ÌâÄ¿ÄÚÈÝ

3£®ÈçͼËùʾ£¬ÔÚÉý½µ»úÖУ¬ÓÃF=20NµÄÁ¦½«ÖØÎª2NµÄÎï¿éѹÔÚÊúÖ±µÄǽ±ÚÉÏ£¬m×ÜÊÇÏà¶ÔÉý½µ»ú±£³Ö¾²Ö¹£¬µ±Éý½µ»úÒÔ2m/sµÄËÙ¶ÈÔÈËÙÉÏÉýʱ£¬Éý½µ»úµÄ±Ú¶ÔmµÄĦ²ÁÁ¦Îªf1£¬µ±Éý½µ»úÒÔ4.9m/s2µÄ¼ÓËٶȳ¯Ï¼ÓËÙÔ˶¯Ê±£¬±Ú¶ÔmµÄĦ²ÁÁ¦´óСΪf2£¬Ôò£¨g=9.8m/s2£©£¨¡¡¡¡£©
A£®f1=2N£¬f2=2N£¬f2µÄ·½ÏòÏòÉÏB£®f1=f2=3N£¬f2µÄ·½ÏòÏòÏÂ
C£®f1=2N£¬f2=1N£¬f2µÄ·½ÏòÏòÉÏD£®f1=2N£¬f2=3N£¬f2µÄ·½ÏòÏòÏÂ

·ÖÎö µ±Éý½µ»úÔÈËÙÉÏÉýʱ£¬¸ù¾ÝƽºâÇó³öÎï¿éËùÊܵľ²Ä¦²ÁÁ¦´óС£¬µ±Éý½µ»ú¼ÓËÙϽµÊ±£¬¸ù¾ÝÅ£¶ÙµÚ¶þ¶¨ÂÉÇó³öĦ²ÁÁ¦µÄ´óСºÍ·½Ïò£®

½â´ð ½â£ºµ±Éý½µ»úÔÈËÙÉÏÉýʱ£¬¸ù¾ÝƽºâÓУºf1=mg=2N£¬·½ÏòÊúÖ±ÏòÉÏ£®
µ±Éý½µ»ú¼ÓËÙϽµÊ±£¬¸ù¾ÝÅ£¶ÙµÚ¶þ¶¨ÂÉÓУºmg-f2=ma£¬½âµÃ${f}_{2}=mg-ma=2-\frac{2}{9.8}¡Á4.9N=1N$£¬·½ÏòÊúÖ±ÏòÉÏ£®
¹ÊÑ¡£ºC£®

µãÆÀ ±¾Ì⿼²éÁËÅ£¶ÙµÚ¶þ¶¨Âɺ͹²µãÁ¦Æ½ºâµÄ»ù±¾ÔËÓã¬ÖªµÀÎï¿éËùÊܵÄĦ²ÁÁ¦Îª¾²Ä¦²ÁÁ¦£¬²»ÒªÊܵ½ÍÆÁ¦FµÄ¸ÉÈÅ£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø