B\u00e0i 26 : Th\u1ebf n\u0103ng<\/h2>\n\n\n\n

C1<\/strong> (trang 137 sgk V\u1eadt L\u00fd 10): Ch\u1ee9ng t\u1ecf \nr\u1eb1ng trong tr\u1ecdng tr\u01b0\u1eddng \u0111\u1ec1u m\u1ecdi v\u1eadt (n\u1ebfu kh\u00f4ng ch\u1ecbu t\u00e1c d\u1ee5ng c\u1ee7a m\u1ed9t l\u1ef1c\n n\u00e0o kh\u00e1c) s\u1ebd chuy\u1ec3n \u0111\u1ed9ng v\u1edbi c\u00f9ng m\u1ed9t gia t\u1ed1c g, g\u1ecdi l\u00e0 gia t\u1ed1c tr\u1ecdng \ntr\u01b0\u1eddng.<\/p>\n\n\n\n

Tr\u1ea3 l\u1eddi:<\/strong><\/ins><\/p>\n\n\n\n

Trong tr\u1ecdng tr\u01b0\u1eddng \u0111\u1ec1u, t\u1ea1i m\u1ecdi \u0111i\u1ec3m, 1 v\u1eadt lu\u00f4n ch\u1ecbu t\u00e1c d\u1ee5ng c\u1ee7a \nvector tr\u1ecdng l\u1ef1c P l\u00e0 nh\u01b0 nhau (c\u00f9ng ph\u01b0\u01a1ng, c\u00f9ng chi\u1ec1u, c\u00f9ng \u0111\u1ed9 l\u1edbn).<\/p>\n\n\n\n

\u00c1p d\u1ee5ng \u0111\u1ecbnh lu\u1eadt II Niu-t\u01a1n: <\/p>\n\n\n\n

<\/figure>\n\n\n\n

C2<\/strong> (trang 138 sgk V\u1eadt L\u00fd 10): T\u00ecm hai v\u00ed \nd\u1ee5 ch\u1ee9ng t\u1ecf r\u1eb1ng m\u1ed9t v\u1eadt c\u00f3 kh\u1ed1i l\u01b0\u1ee3ng m khi \u0111\u01b0a l\u00ean v\u1ecb tr\u00ed c\u00e1ch m\u1eb7t \u0111\u1ea5t\n \u0111\u1ed9 cao z th\u00ec l\u00fac r\u01a1i xu\u1ed1ng c\u00f3 th\u1ec3 sinh c\u00f4ng.<\/p>\n\n\n\n

Tr\u1ea3 l\u1eddi:<\/strong><\/ins><\/p>\n\n\n\n

V\u00ed d\u1ee5 1: B\u00faa m\u00e1y t\u1eeb \u0111\u1ed9 cao z, khi r\u01a1i xu\u1ed1ng c\u00f3 th\u1ec3 \u0111\u00f3ng c\u1ecdc ng\u1eadp s\u00e2u v\u00e0o l\u00f2ng \u0111\u1ea5t- sinh c\u00f4ng.<\/p>\n\n\n\n

V\u00ed d\u1ee5 2: D\u00f2ng n\u01b0\u1edbc t\u1eeb \u0111\u1ed9 cao z \u0111\u1ed5 xu\u1ed1ng l\u00e0m quay tuabin c\u1ee7a m\u00e1y ph\u00e1t \u0111i\u1ec7n- nh\u00e0 m\u00e1y th\u1ee7y \u0111i\u1ec7n.<\/p>\n\n\n\n

C3<\/strong> (trang 138 sgk V\u1eadt L\u00fd 10): N\u1ebfu ch\u1ecdn m\u1ed1c th\u1ebf n\u0103ng t\u1ea1i v\u1ecb tr\u00ed O (\u0111\u1ed9 cao = 0, H\u00ecnh 26.2) th\u00ec t\u1ea1i \u0111i\u1ec3m n\u00e0o <\/p>\n\n\n\n

