Question

2£®Èçͼ¼×Ëùʾ£¬ÓÃÒ»ÇáÖÊÉþË©×ÅÒ»ÖÊÁ¿ÎªmµÄСÇò£¬ÔÚÊúֱƽÃæÄÚ×öÔ²ÖÜÔ˶¯£¨²»¼ÆÒ»ÇÐ×èÁ¦£©£¬Ð¡ÇòÔ˶¯µ½×î¸ßµãʱÉþ¶ÔСÇòµÄÀ­Á¦ÎªF£¬Ð¡ÇòÔÚ×î¸ßµãµÄËٶȴóСΪv£¬ÆäF-v²Í¼ÏóÈçͼÒÒËùʾ£¬Ôò
£¨¡¡¡¡£©

• A£® ÇáÖÊÉþ³¤Îª$\frac{b}{a}$
• B£® µ±µØµÄÖØÁ¦¼ÓËÙ¶ÈΪ$\frac{a}{m}$
• C£® µ±v²=cʱ£¬ÇáÖÊÉþµÄÀ­Á¦´óСΪ$\frac{ac}{b}$+a
• D£® Ö»Òªv²¡Ýb£¬Ð¡ÇòÔÚ×îµÍµãºÍ×î¸ßµãʱÉþµÄÀ­Á¦²î¾ùΪ6a

Answer 1

·ÖÎö ÔÚ×î¸ßµã£¬¸ù¾ÝÅ£¶ÙµÚ¶þ¶¨ÂɵóöFÓëv²µÄ¹Øϵʽ£¬½áºÏͼÏßµÄбÂʺͽؾàÇó³öµ±µØµÄÖØÁ¦¼ÓËٶȺÍÇáÖÊÉþ³¤£®¸ù¾Ý¶¯Äܶ¨ÀíºÍÅ£¶ÙµÚ¶þ¶¨ÂÉÇó³ö×îµÍµãÀ­Á¦µÄ±í´ïʽ£¬½áºÏ×î¸ßµãµÄÀ­Á¦±í´ïʽÇó³öÀ­Á¦Ö®²î£®

½â´ð ½â£ºA¡¢ÔÚ×î¸ßµã£¬¸ù¾ÝÅ£¶ÙµÚ¶þ¶¨Âɵã¬$F+mg=m\frac{{v}^{2}}{L}$£¬½âµÃF=$m\frac{{v}^{2}}{L}-mg$£¬
¸ù¾Ý×ÝÖá½Ø¾àÖª£¬a=mg£¬ÔòÖØÁ¦¼ÓËÙ¶Èg=$\frac{a}{m}$£¬Í¼ÏßµÄбÂÊk=$\frac{m}{L}=\frac{a}{b}$£¬½âµÃÇáÖÊÉþ³¤L=$\frac{mb}{a}$£¬¹ÊA´íÎó£¬BÕýÈ·£®
C¡¢µ±v²=cʱ£¬Éþ×ÓµÄÀ­Á¦F=$m\frac{{v}^{2}}{L}-mg$=$\frac{mc}{L}-mg=\frac{ac}{b}-a$£¬¹ÊC´íÎó£®
D¡¢ÔÚ×î¸ßµã£¬F=$m\frac{{v}^{2}}{L}-mg$£¬Éè×îµÍµãµÄËÙ¶ÈΪv¡ä£¬¸ù¾Ý¶¯Äܶ¨ÀíµÃ£¬$mg•2L=\frac{1}{2}mv{¡ä}^{2}-\frac{1}{2}m{v}^{2}$£¬¸ù¾ÝÅ£¶ÙµÚ¶þ¶¨Âɵã¬$F¡ä-mg=m\frac{v{¡ä}^{2}}{L}$£¬½âµÃ$F¡ä=mg+m\frac{v{¡ä}^{2}}{L}$£¬
À­Á¦²î$¡÷F=F¡ä-F=2mg+\frac{mv{¡ä}^{2}-m{v}^{2}}{L}$=6mg=6a£¬¹ÊDÕýÈ·£®
¹ÊÑ¡£ºBD£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁËÔ²ÖÜÔ˶¯ÏòÐÄÁ¦¹«Ê½ºÍ¶¯Äܶ¨ÀíµÄ×ÛºÏÓ¦Ó㬹ؼüÒªÕýÈ··ÖÎöСÇòÔ²ÖÜÔ˶¯µÄÏòÐÄÁ¦£¬¶ÔÓÚͼÏßÎÊÌ⣬һ°ãµÄ½âÌâ˼·ÊǵóöÎïÀíÁ¿Ö®¼äµÄ¹Øϵʽ£¬½áºÏͼÏßµÄбÂʺͽؾà½øÐÐÇó½â£®