题目

Problem

A rough straight ramp is fixed to horizontal ground. The ramp has length $15\,\mathrm{m}$ and is inclined at an angle $\alpha$ to the ground, where $\tan\alpha=\frac5{12}$ . The line $AB$ is a line of greatest slope of the ramp, where $A$ is at the bottom of the ramp, and $B$ is at the top of the ramp, as shown in Figure 3.

A particle $P$ of mass $6\,\mathrm{kg}$ is projected up the ramp with speed $14\,\mathrm{m\,s^{-1}}$ from $A$ in a straight line towards $B$ . The coefficient of friction between $P$ and the ramp is $0.25$

(a) Find the work done against friction as $P$ moves from $A$ to $B$ .

(3)

At the instant $P$ reaches $B$ , the speed of $P$ is $v\,\mathrm{m\,s^{-1}}$ . After leaving the ramp at $B$ , the particle $P$ moves freely under gravity until it hits the horizontal ground at the point $C$ . Immediately before hitting the ground at $C$ , the speed of $P$ is $w\,\mathrm{m\,s^{-1}}$

(b) Use the work-energy principle to find

(i) the value of $v$ ,

(ii) the value of $w$ .

(7)

题目中文翻译

一个粗糙的直斜坡固定在水平地面上。斜坡长 $15\,\mathrm{m}$ ，与地面成 $\alpha$ 角，其中 $\tan\alpha=\frac5{12}$ 。如图 3 所示， $AB$ 是斜坡的一条最大坡度线， $A$ 位于斜坡底端， $B$ 位于斜坡顶端。

质量为 $6\,\mathrm{kg}$ 的质点 $P$ 从 $A$ 以速率 $14\,\mathrm{m\,s^{-1}}$ 沿直线向 $B$ 顺斜坡向上投射。 $P$ 与斜坡之间的摩擦系数为 $0.25$ 。

(a) 求 $P$ 从 $A$ 运动到 $B$ 时克服摩擦力所做的功。

(3)

$P$ 到达 $B$ 的瞬间，其速率为 $v\,\mathrm{m\,s^{-1}}$ 。质点 $P$ 从 $B$ 离开斜坡后，仅在重力作用下运动，直到在 $C$ 点撞击水平地面。在 $C$ 点撞击地面前一瞬间， $P$ 的速率为 $w\,\mathrm{m\,s^{-1}}$ 。

(b) 利用功—能原理求

(i) $v$ 的值，

(ii) $w$ 的值。

(7)

解答

(a)

解法一

思路

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摩擦功等于摩擦力乘沿斜坡的路程。先求垂直斜面的支持力，再求摩擦力；中间保留精确值。

答题过程

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

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

The normal reaction and friction are

R=6g\cos\alpha=\frac{72g}{13},

F=0.25R=\frac{18g}{13}.

Therefore the work done against friction is

W=15F=\frac{270g}{13} =\boxed{204\text{ J}}\quad(3\text{ s.f.}).

(b)

解法一

思路

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按题意使用功—能原理。上坡阶段，初动能用于增加重力势能、克服摩擦并留下末动能；离开斜坡后不再有摩擦。

答题过程

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(i) The vertical rise from $A$ to $B$ is $15\sin\alpha=75/13\text{ m}$ . By the work–energy principle,

\frac12(6)(14^2) =\frac12(6)v^2+6g\frac{75}{13}+\frac{270g}{13}.

Thus

\begin{align*} 3v^2=&\,588-\frac{720g}{13},\\ v^2=&\,196-\frac{240g}{13}. \end{align*}

Taking the positive square root with $g=9.8$ ,

\boxed{v=3.88}\quad(3\text{ s.f.}).

(ii) The points $A$ and $C$ are at the same height. For the complete motion from $A$ to $C$ , the only loss of mechanical energy is the work against friction on the ramp:

\frac12(6)(14^2)-\frac{270g}{13}=\frac12(6)w^2.

Hence

w^2=196-\frac{90g}{13},

\boxed{w=11.3}\quad(3\text{ s.f.}).

M2 2020 Oct Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程