'commit'

dsLightRag/ST3_DocxToMd.py

@@ -9,7 +9,7 @@ async def main():
     file_path = "ShiTi/Docx/《动能定理》巩固练习.docx"
 
     # 转换docx为md
-    get_docx_content_by_pandoc(file_path,'.')
+    get_docx_content_by_pandoc(file_path,'结果.md')
 
 
     # """

dsLightRag/Util/DocxUtil.py

@@ -46,8 +46,10 @@ def get_docx_content_by_pandoc(docx_file, output_file=None):
                 content += line.strip().replace("**", "") + "\n"
                 content = content.replace("\phantom", "")
     # 将content回写到markdown文件
-    if output_file is None:
+    if output_file is not None:
         with open(output_file, 'w', encoding='utf-8') as f:
             f.write(content)
     print("Conversion completed successfully!")
+    # 删除临时文件
+    os.remove(temp_markdown)
     return content.replace("\n\n", "\n").replace("\\", "")

dsLightRag/结果.md

@@ -0,0 +1,13 @@
+【题型】不定项选择
+【题文】如图所示，固定斜面倾角为*θ*，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*μ*~1~、*μ*~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*θ*、*μ*~1~、*μ*~2~间应满足的关系是(　　)
+![5-40.tif](./static/Images/6855d84768e44d74ad95dfb7033bcf40/media/image1.png){width="0.9847222222222223in"
+height="0.6979166666666666in"}A．$tan\theta\text{＝}\frac{\mu_{1} + 2\mu_{2}}{3}$
+B．$tan\theta\text{＝}\frac{2\mu_{1} + \mu_{2}}{3}$
+C．$tan\theta\text{＝}2\mu_{1} - \mu_{2}$
+D．$tan\theta\text{＝}2\mu_{2} - \mu_{1}$
+【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：
+$$mglsin\theta\text{－}\mu_{1}mg\frac{2l}{3}cos\theta\text{－}\mu_{2}mg\frac{l}{3}cos\theta\text{＝}0$$
+解得$tan\theta\text{＝}\frac{2\mu_{1} + \mu_{2}}{3}$，故B正确．
+【知识点】动能定理
+【难度】中