Monday 14 September 2015

A body of the mass m is dropped from a height h. At half way to the ground, what is the total energy of a body?

Hello!


I suppose that the process takes place near Earth's surface.


The full energy of a body is the sum of its kinetic and potential energy. The kinetic energy is equal to


`(m V^2)/2,` where `V` is the speed.


The potential energy may be computed with respect to any fixed level. If this level is as usual a ground level, then this energy is equal to


`mgh,` where `g` is the gravity acceleration.



To...

Hello!


I suppose that the process takes place near Earth's surface.


The full energy of a body is the sum of its kinetic and potential energy. The kinetic energy is equal to


`(m V^2)/2,` where `V` is the speed.


The potential energy may be computed with respect to any fixed level. If this level is as usual a ground level, then this energy is equal to


`mgh,` where `g` is the gravity acceleration.



To find the full energy, we can find the speed at the half way and use it. But it is simpler to use energy conservation law and note that the full energy will be the same at any moment.


The simplest moment to consider is the initial position. The speed is zero yet and the full energy is only the potential energy m*g*h. This is the answer:)

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