diff --git a/dsLightRag/ShiTi/T3_DocxToMd.py b/dsLightRag/ShiTi/T3_DocxToMd.py index a23cf7fe..15537627 100644 --- a/dsLightRag/ShiTi/T3_DocxToMd.py +++ b/dsLightRag/ShiTi/T3_DocxToMd.py @@ -80,7 +80,6 @@ async def main(): """ 修正一下MinerU生成的Latex中，如果是数字加圆圈的样式 \textcircled{1}，无法在Typora或者PyCharm中显示的问题,改成兼容性更强的 \enclose{circle}{1} """ - # \textcircled path = r'../output/《动能定理》巩固练习/auto' finalName = path + r'/《动能定理》巩固练习.md' with open(finalName, 'r', encoding='utf-8') as f: diff --git a/dsLightRag/Util/LightRagUtil.py b/dsLightRag/Util/LightRagUtil.py index 6c6ed31a..088d8609 100644 --- a/dsLightRag/Util/LightRagUtil.py +++ b/dsLightRag/Util/LightRagUtil.py @@ -209,6 +209,7 @@ def format_exam_content(raw_text, output_path): 要求： - 一道题一道题输出，不要使用表格 +- 原文中有图片信息的，一定要完整保留 - 自动检测数学表达式并用$或$$正确包裹 - 确保公式中的特殊字符正确转义 - 除题目内容外，不要输出其它无关信息 diff --git a/dsLightRag/output/images/099673167e0db6446d08a21c06073c7fa32120ab2b8737a8abb2212871f7deb3.jpg 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b/dsLightRag/output/images/ee9f51d84e51925ee8c9fa332f1fc97ac93093b232f1181a2ed0dcaae72525a7.jpg differ diff --git a/dsLightRag/output/物理Docx整理后的结果.md b/dsLightRag/output/物理Docx整理后的结果.md index 4bca50b8..ba22a8b2 100644 --- a/dsLightRag/output/物理Docx整理后的结果.md +++ b/dsLightRag/output/物理Docx整理后的结果.md @@ -1,80 +1,129 @@ ### 题目1 **题目序号**: 1 -**题目内容**: 如图所示，固定斜面倾角为 $\theta$，整个斜面分为 $AB$、$BC$ 两段，$AB = 2 BC$。小物块 $P$（可视为质点）与 $AB$、$BC$ 两段斜面间的动摩擦因数分别为 $\mu_1$、$\mu_2$。已知 $P$ 由静止开始从 $A$ 点释放，恰好能滑动到 $C$ 点而停下，那么 $\theta$、$\mu_1$、$\mu_2$ 间应满足的关系是(　　) +**题目内容**: +如图所示，固定斜面倾角为 $\theta$，整个斜面分为 $AB$、$BC$ 两段，$AB = 2BC$。小物块 $P$（可视为质点）与 $AB$、$BC$ 两段斜面间的动摩擦因数分别为 $\mu_1$、$\mu_2$。已知 $P$ 由静止开始从 $A$ 点释放，恰好能滑动到 $C$ 点而停下，那么 $\theta$、$\mu_1$、$\mu_2$ 间应满足的关系是(　　) + **选项**: A．$\tan \theta =$ B．$\tan \theta =$ C．$\tan \theta = 2 \mu_1 - \mu_2$ D．$\tan \theta = 2 \mu_2 - \mu_1$ + **答案**: B -**解析**: 设斜面的长度为 $l$，小物块从斜面顶端下滑到斜面底端的全过程由动能定理得： -$m g l \sin \theta - \mu_1 m g \cos \theta \cdot \frac{2l}{3} - \mu_2 m g \cos \theta \cdot \frac{l}{3} = 0$， -解得 $\tan \theta = \frac{2 \mu_1 + \mu_2}{3}$，故B正确。 +**解析**: +设斜面的长度为 $l$，小物块从斜面顶端下滑到斜面底端的全过程由动能定理得： +$m g l \sin \theta - \mu_1 m g \cos \theta - \mu_2 m g \cos \theta = 0$， +解得 $\tan \theta =$，故B正确。 --- ### 题目2 **题目序号**: 2 -**题目内容**: 在有大风的情况下，一小球自 $A$ 点竖直上抛，其运动轨迹如图 9 所示（小球的运动可看做竖直方向的竖直上抛运动和水平方向的初速度为零的匀加速直线运动的合运动），小球运动轨迹上的 $A$、$B$ 两点在同一水平直线上，$M$ 点为轨迹的最高点。