Question

11£®ÈçͼËùʾ£¬°ë¾¶R=2mµÄËÄ·ÖÖ®Ò»´Ö²ÚµÄÔ²»¡¹ìµÀABÓÚÊúֱƽÃæÄÚ£¬¹ìµÀB¶ËÇÐÏßˮƽ£¬ÇÒ¾àˮƽµØÃæµÄ¸ß¶Èh=1.25m£®ÏÖ½«Ò»ÖÊÁ¿m=0.2kgµÄСÇò£¨¿ÉÊÓΪÖʵ㣩´ÓAµãÓɾ²Ö¹ÊÍ·Å£¬Ð¡»¬¿éÑØÔ²¹ìµÀÔ˶¯ÖÁBµãÒÔv=5m/sµÄËÙ¶ÈÅ׳ö£¨g=10m/s²£©£¬Çó£º
£¨1£©Ð¡»¬¿é´ÓBµãÅ׳öÖÁµØÃæÔ˶¯µÄˮƽλÒÆ´óС£»
£¨2£©Ð¡»¬¿é×ŵØʱµÄËٶȴóС£»
£¨3£©Ð¡»¬¿éÑØÔ²¹ìµÀÔ˶¯µÄ¹ý³ÌÖÐËùÊÜĦ²ÁÁ¦×öµÄ¹¦ÊǶàÉÙ£¿

Answer 1

·ÖÎö £¨1£©À뿪Bµãºó×öƽÅ×Ô˶¯£¬¸ù¾Ý·ÖλÒƹ«Ê½ÁÐʽÇó½âˮƽ·ÖλÒÆ£»
£¨2£©¶ÔƽÅ×¹ý³Ì£¬¸ù¾Ý¶¯Äܶ¨ÀíÁÐʽÇó½âÂäµØËٶȣ»
£¨3£©¶Ô´ÓAµ½BµÄ¹ý³Ì£¬¸ù¾Ý¶¯Äܶ¨ÀíÁÐʽÇó½âĦ²ÁÁ¦×öµÄ¹¦£®

½â´ð ½â£º£¨1£©À뿪Bµãºó×öƽÅ×Ô˶¯£¬¹Ê£º
x=vt
h=$\frac{1}{2}g{t}^{2}$
ÁªÁ¢½âµÃ£º
x=v$\sqrt{\frac{2h}{g}}$=5¡Á$\sqrt{\frac{2¡Á1.25}{10}}$=2.5m
£¨2£©¶ÔƽÅ×¹ý³Ì£¬¸ù¾Ý¶¯Äܶ¨Àí£¬ÓУº
mgh=$\frac{1}{2}mv{¡ä}^{2}-\frac{1}{2}m{v}^{2}$
½âµÃ£º
v¡ä=$\sqrt{{v}^{2}+2gh}$=$\sqrt{{5}^{2}+2¡Á10¡Á1.25}=5\sqrt{2}$m/s
£¨3£©¶Ô´ÓAµ½BµÄ¹ý³Ì£¬¸ù¾Ý¶¯Äܶ¨Àí£¬ÓУº
mgR+W_f=$\frac{1}{2}m{v}^{2}$
½âµÃ£ºW_f=$\frac{1}{2}m{v}^{2}-mgR$=$\frac{1}{2}¡Á0.2¡Á{5}^{2}-0.2¡Á10¡Á2$=-1.5J
´ð£º£¨1£©Ð¡»¬¿é´ÓBµãÅ׳öÖÁµØÃæÔ˶¯µÄˮƽλÒÆ´óСΪ2.5m£»
£¨2£©Ð¡»¬¿é×ŵØʱµÄËٶȴóСΪ5$\sqrt{2}$m/s£»
£¨3£©Ð¡»¬¿éÑØÔ²¹ìµÀÔ˶¯µÄ¹ý³ÌÖÐËùÊÜĦ²ÁÁ¦×öµÄ¹¦ÊÇ-1.5J£®

µãÆÀ ±¾Ìâ¹Ø¼üÊÇÃ÷È·»¬¿éµÄÔ˶¯Çé¿ö£¬½áºÏ¶¯Äܶ¨ÀíºÍƽÅ×Ô˶¯µÄ·ÖÔ˶¯¹«Ê½ÁÐʽÇó½â£»¶¯Äܶ¨ÀíµÄÓŵãÔÚÓÚÊÊÓÃÈκÎÔ˶¯°üÀ¨ÇúÏßÔ˶¯£®¶¯Äܶ¨ÀíµÄÓ¦Ó÷¶Î§ºÜ¹ã£¬¿ÉÒÔÇóËٶȡ¢Á¦¡¢¹¦µÈÎïÀíÁ¿£¬ÌرðÊÇ¿ÉÒÔÈ¥Çó±äÁ¦¹¦£®