Đáp án:
\(\begin{align}
& a)v=3m/s \\
& b){{h}_{1}}=2,5m;{{v}_{1}}=5\sqrt{2}m/s;S=5m \\
& c){{h}_{2}}=5m>h=4,55m \\
& \\
\end{align}\)
Giải thích các bước giải:
\({{v}_{0}}=4,55m/s;h=4,55m;\alpha ={{30}^{0}}\)
1) bỏ qua ma sát
a) Bảo toàn cơ năng
\(\begin{align}
& \text{W}={{\text{W}}_{d}}+{{\text{W}}_{t}} \\
& \Leftrightarrow \dfrac{1}{2}.m.v_{0}^{2}=\dfrac{1}{2}.m.{{v}^{2}}+m.g.h \\
& \Leftrightarrow \dfrac{1}{2}{{.10}^{2}}=\dfrac{1}{2}.{{v}^{2}}+10.4,55 \\
& \Rightarrow v=3m/s \\
\end{align}\)
b) Wd=Wt
độ cao
\(\begin{align}
& \text{W}={{\text{W}}_{d}}+{{\text{W}}_{t}}=2{{W}_{t}} \\
& \Leftrightarrow \dfrac{1}{2}.m.v_{0}^{2}=m.g.{{h}_{1}} \\
& \Leftrightarrow \dfrac{1}{2}{{.10}^{2}}=2.10.{{h}_{1}} \\
& \Rightarrow {{h}_{1}}=2,5m \\
\end{align}\)
vị trí cách chân dốc
\(S=\dfrac{{{h}_{1}}}{\sin \alpha }\Rightarrow S=\dfrac{2,5}{\sin 30}=5m\)
Vận tốc:
\(\begin{align}
& \text{W=}{{\text{W}}_{d}}+{{\text{W}}_{t}}=2{{\text{W}}_{d}} \\
& \Leftrightarrow \dfrac{1}{2}.m.v_{0}^{2}=2.\dfrac{1}{2}.m.v_{1}^{2} \\
& \Rightarrow {{10}^{2}}=2.v_{1}^{2} \\
& \Rightarrow {{v}_{1}}=5\sqrt{2}m/s \\
\end{align}\)
2)
\(\mu =\dfrac{1}{5\sqrt{3}}\)
\(\begin{align}
& {{A}_{ms}}={{\text{W}}_{A}}-{{\text{W}}_{C}} \\
& \mu m.cos\alpha .{{S}_{c}}=\dfrac{1}{2}.m.v_{0}^{2}-m.g.{{h}_{2}} \\
& \dfrac{1}{5\sqrt{3}}.cos30.\sin \alpha .{{h}_{2}}=\dfrac{1}{2}{{.10}^{2}}-10.{{h}_{2}} \\
& \Rightarrow {{h}_{2}}=5m \\
\end{align}\)
=> vật có lên được hết dốc
