Question

4£®ÈçͼËùʾ£¬Ë®Æ½Â·ÃæCDµÄ×ó²àÓÐÒ»¹Ì¶¨µÄƽ̨£¬Æ½Ì¨ÉϱíÃæAB³¤s=3m£®¹â»¬°ëÔ²¹ìµÀAFEÊúÖ±¹Ì¶¨ÔÚƽ̨ÉÏ£¬Ô²¹ìµÀ°ë¾¶R=0.4m£¬×îµÍµãÓëƽ̨ABÏàÇÐÓÚA£¬°å³¤L₁=2m£¬ÉϱíÃæÓëƽ̨µÈ¸ß£¬µ±°åµÄ×ó¶Ë¾àÀëƽ̨L=2mʱ£¬·ÅÔÚ°åµÄ×îÓÒ¶ËÖÊÁ¿m=1kgµÄСÎï¿é£¬Ëæ°åÒ»ÆðÒÔËÙ¶Èv₀=8m/sÏòƽ̨Ô˶¯£®µ±°åÓëƽ̨µÄÊúֱǽ±ÚÅöײºó£¬°åÁ¢¼´Í£Ö¹Ô˶¯£¬Îï¿éÔÚ°åÉϼÌÐø»¬¶¯£®¼ºÖª°åÓë·ÃæµÄ¶¯Ä¦²ÁÒòÊý¦Ì₁=0.05£¬Îï¿éÓë°åÉϱíÃæ¼°¹ìµÀABµÄ¶¯Ä¦²ÁÒòÊý¦Ì₂=0.1£¬È¡g=10m/s²£®
£¨1£©ÇóÎï¿é½øÈëÔ²¹ìµÀʱ¶Ô¹ìµÀÉÏAµãµÄѹÁ¦£»
£¨2£©ÅжÏÎï¿éÄÜ·ñµ½´ïÔ²¹ìµÀµÄ×îó{µãEÈç¹ûÄÜ£¬ÇóÎï¿éÀ뿪EºóÔÚƽ̨ÉϵÄÂäµãµ½AµÄ¾àÀ룻Èç¹û²»ÄÜ£¬Ôò˵Ã÷ÀíÓÉ£®

Answer 1

·ÖÎö £¨1£©¶Ô°åºÍÎï¿éÕûÌåÑо¿£¬½áºÏÅ£¶ÙµÚ¶þ¶¨ÂÉÇó³öÔȼõËÙÖ±ÏßÔ˶¯µÄ¼ÓËٶȴóС£¬½áºÏËÙ¶ÈλÒƹ«Ê½Çó³ö°åµ½´ïBµãµÄËٶȣ¬¸ù¾ÝÅ£¶ÙµÚ¶þ¶¨ÂÉÇó³öÎï¿éµÄ¼ÓËٶȴóС£¬½áºÏËÙ¶ÈλÒƹ«Ê½Çó³öÎï¿éµ½´ïAµãµÄËٶȣ¬¸ù¾ÝÅ£¶ÙµÚ¶þ¶¨ÂÉÇó³öAµãµÄÖ§³ÖÁ¦´óС£¬´Ó¶øµÃ³öѹÁ¦µÄ´óС£®
£¨2£©¸ù¾Ý¶¯Äܶ¨ÀíÇó³öÎï¿éµ½´ï×î¸ßµãµÄËٶȣ¬Óë×î¸ßµãµÄÁÙ½çËٶȱȽϣ¬ÅжÏÊÇ·ñÄܹ»µ½´ï×î¸ßµã£¬ÈôÄܵ½´ï×î¸ßµã£¬¸ù¾Ý¸ß¶ÈÇó³öÔ˶¯µÄʱ¼ä£¬Í¨¹ýCµãµÄËٶȺÍʱ¼äÇó³öˮƽλÒÆ£®

½â´ð ½â£º£¨1£©¶Ô°åºÍÎï¿éÕûÌå·ÖÎö£¬ÕûÌåÔȼõËÙÖ±ÏßÔ˶¯µÄ¼ÓËٶȴóСa=$\frac{{¦Ì}_{1}£¨M+m£©g}{M+m}={¦Ì}_{1}g=0.5m/{s}^{2}$£¬
¸ù¾Ý${{v}_{0}}^{2}-{{v}_{1}}^{2}=2aL$µÃ£¬
${v}_{1}=\sqrt{{{v}_{0}}^{2}-2aL}$=$\sqrt{64-2¡Á0.5¡Á2}=\sqrt{62}$m/s£¬
ÔòÎï¿éµ½´ïAµãµÄËÙ¶È${v}_{A}=\sqrt{{{v}_{1}}^{2}-2{¦Ì}_{2}g£¨{L}_{1}+s£©}$=$\sqrt{62-2¡Á1¡Á5}m/s=\sqrt{52}$m/s£¬
¸ù¾ÝÅ£¶ÙµÚ¶þ¶¨Âɵã¬N-mg=m$\frac{{{v}_{A}}^{2}}{R}$£¬
½âµÃN=mg+m$\frac{{{v}_{A}}^{2}}{R}$=10+$1¡Á\frac{52}{0.4}$N=140N£®
¸ù¾ÝÅ£¶ÙµÚÈý¶¨ÂÉÖª£¬ÎïÌå¶ÔÔ²¹ìµÀAµãµÄѹÁ¦Îª140N£®
£¨2£©¸ù¾Ý¶¯Äܶ¨ÀíµÃ£¬$-mg•2R=\frac{1}{2}m{{v}_{C}}^{2}-\frac{1}{2}m{{v}_{A}}^{2}$£¬
´úÈëÊý¾Ý½âµÃv_C=6m/s£®
×î¸ßµãµÄÁÙ½çËÙ¶È${v}_{C}¡ä=\sqrt{gR}$=$\sqrt{10¡Á0.4}$m/s=2m/s£¬
¿ÉÖªÎïÌåÄܵ½´ï×î¸ßµã£¬
¸ù¾Ý2R=$\frac{1}{2}g{t}^{2}$µÃ£¬t=$\sqrt{\frac{4R}{g}}=\sqrt{\frac{4¡Á0.4}{10}}s=0.4s$£¬
ˮƽλÒÆx=v_Ct=6¡Á0.4m=2.4m£®
´ð£º£¨1£©Îï¿é½øÈëÔ²¹ìµÀʱ¶Ô¹ìµÀÉÏAµãµÄѹÁ¦Îª140N£®
£¨2£©Îï¿éÄܵ½´ïÔ²¹ìµÀµÄ×î¸ßµã£¬Îï¿éÀ뿪EºóÔÚƽ̨ÉϵÄÂäµãµ½AµÄ¾àÀëΪ2.4m£®

µãÆÀ ±¾Ì⿼²éÁËÔ²ÖÜÔ˶¯¡¢Ö±ÏßÔ˶¯¡¢Æ½Å×Ô˶¯ÓëÅ£¶ÙµÚ¶þ¶¨ÂÉ¡¢¶¯Äܶ¨ÀíÒÔ¼°Ô˶¯Ñ§¹«Ê½µÄ×ۺϣ¬ÖªµÀÔ²ÖÜÔ˶¯×î¸ßµãºÍ×îµÍµãÏòÐÄÁ¦µÄÀ´Ô´ÒÔ¼°Æ½Å×Ô˶¯ÔÚˮƽ·½ÏòºÍÊúÖ±·½ÏòÉϵÄÔ˶¯¹æÂÉÊǽâ¾ö±¾ÌâµÄ¹Ø¼ü£®