Say we had an object at height $h$ in a gravitational field, and of mass $m$, its GPE would equal $mgh$, now say we dropped it and it fell a distance of $h$, the work done on the object would be equal to $mgh$, as $W = Fx$, and $F = mg$ and $x = h$, so why is the objects energy not now $2mgh$, but rather $0$?
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$\begingroup$ After the object drops by h its height is now zero, so with your choice of gravitational potential now the potential energy is zero. Is that your question? $\endgroup$– FlatterMannCommented May 3, 2023 at 16:06
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$\begingroup$ Not really, I have learnt that when $F$ and $h$ are in the same direction, the object "gains" energy, as if we had an object flat on ground and apply force, it gains kinetic energy, here I am confused, why does the object not "gain" energy? $\endgroup$– Nav BhatthalCommented May 3, 2023 at 16:10
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1$\begingroup$ The object loses gravitational potential energy and it gains kinetic energy. The total energy in the system stays the same. $\endgroup$– FlatterMannCommented May 3, 2023 at 16:12
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$\begingroup$ Alright, so what when we have an object of KE = $0$, then a force is applied, its not got KE which is non zero. What is the conservation of energy here? Where has this extra energy come from? $\endgroup$– Nav BhatthalCommented May 3, 2023 at 16:13
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$\begingroup$ The conservation of energy in this case is mgh=mv^2/2. The left hand side is the loss in potential energy and the right hand side is the final kinetic energy. From that you can calculate that an object (of any mass) has gained a velocity of v=sqrt(2gh) after falling from a height h. For a drop of 1m (like a typical table), the velocity at floor level will be roughly v=sqrt(2*10m/s^2*1m)=sqrt(20m^2/s^2) or approx. 4.5m/s. $\endgroup$– FlatterMannCommented May 3, 2023 at 16:21
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2 Answers
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so why is the objects energy not now $2mgh$, but rather $0$?
The objects energy is not zero. After release and just prior to impact its kinetic energy is $\frac{1}{2}mv^2$ where $\frac{1}{2}mv^2=mgh$ (ignoring any air resistance)
The potential energy of $mgh$ prior to release, which is possessed by the earth plus object, is converted to kinetic energy of the object of $\frac{1}{2}mv^2$ just prior to impact by the positive work done by gravity .
Hope this helps.
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You state that we had an object at height $h$ in a gravitational field, and of mass $m$, its GPE would equal $mgh$.
One can then interpret that you have chosen the zero of GPE when $h=0$.
So where does the $mgh$ come from?
The GPE is the work done by an external force in changing the configuration of the Earth/mass system from its initial state, $\rm height =0$, to its final state, $\rm height = h$, and in this case the work done is equal to $mgh$.
However, there is another way that GPE is defined.
The GPE is $\rm\color{red} {minus}$ the work done by an internal gravitational force, $mg$, in changing the configuration of the Earth/object system from its initial state $\rm height =0$ to its final state $\rm height = h$ and in this case the work done is again equal to $mgh$.
You wrote, . . . . . and it fell a distance of $h$, the work done on the object would be equal to $mgh$, as $W=Fx$, and $F=mg$ and $x=h$ . . . . .
By implication the work is done by the internal (to the Earth/object system) gravitational force.
So rather than moving upwards when the object moves downwards the change in GPE is $\rm\color{red} {minus}$ the work done by the gravitational force, $\color{red}-mgh$.
Just before the object started to fall the GPE was $mgh$.
The object then fell to the ground with a change of GPE of $-mgh$, thus the object lost gpe and the final gpe was $mgh - mgh = 0$.
And where did the gpe go?
It became the change in the kinetic energy of the object.
