·ÖÎö Ê×ÏÈ×ö³ö´ÓÉÕ¶ÏÏßµ½íÀÂë1Ó뵯»É·ÖÀë¾ÀúÔÚÕâ¶Îʱ¼äÄÚ£¬¸÷íÀÂëºÍíÀÂëÍÐÅ̵ÄÊÜÁ¦Çé¿öÈçͼ1Ëùʾ£ºÍ¼ÖУ¬F±íʾ¡÷t ʱ¼äÄÚÈÎÒâʱ¿Ìµ¯»ÉµÄµ¯Á¦£¬T ±íʾ¸Ãʱ¿Ì¿ç¹ý»¬ÂÖ×éµÄÇáÉþÖеÄÕÅÁ¦£¬mgÎªÖØÁ¦£¬T0ΪÐü¹ÒÍÐÅ̵ÄÉþµÄÀÁ¦£®·ÖÇå¹ý³Ì£¬ºÏÀíÀûÓÃÕûÌåºÍ¸ôÀë·¨¶ÔÑо¿¶ÔÏó½øÐÐÊÜÁ¦·ÖÎö£¬²¢Áз½³Ì£¬½ø¶øÖð²½½øÐÐÇó½â£®
½â´ð ½â£ºÉè´ÓÉÕ¶ÏÏßµ½íÀÂë1Ó뵯»É·ÖÀë¾ÀúµÄʱ¼äΪ¡÷t£¬ÔÚÕâ¶Îʱ¼äÄÚ£¬¸÷íÀÂëºÍíÀÂëÍÐÅ̵ÄÊÜÁ¦Çé¿öÈçͼ1Ëùʾ£ºÍ¼ÖУ¬F±íʾ¡÷t ʱ¼äÄÚÈÎÒâʱ¿Ìµ¯»ÉµÄµ¯Á¦£¬T ±íʾ¸Ãʱ¿Ì¿ç¹ý»¬ÂÖ×éµÄÇáÉþÖеÄÕÅÁ¦£¬mgÎªÖØÁ¦£¬T0ΪÐü¹ÒÍÐÅ̵ÄÉþµÄÀÁ¦£®ÒòDµÄÖÊÁ¿ºöÂÔ²»¼Æ£¬ÓÐ T0=2T £¨1£©
ÔÚʱ¼ä¡÷t ÄÚÈÎһʱ¿Ì£¬·¨Âë1ÏòÉÏÔ˶¯£¬ÍÐÅÌÏòÏÂÔ˶¯£¬íÀÂë2¡¢3ÔòÏòÉÏÉýÆð£¬µ«íÀÂë2¡¢3ÓëÍÐÅÌËٶȵĴóСÊÇÏàͬµÄ£®ÉèÔÚíÀÂë1Ó뵯»É·ÖÀëµÄʱ¿Ì£¬íÀÂë1µÄËÙ¶È´óСΪv1£¬íÀÂë2¡¢3ÓëÍÐÅÌËٶȵĴóС¶¼ÊÇv2£¬Óɶ¯Á¿¶¨Àí£¬ÓÐ
IF-Img=mv1£¨2£©IT-Img=mv2£¨3£©IT-Img=mv2£¨4£©${I}_{F}+{I}_{mg}-{I}_{{T}_{0}}={mv}_{2}$£¨5£©
ʽÖÐIF¡¢Img¡¢IT¡¢IT0·Ö±ð´ú±íÁ¦F¡¢mg¡¢T¡¢T0ÔÚ¡÷t ʱ¼äÄÚ³åÁ¿µÄ´óС£®×¢Ò⵽ʽ£¨1£©£¬ÓÐIT0=2IT£¨6£©
ÓÉ£¨2£©¡¢£¨3£©¡¢£¨4£©¡¢£¨5£©¡¢£¨6£©¸÷ʽµÃ ${v}_{2}=\frac{1}{3}{v}_{1}$ £¨7£©
ÔÚµ¯»ÉÉ쳤¹ý³ÌÖУ¬µ¯»ÉµÄÉ϶ËÓëíÀÂë1Ò»ÆðÏòÉÏÔ˶¯£¬Ï¶ËÓëÍÐÅÌÒ»ÆðÏòÏÂÔ˶¯£®ÒÔ¡÷l1±íʾÔÚ¡÷t ʱ¼äÄÚµ¯»ÉÉ϶ËÏòÉÏÔ˶¯µÄ¾àÀ룬¡÷l2±íʾÆä϶ËÏòÏÂÔ˶¯µÄ¾àÀ룮ÓÉÓÚÔÚµ¯»ÉÉ쳤¹ý³ÌÖÐÈÎÒâʱ¿Ì£¬ÍÐÅ̵ÄËٶȶ¼ÎªíÀÂë1µÄËٶȵÄ1/3£¬¹ÊÓÐ![]()
${¡÷l}_{2}=\frac{1}{3}¡÷{l}_{1}$£¨8£©ÁíÓС÷l1+¡÷l2=l0£¨9£©
ÔÚµ¯»ÉÉ쳤¹ý³ÌÖУ¬»úеÄÜÊØºã£¬µ¯»Éµ¯ÐÔÊÆÄܵļõÉÙµÈÓÚϵͳ¶¯ÄܺÍÖØÁ¦ÊÆÄܵÄÔö¼Ó£¬¼´ÓÐ
$\frac{1}{2}{{kl}_{0}}^{2}=\frac{1}{2}{{mv}_{1}}^{2}+3¡Á\frac{1}{2}{{mv}_{2}}^{2}+{mg¡÷l}_{1}-{mg¡÷l}_{2}+2{mg¡÷l}_{2}$ £¨10£©
ÓÉ£¨7£©¡¢£¨8£©¡¢£¨9£©¡¢£¨10£©Ê½µÃ ${{v}_{1}}^{2}=\frac{3}{2m}£¨\frac{1}{2}{{kl}_{0}}^{2}-{mgl}_{0}£©$£¨11£©
íÀÂë1Ó뵯»É·Ö¿ªºó£¬íÀÂë×÷ÉÏÅ×Ô˶¯£¬ÉÏÉýµ½×î´ó¸ß¶È¾Àúʱ¼äΪt1£¬ÓÐv1=gt1 £¨12£©
íÀÂë2¡¢3ºÍÍÐÅ̵ÄÊÜÁ¦Çé¿öÈçͼ2Ëùʾ£¬ÒÔa±íʾ¼ÓËٶȵĴóС£¬ÓÐmg-T=ma £¨13£©
mg-T=ma £¨14£©T0-mg=ma£¨15£©T0=2T £¨16£©
