Question

8£®ÈçͼËùʾ£¬Î¢Á£AλÓÚÒ»¶¨¸ß¶È´¦£¬ÆäÖÊÁ¿m=1.0¡Á10^-4kg£¬´øµçºÉÁ¿q=+1.0¡Á10^-6C£¬ËÜÁϳ¤·½Ìå¿ÕÐĺÐ×ÓBλÓÚˮƽµØÃæÉÏ£¬ÓëµØÃæ¼äµÄ¶¯Ä¦²ÁÒòÊý¦Ì=0.1£®BÉϱíÃæµÄÏ·½´æÔÚ×ÅÊúÖ±ÏòÉϵÄÔÈÇ¿µç³¡£¬³¡Ç¿´óСE=2¡Á10³N/C£¬BÉϱíÃæµÄÉÏ·½´æÔÚ×ÅÊúÖ±ÏòϵÄÔÈÇ¿µç³¡£¬³¡Ç¿´óСΪ0.5E£¬BÉϱíÃ濪ÓÐһϵÁÐÂÔ´óÓÚAµÄС¿×£¬¿×¼ä¾àÂú×ãÒ»¶¨µÄ¹Øϵ£¬Ê¹µÃA½ø³öBµÄ¹ý³ÌÖÐʼÖÕ²»ÓåB½Ó´¥£¬µ±AÒÔv₁=1m/sµÄËٶȴӿ×1ÊúÖ±ÏòϽøÈëBµÄ˲¼ä£¬BÇ¡ÒÔv₂=0.6m/sµÄËÙ¶ÈÏòÓÒ»¬ÐУ¬ÉèB×ã¹»³¤£¬×ã¹»¸ßÇÒÉϱíÃæµÄºñ¶È²»¼Æ£¬È¡g=10m/s²£¬AÇ¡ÄÜ˳´Î´Ó¸÷¸öС¿×½ø³öB£¬ÊÔÇó£º

£¨1£©´ÓAµÚÒ»´Î½øÈëBÖÁBÍ£Ö¹Ô˶¯µÄ¹ý³ÌÖУ¬Bͨ¹ýµÄ×Ü·³Ìs£»
£¨2£©BÉÏÖÁÉÙÒª¿ª¶àÉÙ¸öС¿×£¬²ÅÄܱ£Ö¤AʼÖÕ²»ÓëB½Ó´¥£»
£¨3£©´ÓÓÒµ½×ó£¬BÉϱíÃæ¸÷ÏàÁÚС¿×Ö®¼äµÄ¾àÀë·Ö±ðΪ¶à´ó£¿

Answer 1

·ÖÎö £¨1£©¸ù¾ÝBËùÊܺÏÍâÁ¦£¬ÇóµÃ¼ÓËٶȣ¬È»ºóÓÉÔȼõËÙλÒƹ«Ê½ÇóµÃ×Ü·³Ì£»
£¨2£©¸ù¾ÝAÇòÔÚÉϱíÃæÉÏ·½ºÍÏ·½µÄÊÜÁ¦ÇóµÃ¼ÓËٶȣ¬È»ºó¸ù¾ÝÔȱäËÙÔ˶¯¹æÂÉÇóµÃÔ˶¯Ê±¼ä£¬·Ö±ðÇóµÃÿ´Îÿ´Î¾­¹ý¿×µÄʱ¼ä£»ÔÙÓÉ£¨1£©ÇóµÃBÔ˶¯µÄʱ¼ä£¬¼´¿ÉÁªÁ¢Çó½â¿×µÄÊýÁ¿£»
£¨3£©ÓÉ£¨2£©µÃµ½¾­¹ýÿ¸ö¿×µÄʱ¼ä£¬¸ù¾ÝB×öÔȼõËÙÔ˶¯£¬ÓÉλÒƹ«Ê½ÇóµÃ¿×¼ä¾à£®

