一根弹性轻绳的原长为 $2a$，弹性模量为 $2mg$。

弹性绳的一端连接于固定点 $O$，另一端连接于质量为 $\dfrac12m$ 的质点 $P$。

$A$ 点位于 $O$ 正下方，$OA=4a$。

将质点 $P$ 保持在 $A$ 后由静止释放。$P$ 向上移动距离 $a$ 的瞬间的速率为 $\sqrt{3ag}$。

将 $P$ 运动所受空气阻力的大小视为 $kmg$，其中 $k$ 为常数。

利用此模型与功—能原理，

(a) 证明 $k=\dfrac14$。

现在将质点 $P$ 保持在 $O$ 后由静止释放。$P$ 向下运动时，在经过 $B$ 点时达到最大速率。

(b) 求距离 $OB$。

按功—能原理，弹性势能减少量等于重力势能增加、动能增加和空气阻力耗能之和。粒子质量是 m/2，不能按 m 计算后两项。

Initially the string has extension $4a-2a=2a$. After the particle rises a distance $a$, the extension is $3a-2a=a$. The initial and final elastic energies are

The gain in gravitational potential energy is $(m/2)ga$, and the final kinetic energy is

Air resistance acts downwards while the particle rises, doing negative work of magnitude $kmg(a)$. By the work–energy principle,

Dividing by $mga$,

从 O 下落时，绳先松弛后绷紧。以总下落距离为变量写动能，绷紧后得到开口向下的二次式；其顶点给出最大速率的位置。

Let $y=OP$ be the downward distance travelled from $O$. The air resistance is now upwards and has magnitude $mg/4$. For $0\leq y\leq2a$, the string is slack. Work–energy gives

so $v^2=gy$, which increases with $y$. For $y\geq2a$, the extension is $y-2a$. Work–energy from $O$ gives

Therefore

Differentiating with respect to $y$,

This is zero when $y-2a=a/4$, so $y=9a/4$. Since the coefficient of $y^2$ is negative, this is a maximum. It follows after the increasing-speed slack-string phase and is reached with positive speed. Hence

也可直接找下降过程中合力从向下变向上的位置。绳张力随伸长增大，速率极大时加速度为零。

Before the string becomes taut, the resultant downward force is

Afterwards, if the extension is $e$, the tension is

The downward resultant is $mg/4-mge/a$. It decreases as $e$ increases, so the speed is greatest when this resultant is zero:

Adding the natural length,