A ball is released from a height of 12 m and weighs 37 N. What is the magnitude | Δ U | of the total change in potential energy when the ball falls to the ground?

Answer 1

If a ball weighs 37 N thane that is its mass time "g" [Since F = ma, and since the acceleration due to gravity is "g", then the Force due to gravity (the object's weight) is mg]

Potential Energy = mass times "g" time height or

PE = mgh

mg = 37 N

h =12 m

So when the object is at a height of 12 meters, it has a PE = (37)(12) = 444 joules. We need to only use 2 significant figures though because 2 are given, so the answer would be 440 joules.

When the ball reaches the ground, the height will be zero so the potential energy on the ground is zero. That means the magnitude of the change in potential energy is 440 joules