Energy [Thx for help!]

2007-04-02 9:29 am
A soft rubber ball is dropped from a height on the ground. It then rebounds to 1/16 of its original height. Which of the following statements about the rubber ball is/are correct if air resistance is negligible?

回答 (2)

2007-04-04 12:50 am
✔ 最佳答案
In the example, you have to apply the law of conversation of energy.
Remember when there is no external force (except gravity),
PE + KE = constant or PE gain = KE loss

Let its original height be h, then the potential energy is mgh.
During the free falling of a ball, the total energy of the ball (= PE + KE) is a constant. The PE of the ball is converted to KE until it hits the ground where all PE is converted to KE.

After the rebounds, as the ball goes up, KE is converted to PE until it reaches its highest point. As the height is h/16.

So,
when the ball is at its original height before rebound: PE = mgh, KE = 0
when the ball is just before collision: PE = 0, KE = mgh
when the ball is at its highest point after rebound: PE = mgh/16, KE = 0
when the ball is just after collision: PE = 0, KE = mgh/16

And you can find the speed by the formula, 1/2 m v^2 = KE

So, (1) and (2) are false, (3) is true.
2007-04-02 9:44 am
Kinetic energy of rigid bodies
In classical mechanics, the kinetic energy of a "point object" (a body so small that its size can be ignored), or a rigid non rotating body, is given by the equation
Ek = 0.5mv*2 where m is the mass and v is the speed of the body.

For example - one would calculate the kinetic energy of an 80 kg mass traveling at 18 meters per second (40 mph) as . 0.5(80)(18*2) = 12960 joules

Note that the kinetic energy increases with the square of the speed. This means, for example, that if you are traveling twice as fast, you will have four times as much kinetic energy. As a result of this, a car traveling twice as fast requires four times as much distance to stop (See mechanical work).

Thus, the kinetic energy can be calculated using the formula:

Ek = 0.5 mv*2

For the translational kinetic energy of a body with constant mass m, whose center of mass is moving in a straight line with speed v, as seen above is equal to

Et = 0.5mv*2

where:

m is mass of the body
v is speed of the center of mass of the body.

Thus kinetic energy is a relative measure and no object can be said to have a unique kinetic energy. A rocket engine could be seen to transfer its energy to the rocket ship or to the exhaust stream depending upon the chosen frame of reference. But the total energy of the system, i.e. kinetic energy, fuel chemical energy, heat energy etc, will be conserved regardless of the choice of measurement frame.

The kinetic energy of an object is related to its momentum by the equation:

Ek = (p*2)/(2m)


收錄日期: 2021-04-12 16:28:40
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070402000051KK00308

檢視 Wayback Machine 備份