质量分别为 $6.5\,\mathrm{kg}$ 和 $m\,\mathrm{kg}$ 的两个质点 $A$ 和 $B$ 由一根不可伸长的轻绳连接，绳子绕过一个光滑滑轮。滑轮固定在粗糙斜面的顶端，斜面与水平地面的夹角为 $\alpha$，其中 $\tan\alpha=\frac{5}{12}$。$A$ 位于粗糙斜面上，$B$ 悬挂在滑轮下方（见图）。斜面与 $A$ 之间的摩擦系数为 $0.4$。

(a) 已知系统处于平衡状态，求 $m$ 的所有可能值组成的集合。

(b) 改为已知 $m=12$，绳子绷紧，两质点从静止释放。

使用能量法求每个质点均运动 $0.6\,\mathrm{m}$ 时的速率。可假设此时 $A$ 尚未到达滑轮，$B$ 尚未到达地面。

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 $10\,\mathrm{m\,s^{-2}}$。

静摩擦力可以在正负极限之间变化，不能一开始就固定方向。先求支持力，再用所需摩擦力的绝对值不超过最大值。

Since $\tan\alpha=5/12$,

For equilibrium of $B$, $T=mg$. The normal reaction on $A$ is

so the maximum static friction is $0.4N=24\,\mathrm{N}$.

The downslope weight component of $A$ is $6.5(10)(5/13)=25\,\mathrm{N}$. Thus equilibrium requires a signed friction force of $25-10m$ up the slope. It is available precisely when

Hence the set of possible values is $\boxed{0.1\le m\le4.9}$.

按要求用能量法。B下降损失的势能，扣除A上升增加的势能及克服摩擦的功，等于两粒子总动能增加。

For $m=12$, $B$ moves down and $A$ up the slope. The friction on $A$ is $0.4(60)=24\,\mathrm{N}$, down the slope.

Over a distance of $0.6\,\mathrm{m}$, the loss in potential energy of $B$, gain in potential energy of $A$ and work against friction are

Starting from rest, the work-energy equation for the system is

Thus

也可分别对两粒子写功—能方程，再消去内部张力。这样能看清绳张力在两粒子上做的功符号相反。

Let $T$ be the tension during this motion, and $v$ the common speed after travelling $0.6\,\mathrm{m}$. The forces are constant, so $T$ is constant.

For $B$, the work-energy equation is

For $A$, which moves up the slope against weight component $25\,\mathrm{N}$ and friction $24\,\mathrm{N}$,

Adding eliminates $T$:

Hence $v^2=852/185$, giving $\boxed{v\approx2.15\,\mathrm{m\,s^{-1}}}$ to $3$ significant figures.

另一条路线是先用动力学求运动中的张力，再对B应用能量法。平衡小题的张力不能用于这里。

Let the common acceleration magnitude be $a$ and the tension be $T$. Newton’s second law gives

Adding gives $71=18.5a$, so $a=142/37$. Then

Applying the work-energy equation to $B$ over the downward distance of $0.6\,\mathrm{m}$,

Thus $v^2=852/185$, and $\boxed{v\approx2.15\,\mathrm{m\,s^{-1}}}$ to $3$ significant figures.