高考物理問題, 多多指教^^^^

用戶52666

2008-02-24 6:54 am

我想問 06年al physics paper1b, 第6題d..
A light horizontal platform is supported by a light spring fixed vertically on the ground. A coin of mas 0.01kg is released from rest on the platform. The force constant of the spring is 5Nm^-1. The downwadrd direction is taken to be positive and the coin's downward displacement form its initial position is denoted be x. (neglect air resistance)

回答 (1)

Audrey Hepburn

2008-02-24 7:19 am

✔ 最佳答案

Take downward direction be positive.

At the equilibrium position,

mg = kx

(0.01)(10) = 5x

Equilibrium position, x = 0.02 m

By energy conservation equation:

Loss of G.P.E. = Gain of E.P.E.

mgh = 1/2 kh2

(0.01)(10) = 1/2 (5)h2

Maximum displacement, h = 0.04 m

Since the restoring force is always pointing to a fixed point, namely equilibrium position (characteristics of simple harmonic motion). So at the lowest point, the coin experiences the maximum supporting force.

Denote R be the supporting force

R – mg = kx

R = mg + kx

R = (0.01)(10) + 5(0.04 – 0.02) (Be careful of where the position x = 0 is defined!!!)

Supporting force, R = 0.2 N

參考: Myself~~~

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