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This shows conservation of energy because Bill's potential energy is being comerted into kinetic energy of rolling down the hill. No energy is lost.

2007-06-14 13:24:24 · answer #1 · answered by curbionicle 2 · 0 0

When a ball is at h above some reference point, the bottom of a hill of h height for example, it has potential energy PE = mgh; where m is the mass of the ball, g is acceleration due to gravity (g = 9.81 m/sec^2 on Earth's surface), and h is that height.

If the ball starts to roll downhill, from the pull of gravity, it picks up speed (velocity). And kinetic energy is KE = 1/2 mv^2; where m is the same ball mass and v is its velocity at any given moment rolling down hill.

When the ball reaches the bottom of the hill, all that potential energy has been converted into kinetic energy. In other words, all that PE at the top of the hill is now KE at the bottom of the hill. We can write PE = mgh = 1/2 mv^2 = KE; so that v^2 = 2gh or v = sqrt(2gh), which is how fast that ball will be going when it reaches the bottom of the hill of height h.

Thus, all that potential energy was converted into kinetic energy. None of that energy was destroyed; it was conserved, but changed into kinetic energy. And that demonstrates the conservation of energy.

2007-06-14 13:55:31 · answer #2 · answered by oldprof 7 · 0 0

dont know

2007-06-14 13:42:30 · answer #3 · answered by Anonymous · 0 2

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