Question

14£®ÈçͼËùʾ£¬²ÄÁÏÍêÈ«ÏàͬµÄÎï¿éA£¨¿É¿´×÷Öʵ㣩ºÍ±¡°åB£¨ºñ¶È²»¼Æ£©µÄÖÊÁ¿·Ö±ðΪ2mºÍm£¬AB¼äµÄ¶¯Ä¦²ÁÒòÊýΪ0.1£¬BÓëµØÃæÖ®¼äµÄ¶¯Ä¦²ÁÒòÊýΪ0.2£¬±¡°åBµÄ³¤¶ÈΪL£®¿ªÊ¼Ê±Îï¿éAºÍ±¡°åB¾ù¾²Ö¹ÔÚˮƽµØÃæÉÏ£¬Îï¿éAÔÚ±¡°åB×îÓҶˣ®Ä³Ê±¿ÌÆð£¬ÓÃF=1.2mgµÄˮƽÏòÓҵĺãÁ¦½«±¡°åB´ÓÎï¿éAµÄÏ·½³é³ö gÈ¡10m/s²

£¨1£©ÇóÊ©¼ÓˮƽºãÁ¦µÄ˲¼äÎï¿éAºÍ±¡°åBµÄ¼ÓËٶȣ»
£¨2£©Îï¿éA´Ó±¡°åBµÄ×ó¶ËµôϺ󣬱£³Ö¶Ô±¡°åµÄˮƽÀ­Á¦²»±ä£¬ÇóÎï¿éA¸Õ¸ÕÍ£Ö¹Ô˶¯Ê±¾à±¡°åB×ó¶ËµÄ¾àÀ룮

Answer 1

·ÖÎö £¨1£©ÔÚÊ©¼ÓÍâÁ¦Ë²¼ä£¬¶ÔABÊÜÁ¦·ÖÎö£¬ÓÉÅ£¶ÙµÚ¶þ¶¨ÂÉÇóµÄ¼ÓËٶȣ»
£¨2£©µ±AÔÚBÉÏÏà¶Ô»¬¶¯Ê±£¬ÓÉÅ£¶ÙµÚ¶þ¶¨ÂÉÇóµÄ¸÷×ԵļÓËٶȣ¬»¬Âäʱ¸ù¾ÝÔ˶¯Ñ§¹«Ê½ÇóµÄ·¢ÉúµÄÏà¶ÔλÒÆ£¬ÇóµÄʱ¼ä£¬ÓÉËÙ¶Èʱ¼ä¹«Ê½ÇóµÄ»¬Âäʱ¸÷×ÔµÄËٶȣ¬ÍÑÀëºó·Ö±ðÇóµÄ¸÷×ԵļÓËٶȣ¬ÓÉËÙ¶Èʱ¼ä¹«Ê½ÇóµÄA¼õËÙËùÐèʱ¼ä£¬ÓÉÔ˶¯Ñ§¹«Ê½ÇóµÄÔÚʱ¼ätÄÚABµÄλÒƼ´¿ÉÇóµÃÏà¾à¾àÀë

½â´ð ½â£º£¨1£©Ê©¼ÓˮƽºãÁ¦µÄ˲¼ä£¬B¾²Ö¹²»¶¯£¬¹ÊAÏà¶ÔÓÚB¾²Ö¹£¬¹ÊAÊܵ½µÄºÏÁ¦ÎªÁ㣬¹Ê¼ÓËÙ¶ÈΪ0£¬
¶ÔBÓÉÅ£¶ÙµÚ¶þ¶¨ÂÉ¿ÉÖª
F-¦Ì₂£¨2m+m£©g=ma
a=6m/s²
£¨2£©AB·¢ÉúÏà¶Ô»¬¶¯ºó£¬AµÄ¼ÓËÙ¶ÈΪ${a}_{1}=\frac{2¦Ìmg}{2mg}=1m/{s}^{2}$
BµÄ¼ÓËÙ¶ÈΪ${a}_{2}=\frac{F-{¦Ì}_{1}•2mg-{¦Ì}_{2}£¨2mg+mg£©}{m}=4m/{s}^{2}$
Éè¸ÕºÃµôÏÂʱÓÃʱt£¬Ôò$L=\frac{1}{2}{a}_{2}{t}^{2}-\frac{1}{2}{a}_{1}{t}^{2}$
½âµÃ
t=$\sqrt{\frac{2L}{3}}$
´ËʱABµÄËÙ¶ÈΪ${v}_{1}={a}_{1}t=\sqrt{\frac{2L}{3}}$
BµÄËÙ¶ÈΪ${v}_{2}={a}_{2}t=4\sqrt{\frac{2L}{3}}$
»¬ÂäºóA¼õËÙµ½ÁãËùÐèʱ¼äΪ$t¡ä=\frac{{v}_{1}}{{a}_{2}}=\frac{1}{4}\sqrt{\frac{2L}{3}}$
»¬ÂäºóBµÄ¼ÓËÙ¶ÈΪ${a}_{3}=\frac{{F-¦Ì}_{2}mg}{m}=10m/{s}^{2}$
¹Ê»¬ÂäºóAÇ°½øµÄλÒÆΪ${x}_{1}={v}_{1}t¡ä-\frac{1}{2}{a}_{2}t{¡ä}^{2}$
BÇ°½øµÄλÒÆΪ${x}_{2}={v}_{2}t¡ä+\frac{1}{2}{a}_{3}t{¡ä}^{2}$
¹ÊÁ½ÕßÏà¾à¾àÀëΪ¡÷x=x₂-x₁
ÁªÁ¢½âµÃ$¡÷x=\frac{19L}{24}$
´ð£º£¨1£©Ê©¼ÓˮƽºãÁ¦µÄ˲¼äÎï¿éAºÍ±¡°åBµÄ¼ÓËٶȷֱðΪ0£¬6m/s²£»
£¨2£©Îï¿éA¸Õ¸ÕÍ£Ö¹Ô˶¯Ê±¾à±¡°åB×ó¶ËµÄ¾àÀëΪ$\frac{19L}{24}$£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁËÅ£¶ÙµÚ¶þ¶¨ÂɺÍÔ˶¯Ñ§¹«Ê½£¬¼ÓËÙ¶ÈÊǽâ¾öÎÊÌâµÄÖмäÇÅÁº