–\tTh\u1ebf n\u0103ng = 0?<\/p>\n\n\n\n

–\tTh\u1ebf n\u0103ng > 0?<\/p>\n\n\n\n

–\tTh\u1ebf n\u0103ng < 0 ?<\/p>\n\n\n\n

<\/figure>\n\n\n\n

Tr\u1ea3 l\u1eddi:<\/strong><\/p>\n\n\n\n

– T\u1ea1i m\u1ed1c th\u1ebf n\u0103ng O, th\u1ebf n\u0103ng b\u1eb1ng 0: W_{t<\/sub>(O) = 0<\/p>\n\n\n\n}

– T\u1ea1i A th\u1ebf n\u0103ng d\u01b0\u01a1ng: W_{t<\/sub>(A) > 0<\/p>\n\n\n\n}

– T\u1ea1i B th\u1ebf n\u0103ng \u00e2m: W_{t<\/sub>(B) < 0<\/p>\n\n\n\n}

T\u1eeb \u0111\u00f3 suy ra:<\/p>\n\n\n\n

– T\u1eeb m\u1ed1c th\u1ebf n\u0103ng l\u00ean cao, th\u1ebf n\u0103ng d\u01b0\u01a1ng.<\/p>\n\n\n\n

– T\u1eeb m\u1ed1c th\u1ebf n\u0103ng xu\u1ed1ng d\u01b0\u1edbi, th\u1ebf n\u0103ng \u00e2m.<\/p>\n\n\n\n

C4<\/strong> (trang 139 sgk V\u1eadt L\u00fd 10): Ch\u1ee9ng minh\n r\u1eb1ng, hi\u1ec7u th\u1ebf n\u0103ng c\u1ee7a m\u1ed9t v\u1eadt chuy\u1ec3n \u0111\u1ed9ng trong tr\u1ecdng tr\u01b0\u1eddng kh\u00f4ng \nph\u1ee5 thu\u1ed9c vi\u1ec7c ch\u1ecdn g\u1ed1c th\u1ebf n\u0103ng.<\/p>\n\n\n\n

Tr\u1ea3 l\u1eddi:<\/strong><\/p>\n\n\n\n

+ Th\u1ebf n\u0103ng t\u1ea1i M: W_{t<\/sub>(M) = mgzM<\/sub><\/p>\n\n\n\n}

+ Th\u1ebf n\u0103ng t\u1ea1i N: W_{t<\/sub>(N) = mgzN<\/sub><\/p>\n\n\n\n}

\u21d2Th\u1ebf n\u0103ng t\u1ea1i M v\u00e0 t\u1ea1i N ph\u1ee5 thu\u1ed9c m\u1ed1c ch\u1ecdn th\u1ebf n\u0103ng.<\/p>\n\n\n\n

+ Hi\u1ec7u th\u1ebf n\u0103ng t\u1ea1i M v\u00e0 N l\u00e0:<\/p>\n\n\n\n

W_{t<\/sub>(M) – Wt<\/sub>(N) = mgzM<\/sub> – mgzN<\/sub> = mg(zM<\/sub> – zN<\/sub>) = mg(\t\u0394\tz)<\/p>\n\n\n\n}

Nh\u01b0 v\u1eady: Hi\u1ec7u th\u1ebf n\u0103ng ch\u1ec9 ph\u1ee5 thu\u1ed9c v\u1ecb tr\u00ed \u0111i\u1ec3m \u0111\u1ea7u (M), \u0111i\u1ec3m cu\u1ed1i \n(N) t\u1ee9c v\u00e0o \u0111\u1ed9 cao t\u1eeb N \u0111\u1ebfn M (theo ph\u01b0\u01a1ng th\u1eb3ng \u0111\u1ee9ng) m\u00e0 kh\u00f4ng ph\u1ee5 \nthu\u1ed9c g\u1ed1c th\u1ebf n\u0103ng ch\u1ecdn \u1edf \u0111\u00e2u.<\/ins><\/p>\n\n\n\n

