物理,force or workdone on spring

用戶91254

2013-06-18 11:58 pm

A 70 kg driver gets into an empty taptap to start the day's work. The springs compress 2.5×10−2 m.
Find the spring constant.

我知道要用F=kx去搵spring constant,但我唔明白點解呢條問題唔係用w=1/2kx^2,而且,係呢條問題上,個driver坐上車唔係即係佢既work done on spring咩?
求解答!!

回答 (1)

天同

2013-06-19 12:38 am

✔ 最佳答案

You could still use W = (1/2)kx^2 if you wish, but the calculation is a bit complicated.

When the driver sits on the taptap, the springs are compressed and oscillate in simple harmonic motion. When the motion dies down, the kinetic energy of oscillation has changed into heat, leaving only the potential energy which is transferred to the energy possessed by the spring.

Hence, only half of the original (gravitational) potential energy of the driver remains (the other half, as explained above, vanishes when the oscillation dies down), which is = (70g x 2.5x10^-2)/2 J = 8.75 J
[Take g = 10 m/s^2]

Therefore, 8.75 = (1/2)k(2.5x10^-2)
k = 2 x 8.75/(2.5x10^-2)^2 N/m = 28 000 N/m
You get the same result on using F = kx

Anyway, the use of F = kx is much simpler.

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