题目

Problem

A fixed rough plane is inclined to the horizontal at an angle $\alpha$ , where $\tan\alpha=\frac5{12}$ A package $P$ of mass $2.6\,\mathrm{kg}$ is placed on the plane, as shown in Figure 1. The package is pushed, with constant speed, up a line of greatest slope of the plane. The force pushing $P$ acts parallel to the path of $P$ .

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

The package is modelled as a particle and air resistance is negligible.

Use the model to find the total work done in pushing $P$ , at a constant speed, a distance of $20\,\mathrm{m}$ up the plane.

(5)

题目中文翻译

一个固定粗糙平面与水平面成 $\alpha$ 角，其中 $\tan\alpha=\frac5{12}$ 。如图 1 所示，质量为 $2.6\,\mathrm{kg}$ 的包裹 $P$ 放在平面上。包裹被以恒定速率沿平面的一条最大坡度线向上推。推动 $P$ 的力平行于 $P$ 的路径。

$P$ 与平面之间的摩擦系数为 $\frac15$ 。

将包裹视为质点，忽略空气阻力。

利用此模型，求以恒定速率将 $P$ 沿平面向上推 $20\,\mathrm{m}$ 所做的总功。

(5)

解答

解法一

思路

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恒速意味着动能不变。推力做的总功分为势能增加和克服摩擦的功；推力沿坡，所以支持力不受它影响。

答题过程

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

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

Resolving perpendicular to the plane gives the normal reaction

R=2.6g\cos\alpha.

The frictional force has magnitude

F=\frac15R=\frac15(2.6g\cos\alpha).

Over a distance of 20 m, the gain in gravitational potential energy is

\Delta U=2.6g(20\sin\alpha),

and the work against friction is

W_F=20F=20\left[\frac15(2.6g\cos\alpha)\right].

Since the speed is constant, the kinetic energy does not change. Thus the total pushing work is

W=\Delta U+W_F.

Substituting,

W=2.6(9.8)(20)\frac5{13}+4(2.6)(9.8)\frac{12}{13}.

This gives $W=290.08$ J, so to 3 significant figures,

\boxed{W=2.90\times10^2\text{ J}}.

M2 2025 Jan Q1

题目

解答

解法一

思路

答题过程