most of the energy gets converted into mechanical energy but there will also be a not neglible amount of heat after the collision.

    Does that answer your question a little bit?
     
  5. PhanthomJay

    PhanthomJay 6,272
    Science Advisor
    Homework Helper
    Gold Member

    If a problem is stated in pounds and feet, please stick with those units when calculating potential energy in foot-pounds.
     
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