Định luật II Newton

Home Lớp 10 Vật lý 10 - Sách kết nối tri thứcBài tập nâng cao

{"common":{"save":0,"post_id":"8478","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"7611","mon_id":"1158790","chapter_id":"1158796","post_id":"8478","question":"M\u1ed9t v\u1eadt có kh\u1ed1i l\u01b0\u1ee3ng 20 kg, b\u1eaft \u0111\u1ea7u chuy\u1ec3n \u0111\u1ed9ng d\u01b0\u1edbi tác d\u1ee5ng c\u1ee7a m\u1ed9t l\u1ef1c kéo, \u0111i \u0111\u01b0\u1ee3c quãng \u0111\u01b0\u1eddng S trong th\u1eddi gian $t_1$<\/span> = 10 s. \u0110\u1eb7t thêm lên nó m\u1ed9t v\u1eadt khác có kh\u1ed1i l\u01b0\u1ee3ng 10 kg. \u0110\u1ec3 \u0111i \u0111\u01b0\u1ee3c quãng \u0111\u01b0\u1eddng S và c\u0169ng v\u1edbi l\u1ef1c kéo nói trên, th\u1eddi gian chuy\u1ec3n \u0111\u1ed9ng ph\u1ea3i b\u1eb1ng bao nhiêu?<\/p>","options":["A.<\/strong> $t_2$<\/span> = 12,25 s.","B.<\/strong> $t_2$<\/span> = 12,50 s.","C.<\/strong> $t_2$<\/span> = 7,07 s.","D.<\/strong> $t_2$<\/span> = 12,95 s."],"correct":"1","level":"3","hint":"V\u1eadn d\u1ee5ng công th\u1ee9c: - F = m.a - S = 0,5.a.t²<\/p>","answer":"Ch\u1ecdn A.<\/strong> $t_2$<\/span> = 12,25 s.<\/span><\/p>+ Gia t\u1ed1c chuy\u1ec3n \u0111\u1ed9ng c\u1ee7a v\u1eadt kh\u1ed1i l\u01b0\u1ee3ng 20 kg là <\/p>$S=0,5.a_1.t_1^2$<\/span> ⇒ $a_1=\\dfrac{2S}{t_1^2}=\\dfrac{2S}{10^2}=\\dfrac{S}{50}$<\/span> (m\/s²).<\/p>+ L\u1ef1c tác d\u1ee5ng lên v\u1eadt này là $F=m_1.a_1=20.\\dfrac{S}{50}=0,4S$<\/span> (N).<\/p>+ Kh\u1ed1i l\u01b0\u1ee3ng sau khi \u0111\u1eb7t thêm v\u1eadt 10 kg lên v\u1eadt ban \u0111\u1ea7u là 20 + 10 = 30 (kg).<\/p>+ Khi tác d\u1ee5ng cùng l\u1ef1c F lên v\u1eadt kh\u1ed1i l\u01b0\u1ee3ng 30 kg thì gia t\u1ed1c v\u1eadt này thu \u0111\u01b0\u1ee3c là<\/p>    $a_2=\\dfrac{F}{m_2}=\\dfrac{0,4S}{30}=\\dfrac{S}{75}$<\/span> (m\/s²).<\/p>+ \u0110\u1ec3 v\u1eadt c\u0169ng \u0111i \u0111\u01b0\u1ee3c quãng \u0111\u01b0\u1eddng là S thì th\u1eddi gian chuy\u1ec3n \u0111\u1ed9ng c\u1ee7a v\u1eadt là<\/p>   $S=0,5.a_2.t_2^2$<\/span> ⇒ $t_2=\\sqrt{\\dfrac{2S}{a_2}}=\\sqrt{\\dfrac{2S}{\\dfrac{S}{75}}}=5\\sqrt{6}\\approx12,25(s)$<\/span>.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-02-21 09:10:56","option_type":"math","len":0},{"id":"7614","mon_id":"1158790","chapter_id":"1158796","post_id":"8478","question":"D\u01b0\u1edbi tác d\u1ee5ng c\u1ee7a m\u1ed9t l\u1ef1c F có \u0111\u1ed9 l\u1edbn không \u0111\u1ed5i theo ph\u01b0\u01a1ng ngang, xe chuy\u1ec3n \u0111\u1ed9ng th\u1eb3ng \u0111\u1ec1u v\u1edbi không v\u1eadn t\u1ed1c \u0111\u1ea7u và \u0111i \u0111\u01b0\u1ee3c quãng \u0111\u01b0\u1eddng 2,5 m trong th\u1eddi gian t. N\u1ebfu \u0111\u1eb7t thêm v\u1eadt có kh\u1ed1i l\u01b0\u1ee3ng 250 g lên xe thì xe ch\u1ec9 \u0111i \u0111\u01b0\u1ee3c quãng \u0111\u01b0\u1eddng 2 m trong th\u1eddi gian t. B\u1ecf qua ma sát. Kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a xe là<\/strong><\/span><\/p>","options":["A.