Question

9£®Èçͼ£¬¾ØÐε¥Ôѵ¼Ïß¿òabcdµÄÖÊÁ¿Îªm£¬µç×èΪR£¬Æäab±ß³¤L₁¡¢bc±ß³¤L₂£¬¿ªÊ¼Ê±Ïß¿ò¾²Ö¹ÓÚÊúֱƽÃæÄÚÇÒab±ßˮƽ£¬Ïß¿òÏ·½Ò»¶¨¾àÀë´¦µÄÒ»¾ØÐÎÇøÓòÄÚ´æÔÚ·½Ïò´¹Ö±ÓÚÏß¿òƽÃæ¡¢´Å¸ÐӦǿ¶È´óСΪBµÄÔÈÇ¿´Å³¡ÉÏ¡¢Ï±߽ç¾ùÓëÏß¿òab±ßƽÐУ¨´Å³¡ÇøÓòÄڵĸ߶ȴóÓÚL₂£©£¬ÏÖ½«Ïß¿òÓɾ²Ö¹ÆðÊÍ·Å£¬ÉèÏß¿òÏÂÂä¹ý³ÌÖÐab±ßÓë´Å³¡±ß½çʼÖÕƽÐÐÇÒ·½Ïòˮƽ£¬ÒÑÖªab±ß¸Õ½øÈë´Å³¡ºÍ¸Õ´©³ö´Å³¡Ê±£¬Ïß¿ò¶¼×÷¼õËÙÖ±ÏßÔ˶¯£¬¼ÓËٶȴóС¾ùΪ$\frac{g}{4}$£¬¶øcd±ß³ö´Å³¡Ç°Ë²¼ä£¬Ïß¿ò¼ÓËÙ¶ÈÇ¡ºÃΪÁ㣬²»¼Æ¿ÕÆø×èÁ¦£¬ÖØÁ¦¼ÓËÙ¶ÈΪg£®Çó£º
£¨1£©ÔÚÏß¿ò½øÈë´Å³¡¹ý³ÌÖУ¬Í¨¹ýÏß¿òµ¼Ïßijһ½ØÃæµÄµçÁ¿q£»
£¨2£©ÔÚcd±ß¸Õ½øÈë´Å³¡Ê±£¬Ïß¿òËٶȴóСv£»
£¨3£©Ïß¿ò´©Ô½´Å³¡ÇøÓò¹ý³ÌÖУ¬Ïß¿òËù²úÉúµÄÈÈÁ¿Q£»
£¨4£©´Å³¡ÇøÓòµÄ¸ß¶ÈH£»
£¨5£©Ïß¿ò´Ó¿ªÊ¼ÏÂÂäµ½À뿪´Å³¡¹ý³ÌËù¾­ÀúµÄʱ¼ät£®

Answer 1

·ÖÎö £¨1£©¸ù¾Ý·¨À­µÚµç´Å¸ÐÓ¦¶¨ÂÉ¡¢Å·Ä·¶¨Âɺ͵çÁ÷µÄ¶¨ÒåʽÇó½âµçÁ¿q£®
£¨2£©¸ù¾ÝÏß¿ò½øÈëºÍ´©³ö´Å³¡¹ý³ÌµÄÏàËÆÐÔÖªµÀ£¬cd±ß½ø´Å³¡Ë²¼ä¼ÓËÙ¶ÈΪÁ㣬ÓÉ°²ÅàÁ¦ÓëËٶȵĹØϵ¼°Æ½ºâÌõ¼þÇóÏß¿òµÄËٶȴóСv£®
£¨3£©¸ù¾ÝÅ£¶ÙµÚ¶þ¶¨ÂɺͰ²ÅàÁ¦ÓëËٶȵĹØϵʽ£¬Çó³öÏß¿ò¸Õ½øÈë´Å³¡Ê±µÄËٶȣ¬ÔÙÓÉÄÜÁ¿Êغ㶨ÂÉÇóÈÈÁ¿Q£®
£¨4£©Ïß¿òÍêÈ«Ôڴų¡ÖÐÔ˶¯Ê±²»²úÉú¸ÐÓ¦µçÁ÷£¬²»ÊÜ°²ÅàÁ¦£¬×ñÊØ»úеÄÜÊغ㣬ÓÉ»úеÄÜÊغ㶨ÂÉÇóH£®
£¨5£©Ñо¿Ïß¿ò½øÈë´Å³¡µÄ¹ý³Ì£¬ÓÉÅ£¶ÙµÚ¶þ¶¨ÂɺͼÓËٶȵĶ¨Òåʽ£¬ÔËÓûý·Ö·¨Çóʱ¼ä£®ÓÉÔȱäËÙÖ±ÏßÔ˶¯µÄλÒÆʱ¼ä¹«Ê½Çó³öÏß¿ò´Ócd±ß¸Õ´Å³¡µ½ab±ß¸Õ³ö´Å³¡µÄʱ¼ä£®´Ó¶ø¿ÉÇóµÃ×Üʱ¼ä£®

