ÌâÄ¿ÄÚÈÝ

ÈçͼËùʾΪһˮƽ´«ËÍ´ø×°ÖõÄÄ£ÐÍʾÒâͼ£¬´«ËÍ´øÁ½¶ËµãAÓëB¼äµÄ¾àÀëL=4.0 m£¬´«ËÍ´øÒÔv=2.5 m/sµÄËÙ¶ÈÏòÓÒ×öÔÈËÙÖ±ÏßÔ˶¯£¬ÓÐÒ»¿ÉÊÓΪÖʵãµÄÎï¿éÓë´«ËÍ´ø¼äµÄ¶¯Ä¦²ÁÒòÊý¦Ì=0.2£¬Îï¿é´ÓA´¦ÒÔv0=5.0m/sµÄˮƽËÙ¶ÈÏòÓÒ»¬ÉÏ´«ËÍ´ø£¬ÖØÁ¦¼ÓËÙ¶Èg=10 m/s2¡£Çó£º
(1)СÎï¿é»¬ÖÁB´¦µÄËÙ¶È´óС£»
(2)СÎï¿éÓÉA»¬µ½BµÄ¹ý³ÌÖÐÏà¶Ô´«ËÍ´ø»¬¶¯µÄ¾àÀë¡£
½â£º(1)¼ÙÉèСÎï¿éÔÚ´«ËÍ´øÉÏÒ»Ö±×öÔȼõËÙÔ˶¯£¬Ð¡Îï¿éµ½BʱËÙ¶ÈΪv1£¬Ô˶¯µÄ¼ÓËÙ¶È´óСΪa£¬ÔòÓÐ
¦Ìmg=ma

½âµÃv1=3 m/s
ÓÉÓÚv1>v£¬ËùÒÔ¼ÙÉèÕýÈ·£¬Ð¡Îï¿éµ½BʱËÙ¶ÈΪv1=3 m/s
(2)ÉèСÎï¿éÓÉA»¬µ½BµÄ¹ý³Ì£¬¾­Àúʱ¼äΪt£¬´«ËÍ´øÎ»ÒÆÎªx£¬Ïà¶ÔÎ»ÒÆÎª¡÷x£¬ÔòÓÐ
v1=v0-at
x=vt
¡÷x=L-x
½âµÃ¡÷x=1.5 m
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø