Đáp án:
\(\begin{align}
& 1)\text{W}=625J \\
& 2)\text{W}=25J \\
& 3)\text{W}=0,625J;{{h}_{max}}=2,5m; \\
& 4)m=5kg;v=3\sqrt{10}m/s \\
& 5)\text{W}=2100J;v=20\sqrt{2}m/s;{{v}_{max}}=10\sqrt{14}m/s; \\
& v=10\sqrt{7}m/s;h=35m \\
\end{align}\)
Giải thích các bước giải:
Bài 1:
\(m=10kg;h=5m;v=18km/h=5m/s\)
Bảo toàn cơ năng tại điểm đó ta có:
\(\begin{align}
& \text{W}={{\text{W}}_{d}}+{{\text{W}}_{t}}=\dfrac{1}{2}.m.{{v}^{2}}+m.g.h \\
& =\dfrac{1}{2}{{.10.5}^{2}}+10.10.5 \\
& =625J \\
\end{align}\)
Bài 2:
\(m=0,5kg;v=36km/h=10m/s\)
Cơ năng lúc vật chạm đất:
\(\text{W}={{\text{W}}_{dmax}}=\dfrac{1}{2}.m.{{v}^{2}}=\dfrac{1}{2}.0,{{5.10}^{2}}=25J\)
Bài 3:
\(m=0,025kg;{{v}_{0}}=4,5m/s;{{h}_{0}}=1,5m\)
a) Tại vị trí ném
\(\begin{align}
& {{\text{W}}_{d}}=\dfrac{1}{2}.m.v_{0}^{2}=\dfrac{1}{2}.0,025.4,{{5}^{2}}=0,25J \\
& {{\text{W}}_{t}}=m.g.{{h}_{0}}=0,025.10.1,5=0,375J \\
& \Rightarrow \text{W}={{\text{W}}_{d}}+{{\text{W}}_{t}}=0,625J \\
\end{align}\)
b)
độ cao cực đại
\(\begin{align}
& \text{W}={{\text{W}}_{tmax}} \\
& \Leftrightarrow 0,625=0,025.10.h \\
& \Rightarrow h=2,5m \\
\end{align}\)
Bài 4:
\(\text{W}=375J;{{\text{W}}_{d}}=\dfrac{3}{2}{{\text{W}}_{t}};\)
bảo toàn cơ năng ta có:
\(\left\{ \begin{align}
& \text{W}={{\text{W}}_{d}}+{{\text{W}}_{t}}=375J \\
& {{\text{W}}_{d}}=\frac{3}{2}{{\text{W}}_{t}} \\
\end{align} \right.\)\(\Leftrightarrow \left\{ \begin{align}
& {{\text{W}}_{d}}=225J \\
& {{\text{W}}_{t}}=150J \\
\end{align} \right.\)
khối lượng vật:
\(\begin{align}
& {{\text{W}}_{t}}=m.g.h \\
& \Rightarrow m=\frac{150}{10.3}=5kg \\
\end{align}\)
vận tốc:
\(\begin{align}
& {{\text{W}}_{d}}=\frac{1}{2}.m.{{v}^{2}} \\
& \Leftrightarrow 225=\frac{1}{2}.5.{{v}^{2}} \\
& \Rightarrow v=3\sqrt{10}m/s \\
\end{align}\)
Câu 5:\(m=3kg;{{v}_{0}}=20m/s;{{h}_{0}}=50m\)
a) tại vị trí ném
\(\begin{align}
& {{\text{W}}_{d}}=\frac{1}{2}.m.v_{0}^{2}=\frac{1}{2}{{.3.20}^{2}}=600J \\
& {{\text{W}}_{t}}=m.g.{{h}_{0}}=3.10.50=1500J \\
& \Rightarrow \text{W}={{\text{W}}_{d}}+{{\text{W}}_{t}}=2100J \\
\end{align}\)
b)
bảo toàn cơ năng
\(\begin{align}
& \text{W=}{{\text{W}}_{d}}+{{\text{W}}_{t}} \\
& \Rightarrow {{\text{W}}_{d}}=\text{W}-{{W}_{t}}=2100-3.10.30=1200J \\
& {{\text{W}}_{d}}=\frac{1}{2}.m.{{v}^{2}} \\
& \Rightarrow v=\sqrt{\frac{2.1200}{3}}=20\sqrt{2}m/s \\
\end{align}\)
c) vận tốc khi chạm đất:
\(\begin{align}
& \text{W}={{\text{W}}_{dmax}} \\
& \Leftrightarrow 2100=\frac{1}{2}.3.v_{max}^{2} \\
& \Rightarrow {{v}_{max}}=10\sqrt{14}m/s \\
\end{align}\)
d) Wd=Wt
\(\text{W}={{\text{W}}_{d}}+{{\text{W}}_{t}}=2{{\text{W}}_{d}}=2{{\text{W}}_{t}}\)
\(\Leftrightarrow \left\{ \begin{align}
& 2100=2.\frac{1}{2}.3.{{v}^{2}} \\
& 2100=2.3.10.h \\
\end{align} \right.\)\(\Leftrightarrow \left\{ \begin{align}
& v=10\sqrt{7}m/s \\
& h=35m \\
\end{align} \right.\)
