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A body of mass $15\,kg$ originally at rest $5\,m$ above the ground falls and penetrates $15\,cm$ into soft earth.

  • Determine the loss of potential energy
  • Determine the average resistance of the earth

Please help explain and solve this. I got 735 Joules as the potential energy $E_{pot}=m\cdot g\cdot h$.

As for the other one I am stumped.

Please help.

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closed as too localized by Mark Eichenlaub, Qmechanic Feb 28 '13 at 7:29

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

    
Too localized? On what grounds? Too Localized should be retired. –  Mob Feb 28 '13 at 7:32
    
Generally we like questions that would be of interest to lots of people, and the answer to your homework is only likely to be of interest to you: hence the Too Localized flag. –  John Rennie Feb 28 '13 at 8:00

1 Answer 1

The second part of the question is poorly worded. I assume that by average resistance of the earth the questioner actually means the force the earth applies to the falling object. To work this out remember that the work done by the earth on the object is force times distance.

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Thanks. Is the first correct though? That is 735 Joules? –  Mob Feb 28 '13 at 7:28
    
As for the second part. What distance should I use? The distance moved by the ball in the earth? How would you go about it? Thanks sir. –  Mob Feb 28 '13 at 7:29
    
The forum rules require we don't answer homework questions or confirm whether answers are correct, but consider this. The force on the falling object is $mg$, and if it falls a distance $h$ then the work done on it is force times distance or $mgh$, which is why you get the formula $mgh$ for the energy change. You can apply exactly the same reasoning to the slowing of the object after it touches the surface of the earth. –  John Rennie Feb 28 '13 at 7:35

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