½â´ð ½â£º£¨1£©BÊܵ½µÄºÏÍâÁ¦¼´Ä¦²ÁÁ¦f=¦Ìmg=1.0¡Á10^-4N£¬ËùÒÔ£¬B×öÔȼõËÙÔ˶¯£¬¼ÓËÙ¶ÈΪ£º$a=\frac{f}{m}=1m/{s}^{2}$£»
ÄÇô£¬´ÓAµÚÒ»´Î½øÈëBÖÁBÍ£Ö¹Ô˶¯µÄ¹ý³ÌÖУ¬Bͨ¹ýµÄ×Ü·³ÌΪ£º$s=\frac{{{v}_{2}}^{2}}{2a}=0.18m$£»
£¨2£©AÔÚBÉϱíÃæÉÏ·½£¬Ö»ÊÜÖØÁ¦Îª£ºG=mg=1.0¡Á10^-3N£¬ºÍµç³¡Á¦Îª£º${F}_{1}=0.5qE=1.0¡Á1{0}^{-3}N$£¬¹ÊºÏÍâÁ¦Îª£º${F}_{ºÏ1}={F}_{1}+G=2.0¡Á1{0}^{-3}N$£¬·½ÏòÊúÖ±ÏòÏ£»
ËùÒÔ£¬AÔÚBÉϱíÃæÉÏ·½µÄ¼ÓËÙ¶ÈΪ£º${a}_{1}=\frac{{F}_{ºÏ1}}{m}=20m/{s}^{2}$£»ÄÇô£¬Aÿ´Î´©³öBµ½½øÈëBµÄ¹ý³Ì£¬ÔÚBÉÏ·½Ô˶¯µÄʱ¼äΪ£º${t}_{1}=2\frac{{v}_{1}}{{a}_{1}}=0.1s$£»
AÔÚBÉϱíÃæÏ·½£¬Ö»ÊÜÖØÁ¦G=mg=1.0¡Á10^-3N£¬ºÍµç³¡Á¦${F}_{2}=qE=2.0¡Á1{0}^{-3}N$£¬¹ÊºÏÍâÁ¦${F}_{ºÏ2}={F}_{2}-G=1.0¡Á1{0}^{-3}N$£¬·½ÏòÊúÖ±ÏòÉÏ£»
ËùÒÔ£¬AÔÚBÉϱíÃæÏ·½µÄ¼ÓËÙ¶È${a}_{2}=\frac{{F}_{ºÏ2}}{m}=10m/{s}^{2}$£»ÄÇô£¬Aÿ´Î½øÈëBµ½´©³öBµÄ¹ý³Ì£¬ÔÚBÖÐÔ˶¯µÄʱ¼ä${t}_{2}=2\frac{{v}_{1}}{{a}_{2}}=0.2s$£»
ÄÇôÉèA´Ó¿×1½øÈëBʱµÄʱ¼äΪt£¬ÔòµÚn´Î´Ó¿×2n-1½øÈëBµÄʱ¼äΪt_2n-1=£¨n-1£©£¨t₁+t₂£©=0.3£¨n-1£©s=0.3n-0.3£¨s£©£¬n=1£¬2£¬3£¬¡­
AµÚn´Î´ÓµÚ2n¸ö¿×³öÀ´µÄʱ¼äΪt_2n=t₂+£¨n-1£©£¨t₁+t₂£©=0.3n-0.1£¬n=1£¬2£¬3£¬¡­
ÄÇô£¬ÓÉ£¨1£©¿ÉÖª£¬BÔ˶¯µÄʱ¼ä$t=\frac{{v}_{2}}{a}=0.6s$=0.3¡Á£¨3-1£©s£»
¹ÊA¿É½øÈëB3´Î£¬À뿪B2´Î£¬ËùÒÔ£¬Òª±£Ö¤Ð¡ÇòʼÖÕ²»ÓëBÏàÅö£¬BÉϵÄС¿×¸öÊýÖÁÉÙΪ5£»
£¨3£©ÓÉÓÚBÏòÓÒ×öÔȼõËÙÖ±ÏßÔ˶¯£¬¾­0.6sËٶȼõΪÁ㣬
ÄÇô£¬¿×3¾à¿×1µÄ¾àÀëΪ${S}_{3}=0.6¡Á0.3-\frac{1}{2}¡Á1¡Á{0.3}^{2}£¨m£©=0.135m$£¬¿×5¾à¿×1µÄ¾àÀëΪ${S}_{5}=0.6¡Á0.6-\frac{1}{2}¡Á1¡Á{0.6}^{2}£¨m£©=0.18m$
¿×2¾à¿×1µÄ¾àÀëΪ${S}_{2}=0.6¡Á0.2-\frac{1}{2}¡Á1¡Á{0.2}^{2}£¨m£©=0.1m$£¬¿×4¾à¿×1µÄ¾àÀëΪ${S}_{4}=0.6¡Á0.5-\frac{1}{2}¡Á1¡Á{0.5}^{2}£¨m£©=0.175m$
ËùÒÔ£¬´ÓÓÒµ½×ó£¬BÉϱíÃæ¸÷ÏàÁÚС¿×Ö®¼äµÄ¾àÀë·Ö±ðΪs₁=S₂=0.1m£¬s₂=S₃-S₂=0.035m£¬s₃=S₄-S₃=0.04m£¬s₄=S₅-S₄=0.005m£»
´ð£º£¨1£©´ÓAµÚÒ»´Î½øÈëBÖÁBÍ£Ö¹Ô˶¯µÄ¹ý³ÌÖУ¬Bͨ¹ýµÄ×Ü·³ÌsΪ0.18m£»
£¨2£©BÉÏÖÁÉÙÒª¿ª5¸öС¿×£¬²ÅÄܱ£Ö¤AʼÖÕ²»ÓëB½Ó´¥£»
£¨3£©´ÓÓÒµ½×ó£¬BÉϱíÃæ¸÷ÏàÁÚС¿×Ö®¼äµÄ¾àÀë·Ö±ðΪ0.1m£¬0.035m£¬0.04m£¬0.005m£®

µãÆÀ ÎïÌåÔ˶¯Ñ§ÎÊÌâÖУ¬Ò»°ã·ÖÎöÎïÌåµÄÔ˶¯×´Ì¬µÃµ½¼ÓËٶȵıí´ïʽ£¬È»ºó¶ÔÎïÌå½øÐÐÊÜÁ¦·ÖÎöÇóµÃºÏÍâÁ¦£¬×îºóÀûÓÃÅ£¶ÙµÚ¶þ¶¨ÂÉÁªÁ¢Çó½â¼´¿É£®