Question

9£®ÈçͼËùʾ£¬´Ö²ÚˮƽµØÃæABÓë°ë¾¶R=0.9mµÄ¹â»¬°ëÔ²¹ìµÀBCDÏàÁ¬½Ó£¬ÇÒÔÚͬһÊúֱƽÃæÄÚ£¬OÊÇBCDµÄÔ²ÐÄ£¬BODÔÚͬһÊúÖ±ÏßÉÏ£®ÖÊÁ¿m=2kgµÄСÎïÌåÔÚˮƽºãÁ¦F=20NµÄ×÷ÓÃÏ£¬´ÓAµãÓɾ²Ö¹¿ªÊ¼×öÔȼÓËÙÖ±ÏßÔ˶¯£¬ÒÑÖªx_AB=10m£¬Ð¡Îï¿éÓëˮƽµØÃæ¼äµÄ¶¯Ä¦²ÁÒòÊýΪ¦Ì=0.5£®µ±Ð¡Îï¿éÔ˶¯µ½Bµãʱ³·È¥Á¦F£®È¡ÖØÁ¦¼ÓËÙ¶Èg=10m/s²£®Çó£º
£¨1£©Ð¡Îï¿éµ½´ïBµãʱËٶȵĴóСv_B£»
£¨2£©Ð¡Îï¿éÔ˶¯µ½Dµãʱ¹ìµÀ¶ÔÎïÌåµÄѹÁ¦N£®

Answer 1

·ÖÎö £¨1£©¶ÔABÔ˶¯¹ý³ÌÓ¦Óö¯Äܶ¨Àí¼´¿ÉÇó½â£»
£¨2£©¶Ô´ÓBµ½DµÄÔ˶¯¹ý³ÌÓ¦ÓûúеÄÜÊغãÇóµÃÔÚD´¦µÄËٶȣ¬È»ºóÓÉÅ£¶ÙµÚ¶þ¶¨Âɼ´¿ÉÇóµÃѹÁ¦£®

½â´ð ½â£º£¨1£©Îï¿é´ÓAµ½B×öÔȼÓËÙÔ˶¯£¬Ö»ÓÐFºÍĦ²ÁÁ¦×ö¹¦£¬¸ù¾Ý¶¯Äܶ¨ÀíÓУº$F{x}_{AB}-¦Ìmg{x}_{AB}=\frac{1}{2}m{{v}_{B}}^{2}$
¹ÊÓУº${v}_{B}=\sqrt{\frac{2£¨F-¦Ìmg£©{x}_{AB}}{m}}=10m/s$£»
£¨2£©Îï¿é´ÓBµ½DÖ»ÓÐÖØÁ¦×ö¹¦£¬¹Ê¸ù¾Ý»úеÄÜÊغ㶨ÂÉÓУº$\frac{1}{2}m{{v}_{B}}^{2}=2mgR+\frac{1}{2}m{{v}_{D}}^{2}$£»
Îï¿éÔÚD´¦Êܵ½µÄÁ¦¶¼ÔÚÊúÖ±·½ÏòÉÏ£¬¹ÊºÏÍâÁ¦ÌṩÏòÐÄÁ¦£¬ÓУº${F}_{Ïò}=\frac{m{{v}_{D}}^{2}}{R}=\frac{m{{v}_{B}}^{2}}{R}-4mg$=$\frac{1280}{9}N$£»
¹ÊСÎï¿éÔ˶¯µ½Dµãʱ¹ìµÀ¶ÔÎïÌåµÄѹÁ¦Îª£º$N={F}_{Ïò}-mg=\frac{1100}{9}N$£»
´ð£º£¨1£©Ð¡Îï¿éµ½´ïBµãʱËٶȵĴóСv_BΪ10m/s£»
£¨2£©Ð¡Îï¿éÔ˶¯µ½Dµãʱ¹ìµÀ¶ÔÎïÌåµÄѹÁ¦NΪ$\frac{1100}{9}N$£®

µãÆÀ ¾­µäÁ¦Ñ§ÎÊÌâÒ»°ãÏȶÔÎïÌå½øÐÐÊÜÁ¦·ÖÎö£¬ÇóµÃºÏÍâÁ¦¼°Ô˶¯¹ý³Ì×ö¹¦Çé¿ö£¬È»ºó¸ù¾ÝÅ£¶Ù¶¨ÂÉ¡¢¶¯Äܶ¨Àí¼°¼¸ºÎ¹ØϵÇó½â£®