Dynamatic 問題一問

用戶19771

2011-02-05 4:42 am

A light spring is fixed to the bottom of a vertical tube.A ball is released from rest at a height h above the upper end of the spring. After rebounding serval times the ball eventually comes to rest and stays on the top of the spring. Assuem that all contact surface friction=0:

which of the statement(s) is/are true?

(1) The compression of the spring is proportional to the mass of the ball
(2) The compression of the spring is independent of the height h
(3) The GPE lost by the ball is equal to the elastic potential energy stored in the spring.
A (1) only
B (3) only
C (1) & (2) only
D(2) & (3) only

答案係C
但係我唔明點解,
各位PHYSICS高手可唔可以解釋下,
小弟感激萬分,謝~

更新1:

The loss of gravitational potential energy of the ball = mg(h + x) The gain in elastic potential energy of the spring = (1/2)kx^2 multiply both sides by (x/2) hence, (1/2)kx^2 = (1/2)mgx 咁樣mg(h + x)=(1/2)kx^2=mgx?

回答 (1)

天同

2011-02-05 5:02 am

✔ 最佳答案

The compression of the spring x when the ball finally is resting on it is given by,
mg = k.x ------------------------ (1)
where m is the mass of the ball,g is the acceleration due to gravityk is the force constant of the spring

i.e. x = mg/k
Thus statement (1) is correct.

Also, equation (1) doesn't involve h, statement (2) is correct.The loss of gravitational potential energy of the ball= mg(h + x)The gain in elastic potential energy of the spring= (1/2)kx^2From (1),
mg = kx
multiply both sides by (x/2)
hence, (1/2)kx^2 = (1/2)mgx, which is clearly not equal to mg(h + x)
Therefore, statement (3) is wrong

2011-02-05 00:21:32 補充：
咁樣mg(h + x)=(1/2)kx^2=mgx? is clearly wrong. How could mg(h+x) = mgx, thus statement (3) is not correct.

2011-02-05 00:24:54 補充：
In fact, only the potential energy mgx/2 is transferred to the elastic energy of the spring. The remaining energy, (mgh + mgx/2) is dissipated as heat by the damping force of the oscillation.

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