C5<\/strong> (trang 139 sgk V\u1eadt L\u00fd 10): Ch\u1ee9ng minh\n r\u1eb1ng khi m\u1ed9t v\u1eadt chuy\u1ec3n \u0111\u1ed9ng t\u1eeb M \u0111\u1ebfn N trong tr\u1ecdng tr\u01b0\u1eddng theo nh\u1eefng \n\u0111\u01b0\u1eddng kh\u00e1c nhau th\u00ec c\u00f4ng c\u1ee7a tr\u1ecdng l\u1ef1c theo c\u00e1c \u0111\u01b0\u1eddng \u1ea5y l\u00e0 nh\u01b0 nhau. <\/p>\n\n\n\n

Tr\u1ea3 l\u1eddi:<\/strong><\/p>\n\n\n\n

<\/figure>\n\n\n\n

X\u00e9t tam gi\u00e1c vu\u00f4ng MHN c\u00f3: MN.cos\u03b1 = MH<\/p>\n\n\n\n

\u0110\u1eb7t MN = S \u21d2 S.cos\u03b1 = MH <\/p>\n\n\n\n

\u21d2 c\u00f4ng c\u1ee7a tr\u1ecdng l\u1ef1c l\u00e0m v\u1eadt di chuy\u1ec3n trong tr\u1ecdng tr\u01b0\u1eddng t\u1eeb \u0111\u1ed9 cao z_{M<\/sub> \u0111\u1ebfn \u0111\u1ed9 cao zN<\/sub> l\u00e0: <\/p>\n\n\n\n}

   A = P.S.cos\u03b1 = P(z_{M<\/sub> – zN<\/sub>)<\/p>\n\n\n\n}

\u21d2 C\u00f4ng A ch\u1ec9 ph\u1ee5 thu\u1ed9c hi\u1ec7u (z_{M<\/sub> – zN<\/sub>) m\u00e0 kh\u00f4ng ph\u1ee5 thu\u1ed9c d\u1ea1ng \u0111\u01b0\u1eddng \u0111i t\u1eeb M \u0111\u1ebfn N (h\u00ecnh v\u1ebd)<\/p>\n\n\n\n}

B\u00e0i 2 (trang 141 SGK V\u1eadt L\u00fd 10) :<\/strong> Khi m\u1ed9t v\u1eadt t\u1eeb \u0111\u1ed9 cao z, v\u1edbi c\u00f9ng v\u1eadn t\u1ed1c \u0111\u1ea7u, bay xu\u1ed1ng \u0111\u1ea5t theo nh\u1eefng con \u0111\u01b0\u1eddng kh\u00e1c nhau th\u00ec <\/ins><\/p>\n\n\n\n

A. \u0110\u1ed9 l\u1edbn v\u1eadn t\u1ed1c ch\u1ea1m \u0111\u1ea5t b\u1eb1ng nhau<\/p>\n\n\n\n

B. Th\u1eddi gian r\u01a1i b\u1eb1ng nhau<\/p>\n\n\n\n

C. c\u00f4ng c\u1ee7a tr\u1ecdng l\u1ef1c b\u1eb1ng nhau<\/p>\n\n\n\n

D. gia t\u1ed1c r\u01a1i b\u1eb1ng nhau<\/p>\n\n\n\n

H\u00e3y ch\u1ecdn c\u00e2u sai.<\/p>\n\n\n\n

L\u1eddi gi\u1ea3i:<\/strong><\/p>\n\n\n\n

Ch\u1ecdn B.\n<\/p>\n\n\n\n

C\u00f4ng A ch\u1ec9 ph\u1ee5 thu\u1ed9c hi\u1ec7u \u0111\u1ed9 cao kh\u00f4ng ph\u1ee5 thu\u1ed9c d\u1ea1ng \u0111\u01b0\u1eddng \u0111i n\u00ean theo \u0111\u1ecbnh l\u00fd bi\u1ebfn thi\u00ean \u0111\u1ed9ng n\u0103ng ta c\u00f3: \n<\/p>\n\n\n\n

