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範例:牛頓第二定律實驗

在光滑水平桌面上有一質量m=10kg物體,接以一線,繞過桌邊之定滑輪,再連接一極輕之彈簧稱的掛鉤上,稱的另一端懸質量M的砝碼,m作等加速度運動時,稱上指示力量是20nt,(A)Mm之加速度均為2 m/sec2 (B)M之質量為2.5kg (C)M,m互掉時彈簧稱上之指示力量應不變 (D)若欲m之加速度增為原來之2,則彈簧稱上之指示力量應變為40nt (E)(D)此時M的質量應為5 kg (g=10 m/s2)


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