一根弹性轻绳的原长为 $5\,\mathrm{m}$，弹性模量为 $20\,\mathrm{N}$。

绳的两端分别连接于水平天花板上的两个固定点 $A$、$B$，两点相距 $6\,\mathrm{m}$。

质点 $P$ 连接于绳的中点，并悬挂平衡于 $AB$ 下方 $4\,\mathrm{m}$ 处。

(a) 计算 $P$ 的重量。

现在将质点提高至 $AB$ 的中点，然后由静止释放。

(b) 计算 $P$ 下落 $4\,\mathrm{m}$ 后的速率。

设天花板两端的中点为 $M$。两半绳的原长都是 $2.5\,\mathrm m$，平衡时由 $3$–$4$–$5$ 三角形求实际长度和张力的竖直分量。

Let $M$ be the midpoint of $AB$. By symmetry, $P$ is vertically below $M$, with

Consequently,

Each half of the string has natural length $5/2\ \mathrm m$, extension $5-5/2=5/2\ \mathrm m$ and modulus $20\ \mathrm N$. Hooke’s law gives the tension in each half as

If each half makes angle $\beta$ with the horizontal, $\sin\beta=4/5$. Resolving vertically in equilibrium, the weight $W$ of $P$ is

释放时整根绳的长度是 $6\,\mathrm m$，已经有弹性势能，不能漏掉。用重力势能的减少减去弹性势能的增加，得到动能；质量要由重量除以 $g$ 得到。

At release, $P=M$, so the total string length is $6\ \mathrm m$ and its extension is $1\ \mathrm m$. Its initial elastic potential energy is

After $P$ has fallen $4\ \mathrm m$, the total length is $10\ \mathrm m$ and the extension is $5\ \mathrm m$. The elastic potential energy is then

The loss of gravitational potential energy is $32(4)=128\ \mathrm J$. With mass $m=32/9.8\ \mathrm{kg}$, conservation of mechanical energy gives

Thus