Sairah's work is correct as far as it goes. The potential energy of the ball relative to the bottom of the hill is 58,800 Joules.
To address the second part of the question: In order to get ahold of that energy, the ball must be returned to the bottom of the hill. The most efficient way would be to drop it, so that it wouldn't have to scrape along the grass on the way down. But that can only work if there's a sheer cliff on one side of the hill. Otherwise, you just have to roll it down, and accept the fact that it loses some of its energy to friction on the way.
However the ball gets to the bottom, the energy it has left shows up in the form of kinetic energy, and 58,800 joules is a lot of kinetic energy. If somehow the ball could arrive at the bottom with ALL the energy it had at the top, it would be moving at something like 540 miles per hour !
We know that -
P . E = m ∗ g ∗ h
Where,
m = ma ss
g = a cce l er a t i o n d u e t o g r a v i t y
h = h e i g h t
First we convert height into meters.
1 km = 1000 meters
3 km = 1000 * 3 meters = 3000 meters
So, putting the values in the above formula, and by taking 'g' = 9.8 m/s², we get-
P . E . = 2 ∗ 3000 ∗ 9.8
P . E . = 58800 J o u l es
P . E . = 58.8 k J
To find the potential energy of a 2 kg ball at a height of 3 km, we use the formula PE = m g h . Plugging in the values, we find that the potential energy is 58800 Joules. This energy indicates the capacity of the ball to do work if it were to fall to the ground.
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