题目

Problem

A particle of mass $m$ is held at rest at a point $A$ on a rough plane.

The plane is inclined at an angle $\alpha$ to the horizontal, where $\tan\alpha=\frac5{12}$

The coefficient of friction between the particle and the plane is $\frac15$

The points $A$ and $B$ lie on a line of greatest slope of the plane, with $B$ above $A$ , and $AB=d$ , as shown in Figure 1.

The particle is pushed up the line of greatest slope from $A$ to $B$ .

(a) Show that the work done against friction as the particle moves from $A$ to $B$ is $\frac{12}{65}mgd$

(3)

The particle is then held at rest at $B$ and released.

(b) Use the work-energy principle to find, in terms of $g$ and $d$ , the speed of the particle at the instant it reaches $A$ .

(4)

题目中文翻译

质量为 $m$ 的质点被保持静止在粗糙斜面上的 $A$ 点。

斜面与水平面成 $\alpha$ 角，其中 $\tan\alpha=\frac5{12}$ 。

质点与斜面之间的摩擦系数为 $\frac15$ 。

如图 1 所示， $A$ 、 $B$ 两点位于斜面的一条最大坡度线上， $B$ 在 $A$ 上方，且 $AB=d$ 。

质点沿最大坡度线从 $A$ 被推至 $B$ 。

(a) 证明质点从 $A$ 运动至 $B$ 时克服摩擦力所做的功为 $\frac{12}{65}mgd$ 。

(3)

随后将质点保持静止在 $B$ 点，再释放。

(b) 利用功—能原理，用 $g$ 和 $d$ 表示质点到达 $A$ 的瞬间的速率。

(4)

解答

(a)

解法一

思路

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支持力垂直于斜面，摩擦力大小为 $\mu R$ 。功用沿斜面的实际路程 $d$ 计算。

答题过程

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Since $\tan\alpha=5/12$ ,

\sin\alpha=\frac5{13},\qquad\cos\alpha=\frac{12}{13}.

Resolving perpendicular to the plane,

R=mg\cos\alpha=\frac{12}{13}mg.

The frictional force is

F=\mu R=\frac15\times\frac{12}{13}mg=\frac{12}{65}mg.

Therefore the work done against friction is

\boxed{Fd=\frac{12}{65}mgd}.

(b)

解法一

思路

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释放后摩擦力改为沿斜面向上。损失的重力势能减去摩擦耗能等于动能增加；题目指定使用功—能原理。

答题过程

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On moving down from $B$ to $A$ , the particle loses height $d\sin\alpha=5d/13$ . Friction now acts up the slope, with the same magnitude as before.

Let the speed at $A$ be $v$ . By the work-energy principle,

\frac12mv^2=mgd\left(\frac5{13}\right)-\frac{12}{65}mgd.

Hence

\frac12mv^2=\frac15mgd,\qquad v^2=\frac{2gd}{5}.

Taking the positive root, the required speed is

\boxed{\sqrt{\frac{2gd}{5}}}.

M2 2021 Oct Q1

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程