图中轨道 $ABCD$ 位于一个竖直平面内。$AB$ 段是一条与水平面成 $30^\circ$ 角的光滑直线。$BC$ 段是一条粗糙水平直线。$CD$ 段是一条与水平面成 $30^\circ$ 角的粗糙直线。$AB$、$BC$ 和 $CD$ 的长度均为 $2\,\mathrm{m}$。

一个质点从 $A$ 处由静止释放。质点与 $BC$、$CD$ 两段之间的摩擦系数均为 $\mu$。质点经过点 $B$ 或点 $C$ 时，速率不发生变化。

(a) 已知 $\mu=0.1$。

求质点速率为 $1\,\mathrm{m\,s^{-1}}$ 时沿 $CD$ 段向上运动的距离。

(b) 改为已知在另一个 $\mu$ 值下，质点从 $C$ 沿轨道向上运动 $1\,\mathrm{m}$ 后瞬时静止。

求 $\mu$ 的值以及质点第二次经过 $C$ 的瞬间的速率。

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 $10\,\mathrm{m\,s^{-2}}$。

从A到指定位置整体写能量方程，初始势能扣除BC及CD摩擦功、CD势能增加后剩末动能。

Let the mass be $m$ and let the distance up $CD$ be $d$. The friction on $BC$ is $0.1mg=m$, while on $CD$ it is $0.1mg\cos30^\circ=(\sqrt3/2)m$.

The work-energy equation from $A$ to the required point is

Dividing by $m$ and simplifying gives

Thus

This is less than $CD=2\,\mathrm{m}$, as required.

分三段应用动力学和运动学，先求B、C速度平方，再算上坡减速至1所需距离。

On smooth $AB$, the acceleration is $10\sin30^\circ=5$, so

On $BC$, the acceleration is $-0.1g=-1$, so

On the upward journey along $CD$, gravity and friction both oppose motion:

Therefore

giving

前两段用运动学求C处速度，第三段再用能量处理势能增加及摩擦功。

On $AB$, $a=5$, giving $v_B^2=2(5)(2)=20$. On $BC$, $a=-1$, giving $v_C^2=20-4=16$.

If the distance up $CD$ is $d$, the decrease in kinetic energy equals the gain in potential energy plus the work against friction:

Cancelling $m$,

Hence

先按上坡1米停下写总能量式求新的摩擦系数。返回下坡时摩擦反向，势能损失减摩擦功转为动能。

At the turning point $1\,\mathrm{m}$ above $C$ along the track, the speed is zero. The energy equation is

Cancelling $m$ and simplifying,

Thus

On the return down $CD$, friction acts up the slope. Since $\mu<\tan30^\circ$, the particle slides back down. Its speed at $C$ satisfies

Hence

全部用动力学与运动学。C处第一次经过的速度平方由前两段求出，上坡停下建立μ方程，返回时改用重力减摩擦的加速度。

For $AB$, $v_B^2=20$. On $BC$, the acceleration is $-10\mu$, so

The acceleration during ascent on $CD$ is $-5-5\sqrt3\mu$. Since the particle stops after $1\,\mathrm{m}$,

Thus

On descending, the acceleration down the plane is

Starting from rest and travelling $1\,\mathrm{m}$,

还可以从A到第二次经过C整体算能量。BC只走一次，而CD上、下各走1米，总摩擦路程是2米。

From $A$ to the turning point, the energy equation gives

hence

Between $A$ and the second crossing of $C$, the total friction work is $2\mu mg$ on $BC$ and $2\mu mg\cos30^\circ$ on $CD$. The initial potential energy above $C$ is $mg$. Therefore

Thus