<\/figure>\n\n\n\n

(h l\u00e0 hi\u1ec7u \u0111\u1ed9 cao gi\u1eefa hai \u0111i\u1ec3m)\n<\/p>\n\n\n\n

v_{1<\/sub> l\u00e0 v\u1eadn t\u1ed1c \u0111\u1ea7u kh\u00f4ng \u0111\u1ed5i, h = z, n\u00ean theo c\u00e1c con \u0111\u01b0\u1eddng kh\u00e1c nhau th\u00ec \u0111\u1ed9 l\u1edbn v2<\/sub> v\u1eabn b\u1eb1ng nhau v\u00e0 c\u00f4ng c\u1ee7a tr\u1ecdng l\u1ef1c b\u1eb1ng nhau\n<\/p>\n\n\n\n}

\u2192 B sai, th\u1eddi gian s\u1ebd ph\u1ee5 thu\u1ed9c v\u00e0o h\u00ecnh d\u1ea1ng qu\u1ef9 \u0111\u1ea1o \u0111i.\n<\/p>\n\n\n\n

Ch\u1ecdn B.<\/p>\n\n\n\n

B\u00e0i 3 (trang 141 SGK V\u1eadt L\u00fd 10) :<\/strong> M\u1ed9t v\u1eadt kh\u1ed1i l\u01b0\u1ee3ng 1,0 kg c\u00f3 th\u1ebf n\u0103ng 1,0 J \u0111\u1ed1i v\u1edbi m\u1eb7t \u0111\u1ea5t. L\u1ea5y g = 9,8 m\/s^{2<\/sup>. Khi \u0111\u00f3 , v\u1eadt \u1edf \u0111\u1ed9 cao b\u1eb1ng bao nhi\u00eau?<\/ins><\/p>\n\n\n\n}

A. 0,102 m ;         B. 1,0 m <\/p>\n\n\n\n

C. 9,8 m ;         D. 32 m<\/p>\n\n\n\n

L\u1eddi gi\u1ea3i:<\/strong><\/p>\n\n\n\n

– Ch\u1ecdn A<\/strong> <\/p>\n\n\n\n

– \u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c t\u00ednh th\u1ebf n\u0103ng tr\u1ecdng tr\u01b0\u1eddng:<\/p>\n\n\n\n

<\/figure>\n\n\n\n

B\u00e0i 4 (trang 141 SGK V\u1eadt L\u00fd 10) :<\/strong> M\u1ed9t v\u1eadt kh\u1ed1i l\u01b0\u1ee3ng m g\u1eafn v\u00e0o \u0111\u1ea7u m\u1ed9t l\u00f2 xo \u0111\u00e0n h\u1ed3i c\u00f3 \u0111\u1ed9 c\u1ee9ng k, \u0111\u1ea7u kia c\u1ee7a l\u00f2 xo c\u1ed1 \u0111\u1ecbnh. Khi l\u00f2 xo b\u1ecb n\u00e9n l\u1ea1i m\u1ed9t \u0111o\u1ea1n \u0394l (\u0394l < 0) th\u00ec th\u1ebf n\u0103ng \u0111\u00e0n h\u1ed3i b\u1eb1ng bao nhi\u00eau?<\/p>\n\n\n\n

<\/figure>\n\n\n\n

L\u1eddi gi\u1ea3i:<\/strong><\/p>\n\n\n\n

Ch\u1ecdn A. \n<\/p>\n\n\n\n

Th\u1ebf n\u0103ng \u0111\u00e0n h\u1ed3i c\u1ee7a v\u1eadt l\u00e0: \n<\/p>\n\n\n\n

