Question

try this to answer

Answer 1

http://i1134.photobucket.com/albums/m613/kingkos14/IMG_0002.jpg?t=1313691349

Answer 2

For the ball rolling down the hill problem, the total energy in the system is constant. At the bottom of the hill, the ball has 0J potential energy, and 50J kinetic energy. Therefore the total energy in the system is 50J. All other stages of the ball rolling down the hill must add up to 50J.

Likewise for the man jumping from the tower. The total energy in the system is 15000J, and at any point, the potential energy, and kinetic energy will always add up to 15000J.

In fact, this is the case for every unanswered question on the page.

Answer 3

Actually, not so for the one where the weight is held up by a cable, and then breaks. It doesn't ask about kinetic energy, but rather about the work done. It follows the same solution though, the work done in moving an object up or down is equal to the change in potential energy of the object. Since the potential energy changes from 10\(^4\)J to 0J, this difference is the amount of work done.

Answer 4

ow hard to read can u try to solve so i can see what i do< i got a nosebleed in english sory for my bad english

Answer 5

try to answer some so i can manage

Answer 6

Alrighty, take the ball rolling down the hill one. You're told that when the ball is at the bottom of the hill, PE=0, and KE=50J. PE+KE=50J. Because of the principle of conservation of energy, PE+KE will always equal 50J for this scenario.

Next, take the ball part-way up the hill. You're told that PE=25J. Since we know that PE+KE=50J, we can rearrange that to give us \(KE=50J-PE\). In this case:
\[KE=50-25=25J\]

When the ball is at the top of the hill, PE=50J. Again, use: \[KE=50-PE\]\[=50-50\]\[=0J\]