质量分别为 $2.4\,\mathrm{kg}$ 和 $3.3\,\mathrm{kg}$ 的质点 $A$ 和 $B$ 由一根不可伸长的轻绳连接，轻绳绕过固定在粗糙斜面顶端的光滑滑轮。斜面与水平地面成 $\theta^\circ$ 角。质点 $A$ 位于斜面上，$A$ 与滑轮之间的绳段平行于斜面的最大坡度线。质点 $B$ 悬挂在滑轮正下方，距地面 $1\,\mathrm{m}$（见图）。斜面与 $A$ 之间的摩擦系数为 $\mu$。

(a) 已知 $\theta=30$，且系统处于平衡状态，$A$ 恰好将要沿斜面直接向上运动。

证明：$\mu=1.01$，精确到 $3$ 位有效数字。

(b) 若改为已知 $\theta=20$ 且 $\mu=1.01$。系统在绳绷紧的情况下从静止释放。求 $A$ 瞬时静止前经过的总路程。你可以假设 $A$ 不会到达滑轮，且 $B$ 落地后保持静止。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

Where a numerical value for the acceleration due to gravity ($g$) is needed, use $10\,\mathrm{m\,s^{-2}}$.

需要使用重力加速度 $g$ 的数值时，取 $10\,\mathrm{m\,s^{-2}}$。

临界向上时A的摩擦沿坡向下，B静止给张力33牛，法向反力用A重力的余弦分量。

For $B$ in equilibrium, $T=3.3g=33\,\mathrm{N}$. For $A$,

Since $A$ is on the point of moving up the plane, limiting friction acts down the plane. Resolving up the plane,

Thus

用题目新给的角20度与μ等于1.01重算。B落地前共同加速度，落地后绳松弛，A继续上坡受到重力和摩擦共同减速；最后加上先前1米。

With $\theta=20^\circ$ and $\mu=1.01$, the friction on $A$ is

Before $B$ reaches the ground, take $B$ downwards and $A$ up the plane. Newton’s second law for the whole system gives

Thus $a=0.353222\ldots\,\mathrm{m\,s^{-2}}>0$. When $B$ has fallen $1\,\mathrm{m}$, the common speed satisfies

The string then becomes slack. For $A$ still moving up the plane, the deceleration magnitude is

If the further distance is $s$,

Hence the total distance travelled by $A$ is

两段也可全部使用能量法。第一段对系统，第二段只对绳松弛后的A；初始势能损失与各段摩擦功分别计算。

The friction on $A$ is $F=1.01(2.4)(10)\cos20^\circ$. Before $B$ reaches the ground, the net loss of potential energy less friction work gives

Thus $v^2=0.706444\ldots$. After $B$ stops, the string becomes slack and $A$ rises a further distance $s$. Its kinetic energy is used to increase potential energy and overcome friction:

Therefore

The total distance is