½â´ð ½â£º£¨1£©ÔÚÏß¿ò½øÈë´Å³¡¹ý³ÌÖУ¬Í¨¹ýÏß¿òµ¼Ïßijһ½ØÃæµÄµçÁ¿ q=$\overline{I}t$=$\frac{\overline{E}}{R}t$=$\frac{B{L}_{1}\overline{v}t}{R}$=$\frac{B{L}_{1}{L}_{2}}{R}$
£¨2£©¸ù¾ÝÏß¿ò½øÈëºÍ´©³ö´Å³¡¹ý³ÌµÄÏàËÆÐÔÖªµÀ£¬cd±ß½ø´Å³¡Ë²¼ä¼ÓËÙ¶ÈΪÁ㣬ÔòÓÐ
  mg=F_°²£»
ÓÖ°²ÅàÁ¦ F_°²=BIL₁=B$\frac{B{L}_{1}v}{R}$L₁=$\frac{{B}^{2}{L}_{1}^{2}v}{R}$
ÁªÁ¢½âµÃ v=$\frac{mgR}{{B}^{2}{L}_{1}^{2}}$
£¨3£©ÉèÏß¿ò¸Õ½øÈë´Å³¡Ê±µÄËÙ¶ÈΪv₀£®
¸ù¾ÝÅ£¶ÙµÚ¶þ¶¨Âɵãº$\frac{{B}^{2}{L}_{1}^{2}{v}_{0}}{R}$-mg=ma=m$•\frac{g}{4}$
½âµÃ v₀=$\frac{5mgR}{4{B}^{2}{L}_{1}^{2}}$
Ïß¿ò´©Ô½´Å³¡ÇøÓò¹ý³ÌÖУ¬¸ù¾ÝÄÜÁ¿Êغ㶨ÂɵÃ
  Q=2¡Á[£¨$\frac{1}{2}m{v}_{0}^{2}-\frac{1}{2}m{v}^{2}$£©+mgL₁]=$\frac{9{m}^{3}{g}^{2}{R}^{2}}{32{B}^{4}{L}_{1}^{4}}$+2mgL₁
£¨4£©¶ÔÓÚÏß¿òÍêÈ«Ôڴų¡ÖÐÔ˶¯µÄ¹ý³Ì£¬ÓÉÔ˶¯Ñ§¹«Ê½ÓÐ
    v₀²-v²=2g£¨H-2L₂£©
½âµÃ H=$\frac{9{m}^{2}g{R}^{2}}{32{B}^{4}{L}_{1}^{4}}$+2L₂£®
£¨5£©Ïß¿ò×ÔÓÉÏÂÂäµÄʱ¼ä t₁=$\frac{{v}_{0}}{g}$=$\frac{5mR}{4{B}^{2}{L}_{1}^{2}}$
Ñо¿Ïß¿ò½øÈë´Å³¡µÄ¹ý³Ì£¬ÓÉÅ£¶ÙµÚ¶þ¶¨ÂɵÃ
  $\frac{{B}^{2}{L}_{1}^{2}{v}_{0}}{R}$-mg=ma=m$\frac{¡÷v}{¡÷t}$
¼´ $\frac{{B}^{2}{L}_{1}^{2}{v}_{0}}{R}$¡÷t-mg¡÷t=m¡÷v
