小測驗－問題二答案

(甲)

利用牛頓第二運動定律 F = ma 為基礎, 施加同一力 F 到兩個質量為 m1 及 m2 的物體並觀察所產生的加速度.因 F = m1a1 = m2a2, 所以 m1 / m2 = a2/ / a1 可作比較物體之質量。


(乙)
M = 10.0 kg , g = 10 m/s2
T = m ( g - a )
i. a = -1.0 m/s2
T = 10.0 ( 10 - 1.0 )
　= 90.0 N
ii. a = 3.0 m/s2
T = 10.0 ( 10 + 3.0 )
　= 130.0 N
iii. v = 5.0 m/s , a = 0 m/s2
T = 10.0 ( 10 - 0 )
　= 100.0 N

T = 0 當 m ( g + a ) = 0 => a = -g = -10.0 m/s2 (下落)
即是：當升降機以 -10.0 m/s2 的加速度下落(或自由落體)

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