一條AL phys問題(20marks)

2007-12-30 8:35 pm
A spring of force constant 135Nm^-1 is suspended vertically with its top end rigidly clamped. A mass of 0.15kg is attached to the lower end of the unextended spring and then released. The system then oscillates- the oscillations slowly dying away until the mass achieves static equilibrium.
更新1:

(b)When the mass has achieved static equilibrium, calculate the energy stored in the extended spring.

更新2:

(c)Calculate the change in the gravitational potential energy of the mass from the beginning to the end of the motion. Why is this answer different from the answer to (b) above? Please give me full explanation in part (c)

回答 (1)

2007-12-31 3:25 am
✔ 最佳答案
b) mg = ke <--搵到e
U = 0.5ke^2 <---搵 Elastic Potential Energy

c) 計SHM未放手果點同最低點既Difference in Gravitional PE , 之後compare兩個value
你會發現 Difference in Gravitional PE > Elastic Potential Energy

唔知你有冇留意 " The system then oscillates- the oscillations slowly dying away until the mass achieves static equilibrium. " 呢句說話既含意
佢想表達呢個係 Damped Oscillation , 姐係有 Energy Lost.
而Difference in Gravitional PE 同 Elastic Potential Energy 既相差就係 energy lost左幾多.


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