<\/figure>\n\n\n\n

B\u00e0i 5 (trang 141 SGK V\u1eadt L\u00fd 10) :<\/strong> Trong h\u00ecnh 26.5, hai v\u1eadt c\u00f9ng kh\u1ed1i l\u01b0\u1ee3ng n\u1eb1m \u1edf hai v\u1ecb tr\u00ed M v\u00e0 N sao cho MN n\u1eb1m ngang. So s\u00e1nh th\u1ebf n\u0103ng t\u1ea1i M v\u00e0 t\u1ea1i N.<\/ins><\/p>\n\n\n\n

<\/figure>\n\n\n\n

H\u00ecnh 26.5<\/em><\/p>\n\n\n\n

L\u1eddi gi\u1ea3i:<\/strong><\/p>\n\n\n\n

V\u00ec MN n\u1eb1m ngang n\u00ean n\u1ebfu ch\u1ecdn c\u00f9ng 1 m\u1ed1c th\u1ebf n\u0103ng (v\u00ed d\u1ee5 t\u1ea1i O) th\u00ec th\u1ebf n\u0103ng c\u1ee7a v\u1eadt t\u1ea1i M v\u00e0 t\u1ea1i N l\u00e0 nh\u01b0 nhau.<\/p>\n\n\n\n

B\u00e0i 6 (trang 141 SGK V\u1eadt L\u00fd 10) :<\/strong> L\u00f2 xo \nc\u00f3 \u0111\u1ed9 c\u1ee9ng k = 200 N\/m, m\u1ed9t \u0111\u1ea7u c\u1ed1 \u0111\u1ecbnh, \u0111\u1ea7u kia g\u1eafn v\u1edbi v\u1eadt nh\u1ecf. Khi l\u00f2\n xo b\u1ecb n\u00e9n 2 cm th\u00ec th\u1ebf n\u0103ng \u0111\u00e0n h\u1ed3i c\u1ee7a h\u1ec7 b\u1eb1ng bao nhi\u00eau? Th\u1ebf n\u0103ng n\u00e0y\n c\u00f3 ph\u1ee5 thu\u1ed9c kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a v\u1eadt kh\u00f4ng?<\/p>\n\n\n\n

L\u1eddi gi\u1ea3i:<\/strong><\/ins><\/p>\n\n\n\n

\u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c t\u00ednh th\u1ebf n\u0103ng \u0111\u00e0n h\u1ed3i:\n<\/p>\n\n\n\n

<\/figure>\n\n\n\n

Th\u1ebf n\u0103ng n\u00e0y kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o kh\u1ed1i l\u01b0\u1ee3ng c\u1ee7a v\u1eadt.\n<\/ins><\/p>\n","protected":false},"excerpt":{"rendered":"

B\u00e0i 26 : Th\u1ebf n\u0103ng C1 (trang 137 sgk V\u1eadt L\u00fd 10): Ch\u1ee9ng t\u1ecf r\u1eb1ng trong tr\u1ecdng tr\u01b0\u1eddng \u0111\u1ec1u m\u1ecdi v\u1eadt (n\u1ebfu kh\u00f4ng ch\u1ecbu t\u00e1c d\u1ee5ng c\u1ee7a m\u1ed9t l\u1ef1c n\u00e0o kh\u00e1c) s\u1ebd chuy\u1ec3n \u0111\u1ed9ng v\u1edbi c\u00f9ng m\u1ed9t gia t\u1ed1c g, g\u1ecdi l\u00e0 gia t\u1ed1c tr\u1ecdng tr\u01b0\u1eddng. Tr\u1ea3 l\u1eddi: Trong tr\u1ecdng tr\u01b0\u1eddng \u0111\u1ec1u, t\u1ea1i m\u1ecdi \u0111i\u1ec3m, […]<\/p>\n","protected":false},"author":12,"featured_media":44166,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[1640],"tags":[],"yoast_head":"\n