一根轻弹性绳的原长为 $0.5\text{ m}$，弹性模量为 $14\text{ N}$，其一端连接在光滑斜面上的固定点 $A$。斜面与水平面成 $\alpha$ 角，其中 $\tan\alpha=\frac7{24}$。质量为 $2\text{ kg}$ 的质点 $P$ 连接在绳的另一端。绳沿斜面的最大坡度线放置。最初将质点 $P$ 保持在斜面上高于 $A$ 的位置，其中 $AP=0.8\text{ m}$，随后将 $P$ 从静止释放。

求 $P$ 在随后的运动过程中的最大速度。

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 $10\,\mathrm{m\,s^{-2}}$。

质点越过 $A$ 后，绳拉紧时张力开始阻碍它向下运动。速率在张力恰好平衡重力沿斜面的分量时达到最大，找到此处再用能量守恒。

The downslope component of the weight is $2g\sin\alpha=2(10)(7/25)=5.6\,\mathrm{N}$. The particle initially accelerates downslope, first with the string pulling downslope and then with it slack. After passing $A$ and reaching $0.5\,\mathrm{m}$ below it, the string starts to pull upslope. At extension $e$ its tension is $14e/0.5=28e$. The acceleration changes from downslope to upslope at $28e=5.6$, so $e=0.2\,\mathrm{m}$ and the particle is $0.7\,\mathrm{m}$ below $A$. This is the minimum of the potential energy along the line, so the speed there is maximal.

The initial extension is $0.8-0.5=0.3\,\mathrm{m}$, and the downslope displacement to the maximum is $0.8+0.7=1.5\,\mathrm{m}$. Conservation of energy gives

The maximum velocity is $3.02\,\mathrm{m\,s^{-1}}$ down the plane.