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Let's say we have a ball at a height $h$ at rest. The total energy is $mgh$ which is the ball's potential energy. If we drop the ball, given that no external forces act on it, when the ball reaches the ground its kinetic energy becomes:

$$ T= \tfrac12 mv^2 = mgh $$

My understanding of work is that work is how much energy an object receives. In other words, to do work on an object means to give that object energy.

We can use the work energy principle and see that the net work is equal to the change in kinetic energy which in this case is just the final kinetic energy.

This is what I don't understand. If work is how much energy the object receives and in a closed system like this one the total amount of energy is constant. Shouldn't the net work be $0$?

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This is what I don't understand. If work is how much energy the object receives and in a closed system like this one the total amount of energy is constant. Shouldn't the net work be $0$?

The net work done on the ball-earth system is zero. This is consistent with both conservation of mechanical energy and the work energy theorem which states that the net work done on an object or system equals its change in kinetic energy.

For the work energy theorem there is no change in kinetic energy of the center of mass of the ball-earth system since there are no external forces performing net work on the ball-earth system. For conservation of mechanical energy the decrease in gravitational potential energy of the ball-earth system equals the increase in kinetic energy of the ball component of the system.

On the other hand, applying the work energy theorem to the ball alone, the force of gravity (and any external air resistance) are external forces acting on the ball. For zero air resistance, the net force is that of gravity, $mg$, and the net work done on the ball is $mgh$, for a change in kinetic energy of the ball alone of $\frac{1}{2}mv^2$. The positive work of gravity transfers energy to the ball alone at the expense of the gravitational potential energy of the ball-earth system.

Hope this helps.

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Your statement If we drop the ball, given that no external forces act on it . . . . . implies that the system that you are considering is the ball and the Earth.
Although there are no external forces acting there are **internal forces, viz the gravitational attraction of the Earth on the ball and the gravitational attraction of the ball on the Earth - Newton's third law.
One of the forces produces an increase in the speed of the ball and hence its kinetic energy and the other force produces a minute increase in the speed of the Earth and hence its kinetic energy.

Your first paragraph is all to do with the ball and Earth system which can store gravitational potential energy and the forces involved are all internal forces which can change the kinetic energy of the constituents of the system.
A system consisting of the ball alone cannot store gravitational potential energy.

If you consider the system as the ball alone then the external force acting on the ball is the gravitational attraction of the Earth which produces an increase in the speed of the ball and hence its kinetic energy.

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  • $\begingroup$ If we are considering the earth-ball system, does the work-energy theorem indicate that the amount of energy added to that system is whatever the final kinetic energy is? The ball's kinetic energy increases by the final kinetic energy and the earth's kinetic energy increases by the ball's final kinetic energy. $\endgroup$ Oct 21, 2022 at 7:41
  • $\begingroup$ No! Momentum is conserved so the speed of the Earth is much less than that of the ball. Since the kinetic energy depends on the mass and the speed squared the kinetic energy the ball gains is much, much bigger than the gain in kinetic energy of the Earth. $\endgroup$
    – Farcher
    Oct 21, 2022 at 7:49
  • $\begingroup$ Wouldn't the kinetic energy gain be the same in the same way that the force of attraction is the same? Wouldn't it just be that it would have a minuscule effect on its velocity because we are dividing that kinetic energy by the mass? Am I missing something? $\endgroup$ Oct 21, 2022 at 8:08
  • $\begingroup$ I must have misunderstood your previous comment as either way you get the same value for the gain in kinetic energy of the ball given that the Earth has a mass which is so much greater than that of the ball and so the gain in kinetic energy of the Earth is so much smaller. $\endgroup$
    – Farcher
    Oct 21, 2022 at 9:24

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