Question

20£®ÖÊÁ¿Îª2kgÎïÌåA×öƽÅ×Ô˶¯£¬ÂäµØʱˮƽ·½ÏòµÄλÒƺÍÊúÖ±·½ÏòµÄλÒƾùΪL=5m£¬²»¿¼ÂÇ¿ÕÆø×èÁ¦µÄÓ°Ï죬ÖØÁ¦¼ÓËÙ¶ÈgÈ¡10m/s²£¬ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ£¨¡¡¡¡£©

• A£® ÎïÌåAÂäµØʱµÄ¶¯Á¿´óСΪ$10\sqrt{5}$kg•m/s
• B£® ÎïÌåAÂäµØʱµÄ¶¯ÄÜΪ100J
• C£® ÎïÌåAÂäµØʱ£¬ËÙ¶ÈÓëˮƽ·½ÏòµÄ¼Ð½ÇÊÇ45¡ã
• D£® ÎïÌåA×öƽÅ×Ô˶¯ÖкÏÁ¦µÄƽ¾ù¹¦ÂÊΪ125W

Answer 1

·ÖÎö ¸ù¾ÝϽµµÄ¸ß¶ÈÇó³öƽÅ×Ô˶¯µÄʱ¼ä£¬½áºÏˮƽλÒƺÍʱ¼äÇó³ö³õËٶȣ¬½áºÏƽÐÐËıßÐζ¨ÔòÇó³öÂäµØµÄËٶȣ¬´Ó¶ø¸ù¾Ý¶¯Á¿ºÍ¶¯ÄܵĹ«Ê½Çó³öÂäµØʱÎïÌåµÄ¶¯Á¿ºÍ¶¯ÄÜ´óС£®¸ù¾ÝƽÐÐËıßÐζ¨ÔòÇó³öÂäµØʱËٶȷ½ÏòÓëˮƽ·½ÏòµÄ¼Ð½Ç£®¸ù¾Ýƽ¾ù¹¦ÂʵĹ«Ê½Çó³öÎïÌåÔÚƽÅ×Ô˶¯µÄ¹ý³ÌÖкÏÁ¦µÄƽ¾ù¹¦ÂÊ£®

½â´ð ½â£ºA¡¢¸ù¾ÝL=$\frac{1}{2}g{t}^{2}$µÃ£ºt=$\sqrt{\frac{2L}{g}}=\sqrt{\frac{2¡Á5}{10}}s=1s$£¬Æ½Å×Ô˶¯µÄ³õËÙ¶ÈΪ£º${v}_{0}=\frac{L}{t}=\frac{5}{1}m/s=5m/s$£¬ÔòÂäµØµÄËÙ¶ÈΪ£º$v=\sqrt{{{v}_{0}}^{2}+2gL}$=$\sqrt{25+2¡Á10¡Á5}$m/s=$5\sqrt{5}m/s$£¬ÎïÌåAÂäµØµÄ¶¯Á¿Îª£ºP=mv=$2¡Á5\sqrt{5}kg•m/s=10\sqrt{5}kg•m/s$£¬¹ÊAÕýÈ·£®
B¡¢ÎïÌåAÂäµØµÄ¶¯ÄÜΪ£º${E}_{k}=\frac{1}{2}m{v}^{2}=\frac{1}{2}¡Á2¡Á125J=125J$£¬¹ÊB´íÎó£®
C¡¢ÉèÂäµØʱËٶȷ½ÏòÓëˮƽ·½ÏòµÄ¼Ð½ÇΪ¦Á£¬ÓУº$cos¦Á=\frac{{v}_{0}}{v}=\frac{5}{5\sqrt{5}}=\frac{\sqrt{5}}{5}$£¬¿ÉÖª¦Á¡Ù45¡ã£¬¹ÊC´íÎó£®
D¡¢Æ½Å×Ô˶¯µÄ¹ý³ÌÖУ¬ºÏÁ¦×ö¹¦Îª£ºW=mgL=20¡Á5J=100J£¬ÔòºÏÁ¦µÄƽ¾ù¹¦ÂÊΪ£ºP=$\frac{W}{t}=\frac{100}{1}W=100W$£¬¹ÊD´íÎó£®
¹ÊÑ¡£ºA£®

µãÆÀ ½â¾ö±¾ÌâµÄ¹Ø¼üÖªµÀƽÅ×Ô˶¯ÔÚˮƽ·½ÏòºÍÊúÖ±·½ÏòÉϵÄÔ˶¯¹æÂÉ£¬Çó³ö³õËٶȺÍÂäµØËÙ¶ÈÊǹؼü£¬ÕÆÎÕ¶¯ÄÜ¡¢¶¯Á¿¡¢¹¦ÂʵĹ«Ê½£¬²¢ÄÜÁé»îÔËÓã®