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【题型】不定项选择 【题文】如图所示，固定斜面倾角为θ，整个斜面分为AB、BC两段，AB＝2BC.小物块P(可视为质点)与AB、BC两段斜面间的动摩擦因数分别为μ~1~、μ~2~.已知P由静止开始从A点释放，恰好能滑动到C点而停下，那么θ、μ~1~、μ~2~间应满足的关系是(　　) 5-40.tif{width="0.9847222222222223in" height="0.6979166666666666in"}A．tan\theta\text{＝}\frac{\mu_{1} + 2\mu_{2}}{3} B．tan\theta\text{＝}\frac{2\mu_{1} + \mu_{2}}{3} C．tan\theta\text{＝}2\mu_{1} - \mu_{2} D．tan\theta\text{＝}2\mu_{2} - \mu_{1} 【答案】　B 【解析】设斜面的长度为l，小物块从斜面顶端下滑到斜面底端的全过程由动能定理得：

mglsin\theta\text{－}\mu_{1}mg\frac{2l}{3}cos\theta\text{－}\mu_{2}mg\frac{l}{3}cos\theta\text{＝}0

解得tan\theta\text{＝}\frac{2\mu_{1} + \mu_{2}}{3}，故B正确． 【知识点】动能定理 【难度】中