题目

Problem

A particle $P$ of mass $2\,\mathrm{kg}$ is attached to one end of a light elastic spring, of natural length $0.8\,\mathrm{m}$ and modulus of elasticity $12\,\mathrm{N}$ . The other end of the spring is attached to a fixed point $A$ on a rough plane. The plane is inclined at $30^\circ$ to the horizontal. Initially $P$ is held at rest on the plane at the point $B$ , where $B$ is below $A$ , with $AB=0.3\,\mathrm{m}$ and $AB$ lies along a line of greatest slope of the plane. The point $C$ lies on the plane with $AC=1\,\mathrm{m}$ , as shown in Figure 3.

The coefficient of friction between $P$ and the plane is $0.3$

After being released $P$ passes through the point $C$ .

Find the speed of $P$ at the instant it passes through $C$ .

(7)

题目中文翻译

一个质量为 $2\,\mathrm{kg}$ 的质点 $P$ 连接于一根原长为 $0.8\,\mathrm{m}$ 、弹性模量为 $12\,\mathrm{N}$ 的弹性轻弹簧的一端。弹簧另一端固定于粗糙斜面上的 $A$ 点。斜面与水平面成 $30^\circ$ 角。最初将 $P$ 静止保持在斜面上的 $B$ 点， $B$ 位于 $A$ 下方， $AB=0.3\,\mathrm{m}$ ，且 $AB$ 沿斜面的最大坡度线。如图 3 所示， $C$ 点位于斜面上， $AC=1\,\mathrm{m}$ 。

$P$ 与斜面之间的摩擦系数为 $0.3$ 。

释放后， $P$ 经过 $C$ 点。

求 $P$ 经过 $C$ 的瞬间的速率。

(7)

解答

解法一

思路

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初始弹簧受压 0.5 米，末位置弹簧伸长 0.2 米，两者都有弹性势能。下滑的重力势能减少，同时摩擦耗能；弹力随位置变化，不能按匀加速处理。

答题过程

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The distance travelled down the plane is

BC=AC-AB=1-0.3=0.7\text{ m}.

Initially the spring is compressed by $0.8-0.3=0.5$ m. At $C$ it is extended by $1-0.8=0.2$ m. Therefore the initial and final elastic energies are

E_B=\frac{12(0.5)^2}{2(0.8)},\qquad E_C=\frac{12(0.2)^2}{2(0.8)}.

The spring acts along the plane, so the normal reaction is $2g\cos30^\circ$ . The work done against friction is

E_f=0.3(2g\cos30^\circ)(0.7).

The gravitational potential energy lost is $2g(0.7\sin30^\circ)$ . Using the work–energy principle, with initial speed zero and speed $v$ at $C$ ,

\frac{12(0.5)^2}{2(0.8)}+2g(0.7\sin30^\circ)=\frac{12(0.2)^2}{2(0.8)}+0.3(2g\cos30^\circ)(0.7)+\frac12(2)v^2.

Hence

v^2=1.875+6.86-0.3-0.42(9.8)\cos30^\circ=4.8704\ldots.

Thus

\boxed{v=2.21\text{ m s}^{-1}}\quad(3\text{ s.f.}).

M3 2020 Oct Q3

题目

解答

解法一

思路

答题过程