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Would this be a valid equation to calculate kinetic energy created from a drop from a height: $$E_{kinetic} ~=~ v_{vertical}tmg$$ Velocity multiplied by time gives distance. Distance multiplied by gravitational force acting on it provides kinetic energy. Would this equation be valid? |
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No. You are correct that the kinetic energy is equal to the change in the potential energy, $mgh$, where $h$ is the distance fallen, but because the object is accelerating $h$ is not simply velocity times time. If the object starts at rest then (ignoring air resistance): $$h = \frac{1}{2} g t^2 $$ so substituting this into $mgh$ gives: $$E_{kinetic} = \frac{1}{2} m g^2 t^2 $$ Note that $gt$ is just the velocity at time $t$, so this expression is the same as: $$E_{kinetic} = \frac{1}{2} m v^2 $$ which may look familiar :-) Note however that the velocity $v$ is a function of time. |
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For a body of mass m in a uniform specific-force field $g$: $$E_{kinetic} = \frac{1}{2} m g^2 t^2 $$ For two mutually interacting bodies of masses M and m: $$E_{kinetic} = \frac{G(M+m)}{r_1}-\frac{G(M+m)}{r_0} $$ where $r_0$ and $r_1$ are the initial and final separations of the two bodies. Since uniform gravitational fields cannot exist in nature, the first equation is not applicable to this problem. It is a commonly used approximation that fails miserably unless the initial seperation is large, and the change in separation is small. |
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