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【题型】不定项选择
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【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*＝2*BC*.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( )
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{width="0.9847222222222223in"
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height="0.6979166666666666in"}A．$tan\theta\text{＝}\frac{\mu_{1} + 2\mu_{2}}{3}$
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B．$tan\theta\text{＝}\frac{2\mu_{1} + \mu_{2}}{3}$
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C．$tan\theta\text{＝}2\mu_{1} - \mu_{2}$
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D．$tan\theta\text{＝}2\mu_{2} - \mu_{1}$
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【答案】 B
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【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：
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$$mglsin\theta\text{－}\mu_{1}mg\frac{2l}{3}cos\theta\text{－}\mu_{2}mg\frac{l}{3}cos\theta\text{＝}0$$
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解得$tan\theta\text{＝}\frac{2\mu_{1} + \mu_{2}}{3}$，故B正确．
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【知识点】动能定理
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【难度】中
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