Chương IV- Các định luật bảo toàn

Theo bài ta có: \(T_{max}=1,02T_{min}\)

\(\Rightarrow mg\cdot\left(3-2cos\alpha_0\right)=1,02\cdot mg\cdot cos\alpha_0\)

\(\Rightarrow3,02cos\alpha_0=2\Rightarrow cos\alpha_0=\dfrac{100}{151}\Rightarrow\alpha_0=48,53^o\)

Vận tốc vật khi đi qua VTCB: 

\(v=\alpha_0\cdot l\cdot g=0,1\cdot0,5\cdot10=0,5\)m/s

Đổi: \(90g=0,09kg\)

Cơ năng của con lắc:

\(W=\dfrac{1}{2}mgl_{\alpha_0^2}=\dfrac{1}{2}\cdot0,09\cdot9,8\cdot1\cdot\left(\dfrac{6\pi}{180}\right)^2\approx4,8\cdot10^{-3}J\)

`1.`

`a,W_A=W_[t(max)]=mgz_[max]=mgsin \alpha AB`

                         `=1.10.sin30^o .3=15(J)`

`b,W_B=W_A`

`<=>W_[đ(max)]=W_[t(max)]`

`<=>1/2mv_[max] ^2=15`

`<=>1/2 .1.v_[max] ^2=15`

`<=>v_[max]=\sqrt{30} (m//s)`

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`2.`

Theo `Oy` có: `N=P=mg=1.10=10(N)`

  `=>F_[MS]=\mu .N=0,5.10=5(N)`

Theo `Ox` có: `-F_[MS]=ma`

                 `=>-5=1.a<=>a=-5(m//s^2)`

Có:`v_s ^2-v_t ^2=2as`

`<=>0^2 -(\sqrt{30})^2=2.(-5).s`

`<=>s=3(m)`

`@W_t=mgz=2.10.2=40(J)`

   `W_đ=1/2mv^2=1/2 .2.0^2=0(J)`

  `W=W_t+W_đ=40+0=40(J)`

`@W_[(W_đ=2W_t)]=W_[đ(W_đ=2W_t)]+W_[t(W_đ=2W_t)]=40`

    Mà `W_[đ(W_đ=2W_t)]=2W_[t(W_đ=2W_t)]`

   `=>3W_[t(W_đ=2W_t)]=40`

`<=>3mgz_[(W_đ=2W_t)]=40`

`<=>3.2.10.z_[(W_đ=2W_t)]=40`

`<=>z_[(W_đ=2W_t)]~~0,67(m)`

`@W_[đ(max)]=W_[t(max)]=40`

`<=>1/2mv_[max] ^2=40`

`<=>1/2 .2v_[max] ^2=40`

`<=>v_[max]=2\sqrt{10}(m//s)`

`a// W=W_[đ(max)]=1/2mv_[max] ^2=1/2 .0,5.10^2=25(J)`

`b// W_[đ(max)]=W_[t(max)]`

`<=>25=mgz_[max]`

`<=>25=0,5.10.z_[max]`

`<=>z_[max]=5(m)`

`c// W_[(W_đ=3W_t)]=W_[đ(W_đ=3W_t)]+W_[t(W_đ=3W_t)]=25`

    Mà `W_[đ(W_đ=3W_t)]=3W_[t(W_đ=3W_t)]`

 `=>4W_[t(W_đ=3W_t)]=25`

`<=>4mgz_[(W_đ=3W_t)]=25`

`<=>4.0,5.z_[(W_đ=3W_t)]=25`

`<=>z_[(W_đ=3W_t)]=12,5(m)`

    Mà 

a, Áp dụng định lý động năng

\(A=W_{d_B}-W_{d_A}\Rightarrow A_{\overrightarrow{F}}+A_{\overrightarrow{F_{ms}}}=\dfrac{mv^2_B}{2}-\dfrac{mv^2_A}{2}\\ A_F=Fs=4000.1000=4.10^5J\\ A_{\overrightarrow{F_{ms}}}=-F_{ms}s=-\mu mgs=-\mu2.10^6J\)

b, giả sử D là vị trí vận tốc = 0

\(A=W_{d_D}-W_{d_B}\\ \Rightarrow A_{\overrightarrow{P}}+A_{\overrightarrow{F_{ms}}}=\dfrac{mv^2_D}{2}-\dfrac{mv^2_B}{2}\\ A_{\overrightarrow{P}}=-B_xBD=-mg.sin30^o.BD=-10^4BD\left(J\right)\\ A_{F_{ms}}=-F_{ms}BD=-\mu NBD=-\mu mgcos30^oBD=-2000BD\left(J\right)\\ \Rightarrow-10^4BD-2000BD.\dfrac{1}{2}=\dfrac{1}{2}.2000.0-\dfrac{1}{2}.2000.20^2\\ \Rightarrow BD=33,333m\)

 \(\Rightarrow BC>BD\Rightarrow\) xe không lên được đỉnh dốc