Á½±ßÇóºÍµÃ£º$\sum_{\;}^{\;}$£¨$\frac{{B}^{2}{L}_{1}^{2}{v}_{0}}{R}$¡÷t£©-$\sum_{\;}^{\;}$£¨mg¡÷t£©=$\sum_{\;}^{\;}$m¡÷v
¿ÉµÃ $\frac{{B}^{2}{L}_{1}^{2}}{R}$•L₂-mgt₂=m£¨v₀-v£©
½âµÃÏß¿ò½øÈë´Å³¡µÄʱ¼ä t₂=$\frac{{B}^{2}{L}_{1}^{2}{L}_{2}}{mgR}$-$\frac{mR}{{B}^{2}{L}_{1}^{2}}$
Ïß¿ò´Ócd±ß¸Õ´Å³¡µ½ab±ß¸Õ³ö´Å³¡µÄʱ¼äÉèΪt₃£®
Ôò t₃=$\frac{{v}_{0}-v}{g}$=$\frac{mR}{4{B}^{2}{L}_{1}^{2}}$
¹ÊÏß¿ò´Ó¿ªÊ¼ÏÂÂäµ½À뿪´Å³¡¹ý³ÌËù¾­ÀúµÄʱ¼ä t=t₁+2t₂+t₃=$\frac{2{B}^{2}{L}_{1}^{2}{L}_{2}}{mgR}$-$\frac{mR}{2{B}^{2}{L}_{1}^{2}}$£®
´ð£º
£¨1£©ÔÚÏß¿ò½øÈë´Å³¡¹ý³ÌÖУ¬Í¨¹ýÏß¿òµ¼Ïßijһ½ØÃæµÄµçÁ¿qÊÇ$\frac{B{L}_{1}{L}_{2}}{R}$£»
£¨2£©ÔÚcd±ß¸Õ½øÈë´Å³¡Ê±£¬Ïß¿òËٶȴóСvÊÇ$\frac{mgR}{{B}^{2}{L}_{1}^{2}}$£»
£¨3£©Ïß¿ò´©Ô½´Å³¡ÇøÓò¹ý³ÌÖУ¬Ïß¿òËù²úÉúµÄÈÈÁ¿QÊÇ$\frac{9{m}^{3}{g}^{2}{R}^{2}}{32{B}^{4}{L}_{1}^{4}}$+2mgL₁£»
£¨4£©´Å³¡ÇøÓòµÄ¸ß¶ÈHÊÇ$\frac{9{m}^{2}g{R}^{2}}{32{B}^{4}{L}_{1}^{4}}$+2L₂£»
£¨5£©Ïß¿ò´Ó¿ªÊ¼ÏÂÂäµ½À뿪´Å³¡¹ý³ÌËù¾­ÀúµÄʱ¼ätÊÇ$\frac{2{B}^{2}{L}_{1}^{2}{L}_{2}}{mgR}$-$\frac{mR}{2{B}^{2}{L}_{1}^{2}}$£®

µãÆÀ ±¾ÌâÒª·ÖÎöÇå³þÏß¿òµÄÔ˶¯¹ý³Ì£¬ÊìÁ·ÍƵ¼³öµçÁ¿ºÍ°²ÅàÁ¦µÄ±í´ïʽ£¬ÔËÓÃÅ£¶ÙµÚ¶þ¶¨ÂɺͼÓËٶȵĶ¨ÒåʽÇóÏß¿ò½øÈë´Å³¡µÄʱ¼äÊǹؼü£®