ÓÉ£¨14£©¡¢£¨15£©ºÍ£¨16£©Ê½µÃ $a=\frac{1}{3}g$£¨17£©
ÍÐÅ̵ļÓËÙ¶ÈÏòÉÏ£¬³õËÙ¶Èv2ÏòÏ£¬Éè¾Àúʱ¼ät2£¬ÍÐÅÌËٶȱäΪÁ㣬ÓÐv2=at2 £¨18£©
ÓÉ£¨7£©¡¢£¨12£©¡¢£¨17£©ºÍ£¨18£©Ê½£¬µÃ ${t}_{1}={t}_{2}=\frac{{v}_{1}}{g}$£¨19£©¼´íÀÂë1×ÔÓ뵯»É·ÖÀëµ½ËÙ¶ÈΪÁã¾ÀúµÄʱ¼äÓëÍÐÅÌ×Ô·ÖÀëµ½ËÙ¶ÈΪÁã¾ÀúµÄʱ¼äÏàµÈ£®ÓɶԳÆÐÔ¿ÉÖª£¬µ±íÀÂë»Øµ½·ÖÀëλÖÃʱ£¬ÍÐÅÌÒà»Øµ½·ÖÀëλÖ㬼´ÔÙ¾Àút1£¬íÀÂëÓ뵯»ÉÏàÓö£®ÌâÖÐÒªÇóµÄʱ¼ät×Ü=2t1£¨20£©
ÓÉ£¨11£©¡¢£¨12£©¡¢£¨20£©Ê½µÃ ${t}_{×Ü}=\frac{2}{g}\sqrt{\frac{3}{2m}£¨\frac{1}{2}{{kl}_{0}}^{2}-{mgl}_{0}£©}$
µãÆÀ ¸ÃÌâÖÐÉæ¼°»úеÄÜÊØºã£¬¶¯Á¿¶¨ÀíµÈ֪ʶ£¬²¢ÒªÇóѧÉúÊìÖª»¬ÂÖµÄÏà¹ØÖªÊ¶£¬Éæ¼°µÄ¹ý³Ì¸´ÔÓ£¬ÊÜÁ¦·ÖÎö±È½Ï·±Ëö£¬½âÌâÄѶȴó£®
| Äê¼¶ | ¸ßÖÐ¿Î³Ì | Äê¼¶ | ³õÖÐ¿Î³Ì |
| ¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | СÌáÇÙ | B£® | ¸ÖÇÙ | C£® | ÊÖ·çÇÙ |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | ÎÀÐÇÄÚµÄÎïÌåÈÔÊÜÖØÁ¦×÷Ó㬲¢¿ÉÓõ¯»É³ÓÖ±½Ó²â³öËùÊÜÖØÁ¦µÄ´óС | |
| B£® | ÔÚÈκιìµÀÉÏÔ˶¯Ê±£¬µØÇòÇòÐͼÔÚÎÀÐǵĹìµÀÆ½ÃæÄÚ | |
| C£® | ÎÀÐÇÔ˶¯ËÙ¶ÈÒ»¶¨²»³¬¹ý7.9 km/s | |
| D£® | ÎÀÐÇÔËÐÐʱµÄÏòÐļÓËٶȵÈÓÚÎÀÐǹìµÀËùÔÚ´¦µÄÖØÁ¦¼ÓËÙ¶È |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£ºÊµÑéÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | Ëæ×ſƼ¼µÄ½ø²½ÓÀ¶¯»úÒ»¶¨»áÑÐÖÆ³É¹¦ | |
| B£® | Èȹ¦µ±Á¿µÄ²â¶¨ÎªÄÜÁ¿Êغ㶨Âɽ¨Á¢ÁËÀι̵ÄʵÑé»ù´¡ | |
| C£® | ÄÜÁ¿Êغ㶨ÂɾÍÊÇÈÈÁ¦Ñ§µÚÒ»¶¨ÂÉ£¬ÕâÁ½¸ö¶¨ÂÉÊÇµÈ¼ÛµÄ | |
| D£® | ÎïÌåÎüÊÕÈÈÁ¿£¬Í¬Ê±¶ÔÍâ×ö¹¦£¬ÄÚÄÜÒ»¶¨Ôö´ó |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£º¶àÑ¡Ìâ
| A£® | ÖØÁ¦µÄ³åÁ¿µÄ´óС | B£® | µ¯Á¦µÄ³åÁ¿µÄ´óС | ||
| C£® | ºÏÁ¦µÄ³åÁ¿µÄ´óС | D£® | ¸Õµ½´ïµ×¶ËµÄ¶¯Á¿µÄ´óС |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖÐÎïÀí À´Ô´£º ÌâÐÍ£º½â´ðÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¹ú¼ÊѧУÓÅÑ¡ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com