题目

Problem

A rough ramp is fixed to horizontal ground.

The ramp is inclined to the horizontal at an angle $\alpha$ , where $\sin\alpha=\frac37$

The line $AB$ is a line of greatest slope of the ramp, with $B$ above $A$ and $AB=6\,\mathrm{m}$ , as shown in Figure 1.

A block $P$ of mass $2\,\mathrm{kg}$ is pushed, with constant speed, in a straight line up the slope from $A$ to $B$ . The force pushing $P$ acts parallel to $AB$ .

The coefficient of friction between $P$ and the ramp is $\frac13$

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

(a) Use the model to find the total work done in pushing the block from $A$ to $B$ .

(5)

The block is now held at $B$ and released from rest.

(b) Use the model and the work-energy principle to find the speed of the block at the instant it reaches $A$ .

(4)

题目中文翻译

一个粗糙斜坡固定于水平地面上。

斜坡与水平面成 $\alpha$ 角，其中 $\sin\alpha=\frac37$ 。

如图 1 所示， $AB$ 是斜坡的一条最大坡度线， $B$ 在 $A$ 上方，且 $AB=6\,\mathrm{m}$ 。

质量为 $2\,\mathrm{kg}$ 的物块 $P$ 被以恒定速率沿直线从 $A$ 推上斜坡至 $B$ 。推动 $P$ 的力平行于 $AB$ 。

$P$ 与斜坡之间的摩擦系数为 $\frac13$ 。

将物块视为质点，忽略空气阻力。

(a) 利用此模型，求将物块从 $A$ 推至 $B$ 所做的总功。

(5)

现在将物块保持在 $B$ 点，再由静止释放。

(b) 利用此模型及功—能原理，求物块到达 $A$ 的瞬间的速率。

(4)

解答

(a)

解法一

思路

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推力沿坡，不影响支持力。恒速意味着动能不变，推力做功等于克服摩擦的功加重力势能增加；先求支持力和摩擦力。

答题过程

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Since $\sin\alpha=3/7$ and $\alpha$ is acute,

\cos\alpha=\sqrt{1-\frac9{49}}=\frac{2\sqrt{10}}7.

Resolving perpendicular to the ramp,

R=2g\cos\alpha.

The frictional force opposing motion has magnitude

F=\frac13R=\frac23g\cos\alpha.

The gain in gravitational potential energy from $A$ to $B$ is

\Delta U=2g(6\sin\alpha).

The kinetic energy is unchanged because the speed is constant. By work-energy, the pushing work is

W=6F+\Delta U.

Thus

W=6\left(\frac23g\cos\alpha\right)+12g\sin\alpha.

Substituting the exact trigonometric values and $g=9.8$ gives

\boxed{W=85.8\text{ J}}.

(b)

解法一

思路

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下滑时摩擦改为沿坡向上，仍消耗机械能。按题目要求用功—能，把势能损失分成动能增加与摩擦耗能。

答题过程

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During the descent the friction acts up the ramp, with unchanged magnitude

F=\frac23g\cos\alpha.

The block starts from rest. By the work-energy principle,

\frac12(2)v^2=2g(6\sin\alpha)-6\left(\frac23g\cos\alpha\right).

Hence

v^2=12g\frac37-4g\frac{2\sqrt{10}}7.

Taking the positive square root,

\boxed{v=3.87\text{ m s}^{-1}}.

M2 2024 Jan Q3

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程