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What owns gravitational potential energy mgh



1. Suppose you lift a ball of mass m a height h from the floor. You do
this lifting at constant speed (except for a short time at the beginning
and end). The ball starts at rest and ends at rest. You can ignore the
effects of the air.
2. As a result of this process, the gravitational potential energy U has
increased by an amount mgh (assuming h is much less than the radius of the
earth so that you can accurately treat g as constant).
3. Since potential energy is a relative quantity, you let U = 0 when the
ball was on the floor.
4. Thus, U = mgh when the ball is at that height h.
5. This U does NOT belong to the ball. It is a property of the ball-earth
system. In fact, it is stored in the gravitational field of that
system. (This total field was changed when you changed the ball's
position, thus changing the total energy stored in the field.)

If you still believe the ball owns U = mgh:
1. We agree that if you drop the ball from rest at that height h, it will
have kinetic energy K = mgh when it hits the floor (ignoring the effects
of the air). (It has a speed square root of (2gh).)
2. Consider a system composed only of the ball. It has zero initial
kinetic energy. IF you give it initial potential energy mgh, its total
mechanical energy initially is mgh. As it falls, a free body diagram shows
the only force acting on it is its weight mg. This force does work mgh on
it as it falls, giving the ball a total energy of mgh + mgh = 2mgh as it
hits the floor. Since the potential energy is zero at the floor, 2mgh must
be the kinetic energy K of the ball as it hits.
3. As you see, "K = 2mgh" is twice as large as the actual value, K = mgh, so
the belief that the ball has initial potential energy mgh is wrong. The
initial potential energy of the ball is zero.
4. Consider a second system composed of the ball and the earth. In this
system, the initial potential energy is mgh but there is no net external
force doing work on that system as the ball falls. Thus the final kinetic
energy is the correct value K = mgh.

If you worry about where the work went that you did in lifting the
ball:
1. Draw a free-body diagram of the ball as you lift the ball. The
gravitational field exerts a constant downward force of mg on the ball.
The second force is your upward lifting force on the ball, which equals mg
during most of the constant-speed lift. (Your upward force is greater than
mg at the beginning of the lift and less than mg at the end by amounts that
give an average force of mg.) Therefore, the total work done by your
upward force is mgh.
2. In contrast, since the downward gravitational-field force and the
upward displacement are opposite vectors (antiparallel), the work done by
the gravitational-field force is -mgh.
2. The net work done is mgh + (-mgh) = 0. Every bit of work you do is fed
into the gravitational field, increasing its potential energy by mgh
during the lift.
I think this is right. Tom Sandin