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I was solving a true or false question regarding total mechanical energy and the following was the problem.

It is possible for a moving object to have negative total mechanical energy.

This is my claim.

Total mechanical energy is defined as

$$M=K+U$$

where $K=\frac{1}{2}mv^2$ and $U=mgh$.

No matter which direction the object is moving $K$ must be positive due to the $v^2$. (I am assuming that mass is also never negative.)

Potential energy is also positive because $g=9.8$ and height is a distance, which is also non-negative.

And yet, the problems says that the answer is true, as in the total mechanical energy is allowed to be negative.

What am I not seeing here ?

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  • $\begingroup$ Are you sure height is necessarily a non-negative distance? $\endgroup$
    – Kyle Kanos
    Commented Dec 21, 2013 at 3:28
  • $\begingroup$ What ? Are you suggesting that height can be negative? as in, the object still moving after it hits the ground and go inside ? $\endgroup$
    – hyg17
    Commented Dec 21, 2013 at 3:38
  • $\begingroup$ Relative to your head, how far away are your feet? $\endgroup$
    – Kyle Kanos
    Commented Dec 21, 2013 at 3:39
  • $\begingroup$ I see, so relativity is the issue. That makes sense. So, considering the top of a building 1km below a plane, an object that is falling on the ground can have negative mechanical energy while it's still moving. Interesting. $\endgroup$
    – hyg17
    Commented Dec 21, 2013 at 3:54

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As Kyle implies in the comments, mechanical energy is generally defined only up to a constant. Therefore, if you choose your constant as a large, negative number, you could have a total energy that is negative even with a very fast moving particle. Likewise, if you choose your potential energy to equal zero at, say, the top of a cliff, then anything you throw off the cliff will have negative potential energy once it falls below your feet.

This is not strictly a duplicate, but it is probably worthwhile to link to this answer I posted a while back.

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    $\begingroup$ I like using $mgy$ instead of $mgh$. Makes the arbitrary nature of origin more apparent. $\endgroup$
    – BMS
    Commented Dec 21, 2013 at 4:51

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