Đáp án:
Bài 4:
\(\begin{align}
& a)\text{W}=90J \\
& b){{h}_{max}}=45m \\
& c)h=22,5m;v=15\sqrt{2}m/s \\
& d)h'=11,25m;v'=15\sqrt{3}m/s \\
\end{align}\)
Bài 5:
\(\begin{align}
& \text{a)}{{\text{W}}_{t}}=0,25J \\
& b){{v}_{max}}=\frac{\sqrt{10}}{2}m/s \\
& c){{x}_{nem}}=10cm \\
\end{align}\)
Bài 6:
\(\begin{align}
& a){{v}_{max}}=20m/s \\
& b)h=10m \\
& c)h'=5m \\
& d)v=\sqrt{200}m/s \\
\end{align}\)
Giải thích các bước giải:
Bài 4:
$m=0,2kg;{{v}_{0}}=30m/s$
a) cơ năng:
\(\text{W}={{\text{W}}_{dmax}}=\dfrac{1}{2}m.v_{0}^{2}=\dfrac{1}{2}.0,{{2.30}^{2}}=90J\)
b) độ cao cực đại:
\(\begin{align}
& \text{W}={{\text{W}}_{tmax}} \\
& \Leftrightarrow 90=0,2.10.{{h}_{max}} \\
& \Rightarrow {{h}_{max}}=45m \\
\end{align}\)
c) Wd=Wt
độ cao
\(\begin{align}
& \text{W}={{\text{W}}_{d}}+{{\text{W}}_{t}}=2{{\text{W}}_{t}} \\
& \Leftrightarrow 90=2.0,2.10.h \\
& \Rightarrow h=22,5m \\
\end{align}\)
Vận tốc:
\(\begin{align}
& \text{W}={{\text{W}}_{d}}+{{\text{W}}_{t}}=2{{\text{W}}_{d}} \\
& \Leftrightarrow 90=2\dfrac{1}{2}.0,2.{{v}^{2}} \\
& \Rightarrow v=15\sqrt{2}m/s \\
\end{align}\)
d)Wd=3Wt
\(\begin{align}
& \text{W}={{\text{W}}_{d}}+{{\text{W}}_{t}}=4{{\text{W}}_{t}} \\
& \Leftrightarrow 90=4.0,2.10.h' \\
& \Rightarrow h'=11,25m \\
\end{align}\)
vận tốc:
\(\begin{align}
& \text{W}={{\text{W}}_{d}}+{{\text{W}}_{t}}=\dfrac{4}{3}{{\text{W}}_{d}} \\
& \Leftrightarrow 90=\dfrac{4}{3}.\dfrac{1}{2}0,2.v{{'}^{2}} \\
& \Rightarrow h'=15\sqrt{3}m/s \\
\end{align}\)
Bài 5: \(k=50N/m;m=0,2kg;{{x}_{max}}=10cm\)
a) thế năng đàn hồi:
\(\text{W}={{\text{W}}_{tmax}}=\dfrac{1}{2}.k.x_{max}^{2}=\dfrac{1}{2}.50.0,{{1}^{2}}=0,25J\)
b) tại vị trí cân bằng:
\(\begin{align}
& {{\text{W}}_{t}}=0J \\
& \Rightarrow \text{W}={{\text{W}}_{dmax}} \\
& \Leftrightarrow 0,25=\dfrac{1}{2}.0,2.v_{max}^{2} \\
& \Rightarrow {{v}_{max}}=\dfrac{\sqrt{10}}{2}m/s \\
\end{align}\)
c)bảo toàn cơ năng:
\({{x}_{nen}}={{x}_{dan}}=10cm\)
Bài 6:
a) vận tốc khi chạm đất
\(\begin{align}
& {{\text{W}}_{tmax}}={{\text{W}}_{dmax}} \\
& \Leftrightarrow m.g.{{h}_{max}}=\dfrac{1}{2}.m.v_{max}^{2} \\
& \Leftrightarrow 10.20=\dfrac{1}{2}.v_{max}^{2} \\
& \Rightarrow {{v}_{max}}=20m/s \\
\end{align}\)
b) Wd=Wt
độ cao:
\(\begin{align}
& \text{W}={{\text{W}}_{d}}+{{\text{W}}_{t}}=2{{\text{W}}_{t}} \\
& \Leftrightarrow m.g.{{h}_{max}}=2.m.g.h \\
& \Rightarrow h=\dfrac{{{h}_{max}}}{2}=10m \\
\end{align}\)
c)
\(\begin{align}
& {{\text{W}}_{t}}=\dfrac{1}{3}{{\text{W}}_{d}} \\
& \Leftrightarrow \text{W}={{\text{W}}_{d}}+{{\text{W}}_{t}}=4{{\text{W}}_{t}} \\
& \Leftrightarrow m.g.{{h}_{max}}=4.m.g.h' \\
& \Rightarrow h'=\dfrac{h}{4}=5m \\
\end{align}\)
d)
\(\begin{align}
& S=\frac{20}{2}=10m \\
& \Leftrightarrow {{v}^{2}}=2.g.S \\
& \Rightarrow v=\sqrt{2.10.10}=\sqrt{200}m/s \\
\end{align}\)
