如图9所示,在倾角为的长斜面上有一带风帆的滑块,从静止开始沿

如图9所示,在倾角为的长斜面上有一带风帆的滑块,从静止开始沿斜面下滑,滑块质量为m,它与斜面间的动摩擦因数为,帆受到的空气阻力与滑块下滑速度的大小成正比,即

写出滑块下滑加速度的表达式。

写出滑块下滑的最大速度的表达式。

,从静止下滑的速度图象如图所示的曲线,图中直线是t=0时的速度图线的切线,由此求出和k的值。

答案

【小题1】

【小题2】

【小题3】   


解析:

(1)对滑块应用牛顿第二定律有:

   

    滑块下滑加速度表达式为:

   

    (2)由<1>式可知,当滑块的速度增大时,其加速度是减小的,当加速度为零时,滑块的速度达到最大,由<1>式可知最大速度为:

   

    (3)由图可知,当滑块的速度为零时,其加速度为最大加速度,而由<1>式可知当滑块的加速度为零时,它的速度最大,滑块的最大速度为,由<1>式和<2>式有:

   

    将g、m、代入<3>式和<4>式后解得:

   

   

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