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範例:動力學與轉動平衡[92指考]

質量2.0 kg,長、寬、高為5.0 cm×5.0 cm×4.0 cm 的均勻木塊,置放在水平桌面上。在距桌面高3 cm ,施一水平力F 向右, 已知F=5.0 N 時方能拉動靜止的木塊,木塊拉動後,F=2.0 N 即可使之做等速滑動,則下列敘述哪些正確?(A)木塊與桌面間之靜摩擦係數為0.20(B)木塊做等速度滑動時,作用於木塊的合力矩為零(C)木塊做等速度滑動時,桌面施於木塊之正向力,對通過木塊質心(轉軸垂直於紙面)所施的力矩大小為0.06 N-m (D)木塊被拉動後,若F=5.0 N,則木塊的加速度為2.5 m/s2(E)當木塊以v=1.0 m/s 的等速率運動時,若改施以F=4.0 N 的力,2 秒鐘後,木塊速率變為3.0 m/s[92指定考]

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Posted in 歷屆物理考題, 牛頓第二運動定律.

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