<\/strong> 15 kg.","B.<\/strong> 1 kg.","C.<\/strong> 2 kg.","D.<\/strong> 5 kg."],"correct":"2","level":"3","hint":"","answer":"Ch\u1ecdn B.<\/strong> 1 kg.<\/span><\/p>\u0110\u1ed5i 250 g= 0,25 kg.<\/p>G\u1ecdi kh\u1ed1i l\u01b0\u1ee3ng xe là m.<\/p>&diams; Khi ch\u01b0a \u0111\u1eb7t v\u1eadt, gia t\u1ed1c c\u1ee7a xe là $a=\\dfrac{F}{m}$<\/span>.<\/p>Lúc này xe \u0111i \u0111\u01b0\u1ee3c quãng \u0111\u01b0\u1eddng 2,5 m trong th\u1eddi gian t nên <\/p>$2,5=\\dfrac{at^2}{2}$<\/span> ⇒ $\\dfrac{Ft^2}{2m}=2,5$<\/span> (1).<\/p>&diams; Khi \u0111\u1eb7t thêm v\u1eadt 250 g lên xe, gia t\u1ed1c c\u1ee7a xe là $a^\\prime=\\dfrac{F}{m+0,25}$<\/span>.<\/p>Lúc này xe \u0111i \u0111\u01b0\u1ee3c quãng \u0111\u01b0\u1eddng 2 m trong th\u1eddi gian t nên <\/p>$2=\\dfrac{a^\\prime t^2}{2}$<\/span> ⇒ $\\dfrac{Ft^2}{2(m+0,25)}=2$<\/span> (2).<\/p>&diams; L\u1ea5y (1) chia (2), v\u1ebf theo v\u1ebf ta \u0111\u01b0\u1ee3c:<\/p>$\\dfrac{m+0,25}{m}=\\dfrac{2,5}{2}$<\/span> ⇒ 2m + 0,5 = 2,5 m ⇒ m = 1 (kg).<\/p>V\u1eady kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a xe là 1 kg.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-02-21 09:33:15","option_type":"txt","len":0},{"id":"7615","mon_id":"1158790","chapter_id":"1158796","post_id":"8478","question":"M\u1ed9t ô tô ch\u1ea1y v\u1edbi v\u1eadn t\u1ed1c 60 km\/h thì ng\u01b0\u1eddi lái xe hãm phanh, xe \u0111i ti\u1ebfp quãng \u0111\u01b0\u1eddng 50 m r\u1ed3i d\u1eebng l\u1ea1i. L\u1ef1c hãm phanh trong hai tr\u01b0\u1eddng h\u1ee3p b\u1eb1ng nhau. N\u1ebfu ôtô \u0111ang ch\u1ea1y v\u1edbi v\u1eadn t\u1ed1c 120 km\/h thì quãng \u0111\u01b0\u1eddng \u0111i \u0111\u01b0\u1ee3c t\u1eeb lúc hãm phanh \u0111\u1ebfn khi d\u1eebng l\u1ea1i là<\/p>","options":["A.<\/strong> 100 m.","B.<\/strong> 70,7 m.","C.<\/strong> 141 m.","D.<\/strong> 200 m."],"correct":"4","level":"3","hint":"","answer":"Ch\u1ecdn D.<\/strong> 200 m.<\/span><\/p>Vì l\u1ef1c hãm phanh trong hai tr\u01b0\u1eddng h\u1ee3p b\u1eb1ng nhau nên gia t\u1ed1c hãm trong hai tr\u01b0\u1eddng h\u1ee3p b\u1eb1ng nhau.<\/p>Áp d\u1ee5ng công th\u1ee9c:<\/p>$v^2-v^2_0=2as$<\/span> ⇔ $v^2=2as+v^2_0=2as$<\/span> (vì $v_0=0$<\/span>).<\/p>Xét t\u1ec9 s\u1ed1:<\/p>$\\dfrac{v_1^2}{v_2^2}=\\dfrac{s_1}{s_2}$<\/span> ⇔ $\\dfrac{60^2}{120^2}=\\dfrac{50}{s_2}$<\/span> ⇔ $s_2=200$<\/span> (m).<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2025-02-21 09:38:17","option_type":"txt","len":0}]}