如下圖所示，是一輛汽車通過滑輪組將深井中的物體拉至井口的裝置圖。已知井深12m，物體重，汽車重G車=3×104...

問題詳情：

如下圖所示，是一輛汽車通過滑輪組將深井中的物體拉至井口的裝置圖。已知井深12m，物體重，汽車重G車=3×104N，汽車勻速拉繩子時的拉力，汽車受到的阻力為車重的0.1倍。求：

（1）將物體從井底拉至井口的過程中，汽車拉繩子的拉力對滑輪組做了多少功？

（2）滑輪組的機械效率為多少？（保留一位小數）

（3）若汽車運動的速度為3m/s，則將物體由井底拉至井口需要多長時間？

（4）汽車牽引力為多少？牽引力的功率為多少？

【回答】

（1）n=3　　   s=3h=3×12m=36m······································································ (1分)

W=Fs=2.2×103N×36m=7.92×104J·························································· (1分)

（2）η==······················ (2分)

（3）υ物=1/3υ=1/3×3m/s=1m/s···································································· (1分)

t物=h/υ物=12m/1 m/s=12s································································ (1分)[來源:學*科*網]

（4）F牽=F′+f=F′+0.1G車=2.2×103N+0.1×3×104N=5.2×103N·························· (1分)

　 P= F牽υ汽=5.2×103N×3 m/s=1.56×104W (1分)

知識點：力和機械 單元測試

題型：計算題