若风力的大小恒定，方向水平向右，小球在 $A$ 点抛出时的动能为4 J，在 $M$ 点时它的动能为2 J，落回到 $B$ 点时动能记为 $E_{kB}$，小球上升时间记为 $t_1$，下落时间记为 $t_2$，不计其他阻力，则( ) +**题目内容**: +在有大风的情况下，一小球自 $A$ 点竖直上抛，其运动轨迹如图 9 所示（小球的运动可看做竖直方向的竖直上抛运动和水平方向的初速度为零的匀加速直线运动的合运动），小球运动轨迹上的 $A$、$B$ 两点在同一水平直线上，$M$ 点为轨迹的最高点。若风力的大小恒定，方向水平向右，小球在 $A$ 点抛出时的动能为4 J，在 $M$ 点时它的动能为2 J，落回到 $B$ 点时动能记为 $E_{kB}$，小球上升时间记为 $t_1$，下落时间记为 $t_2$，不计其他阻力，则( ) + **选项**: A．$x_1 : x_2 = 1 : 3$ B．$t_1 < t_2$ -C．$E_{kB} = 6 \mathrm{J}$ -D．$E_{kB} = 12 \mathrm{J}$ +C．$E_{kB} = 6 \mathrm{~J}$ +D．$E_{kB} = 12 \mathrm{~J}$ + **答案**: AD -**解析**: 由小球上升与下落时间相等即 $t_1 = t_2$， +**解析**: +由小球上升与下落时间相等即 $t_1 = t_2$， $x_1 : (x_1 + x_2) = 1 : 2^2 = 1 : 4$， -即 $x_1 : x_2 = 1 : 3$。$A \to M$ 应用动能定理得 -$-m g h + W_1 = \frac{1}{2} m v_M^2 - \frac{1}{2} m v^2$， +即 $x_1 : x_2 = 1 : 3$。 +$A \rightarrow M$ 应用动能定理得 +$-m g h + W_1 = m v_M^2 - m v^2$， 竖直方向有 $v^2 = 2 g h$， -联立得 $W_1 = 2 \mathrm{J}$。 -$A \to B$ 风力做功 $W_2 = 4 W_1 = 8 \mathrm{J}$， -$A \to B$ 由动能定理 $W_2 = E_{kB} - E_{kA}$， -可求得 $E_{kB} = 12 \mathrm{J}$，A、D 正确。 +联立得 $W_1 = 2 \mathrm{~J}$。 +$A \rightarrow B$ 风力做功 $W_2 = 4 W_1 = 8 \mathrm{~J}$， +$A \rightarrow B$ 由动能定理 $W_2 = E_{kB} - E_{kA}$， +可求得 $E_{kB} = 12 \mathrm{~J}$，A、D 正确。 --- ### 题目3 **题目序号**: 3 -**题目内容**: 如图所示，一个质量为 $m$ 的圆环套在一根固定的水平长直杆上，环与杆的动摩擦因数为 $\mu$。现给环一个向右的初速度 $v_0$，同时对环施加一个竖直向上的作用力 $F$，并使 $F$ 的大小随环的速度的大小变化，两者关系为 $F = k v$，其中 $k$ 为常数，则环在运动过程中克服摩擦所做的功的大小不可能为(　　) +**题目内容**: +如图所示，一个质量为 $m$ 的圆环套在一根固定的水平长直杆上，环与杆的动摩擦因数为 $\mu$。现给环一个向右的初速度 $v_0$，同时对环施加一个竖直向上的作用力 $F$，并使 $F$ 的大小随环的速度的大小变化，两者关系为 $F = k v$，其中 $k$ 为常数，则环在运动过程中克服摩擦所做的功的大小不可能为(　　) + **选项**: A. B．0 C. D. + **答案**: C -**解析**: 若圆环最终静止，则 $-W_f = 0 - \frac{1}{2} m v_0^2$，$W_f = \frac{1}{2} m v_0^2$，A 可能；若圆环刚开始运动时，$m g = F = k v_0$，圆环一直做匀速运动，克服摩擦所做的功为零，B 可能；若圆环最终做匀速运动，$m g = F = k v$，$v = \frac{m g}{k}$，则 $-W_f = \frac{1}{2} m v^2 - \frac{1}{2} m v_0^2$，化简得 $W_f = \frac{1}{2} m v_0^2 - \frac{m^3 g^2}{2 k^2}$，D 可能，C 不可能。 +**解析**: +若圆环最终静止，则 $-W_f = 0 - \frac{1}{2} m v_0^2$，$W_f = \frac{1}{2} m v_0^2$，A 可能；若圆环刚开始运动时，$m g = F = k v_0$，圆环一直做匀速运动，克服摩擦所做的功为零，B 可能；若圆环最终做匀速运动，$m g = F = k v$，则 $-W_f = \frac{1}{2} m v^2 - \frac{1}{2} m v_0^2$，化简得 $W_f = \frac{1}{2} m v_0^2 - \frac{m^3 g^2}{2 k^2}$，D 可能，C 不可能。 --- ### 题目4 **题目序号**: 4 -**题目内容**: 物体在合外力作用下做直线运动的 $v - t$ 图象如图所示。下列表述正确的是(　　) +**题目内容**: +物体在合外力作用下做直线运动的 $v - t$ 图象如图所示。下列表述正确的是(　　) + **选项**: A．在 $0 \sim 2 \mathrm{s}$ 内，合外力总是做负功 B．在 $1 \sim 2 \mathrm{s}$ 内，合外力不做功 C．在 $0 \sim 3 \mathrm{s}$ 内，合外力做功为零 D．在 $0 \sim 1 \mathrm{s}$ 内比 $1 \sim 3 \mathrm{s}$ 内合外力做功快 + **答案**: CD -**解析**: 根据物体的速度图象和动能定理可知在 $0 \sim 2 s$ 内物体先加速后减速，合外力先做正功后做负功，A 错；根据动能定理得 $0 \sim 3 \mathrm{s}$ 内合外力做功为零，$1 \sim 2 \mathrm{s}$ 内合外力做负功，C 对，B 错；在 $0 \sim 1 \mathrm{s}$ 内比 $1 \sim 3 \mathrm{s}$ 内合外力做功快，D 对。 +**解析**: +根据物体的速度图象和动能定理可知在 $0 \sim 2 \mathrm{s}$ 内物体先加速后减速，合外力先做正功后做负功，A 错；根据动能定理得 $0 \sim 3 \mathrm{s}$ 内合外力做功为零，$1 \sim 2 \mathrm{s}$ 内合外力做负功，C 对，B 错；在 $0 \sim 1 \mathrm{s}$ 内比 $1 \sim 3 \mathrm{s}$ 内合外力做功快，D 对。 --- ### 题目5 **题目序号**: 5 -**题目内容**: 如右图所示，质量为 $m$ 的物体用细绳经过光滑小孔牵引在光滑水平面上做匀速圆周运动，拉力为某个值 $F$ 时，转动半径为 $R$，当拉力逐渐减小到 $\frac{F}{4}$ 时，物体仍做匀速圆周运动，半径为 $2 R$，则外力对物体所做的功大小是(　　) +**题目内容**: +如右图所示，质量为 $m$ 的物体用细绳经过光滑小孔牵引在光滑水平面上做匀速圆周运动，拉力为某个值 $F$ 时，转动半径为 $R$，当拉力逐渐减小到 $\frac{F}{4}$ 时，物体仍做匀速圆周运动，半径为 $2 R$，则外力对物体所做的功大小是(　　) + **选项**: A. B. C. D．0 + **答案**: A -**解析**: 设当绳的拉力为 $F$ 时，小球做匀速圆周运动时线速度为 $v_1$，则有 $F = m \frac{v_1^2}{R}$。 -当绳的拉力减为 $\frac{F}{4}$ 时，小球做匀速圆周运动的线速度为 $v_2$，则有 $\frac{F}{4} = m \frac{v_2^2}{2 R}$。 +**解析**: +设当绳的拉力为 $F$ 时，小球做匀速圆周运动时线速度为 $v_1$，则有 +$F = m \frac{v_1^2}{R}$ +当绳的拉力减为 $\frac{F}{4}$ 时，小球做匀速圆周运动的线速度为 $v_2$，则有 +$\frac{F}{4} = m \frac{v_2^2}{2 R}$ 在绳的拉力由 $F$ 减为 $\frac{F}{4}$ 的过程中，绳的拉力所做的功为 -$W = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2 = -\frac{1}{4} F R$。 +$W = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2 = -\frac{1}{4} F R$ 所以，绳的拉力所做功的大小为 $\frac{1}{4} F R$。 --- -（由于篇幅限制，后续题目将按相同格式继续整理。） \ No newline at end of file +### 题目6 +**题目序号**: 6 +**题目内容**: +一质量为 $M = 2 . 0 \mathrm{kg}$ 的小物块随足够长的水平传送带一起向右匀速运动，被一水平向左飞来的子弹击中，且子弹从小物块中穿过，子弹和小物块的作用时间极短，如图甲所示。地面观察者记录的小物块被击中后的速度随时间变化关系如图乙所示（图中取向右运动的方向为正方向）。已知传送带的速度保持不变，$g$ 取 $10 \mathrm{m/s}^2$。求： + +(1)传送带的速度 $v$ 的大小； +(2)小物块与传送带之间的动摩擦因数 $\mu$； +(3)传送带对小物块所做的功 + +**答案**: +(1) $2 . 0 \mathrm{m/s}$ +(2) 0.2 +(3) $-12 \mathrm{J}$ + +**解析**: +(1)小物块最后与传送带的运动速度相同，从图象上可读出传送带的速度 $v$ 的大小为 $2 . 0 \mathrm{m/s}$。 +(2)由速度图象可得，小物块在滑动摩擦力的作用下做匀变速运动的加速度为 $a = \Delta v / \Delta t = 2 . 0 \mathrm{m/s}^2$ +由牛顿第二定律得 $f = \mu M g = M a$ +得到小物块与传送带之间的动摩擦因数 $\mu = 0 . 2$ +(3)从子弹离开小物块到小物块与传送带一起匀速运动的过程中，设传送带对小物块所做的功为 $W$，由动能定理得： +$W = \Delta E_k = \frac{1}{2} M v^2 - \frac{1}{2} M v_1^2$ +从速度图象可知：$v_1 = 4 . 0 \mathrm{m/s}$，$v_2 = v = 2 . 0 \mathrm{m/s}$ +解得：$W = -12 \mathrm{J}$。 + +--- + +（由于篇幅限制，剩余题目将逐步补充。） \ No newline at end of file