A vehicles of mass 10 quintals climba up a hill 200m high and thenmoves on a level road with a velocity of \({36}{k}{m}/{h}\) .Calculate its total mechanical energy while running on the top of the hill.
Answer
Here, \({m}={10}\) qunitals\(={1000}{k}{g}.\)
\({h}={200}{m}.\)
\({v}={36}{k}{m}/{h}=\frac{{{36}\times{1000}}}{{{60}\times{60}}}{m}/{s}={10}{m}/{s}.\)
Potential energy gained.
\({U}={m}{g}{h}={1000}\times{9.8}\times{200}={1.96}\times{10}^{{{6}}}{J}\)
KE of car, \({k}=\frac{{{1}}}{{{2}}}{m}{v}^{{{2}}}=\frac{{{1}}}{{{2}}}\times{1000}{\left({10}\right)}^{{{2}}}\)
\(={0.5}\times{10}^{{{5}}}{J}\)
Mechanical energy \(={U}+{K}\)
\(={1.96}\times{10}^{{{6}}}+{0.5}\times{10}^{{{5}}}={2.01}\times{10}^{{{6}}}{J}\)
\({h}={200}{m}.\)
\({v}={36}{k}{m}/{h}=\frac{{{36}\times{1000}}}{{{60}\times{60}}}{m}/{s}={10}{m}/{s}.\)
Potential energy gained.
\({U}={m}{g}{h}={1000}\times{9.8}\times{200}={1.96}\times{10}^{{{6}}}{J}\)
KE of car, \({k}=\frac{{{1}}}{{{2}}}{m}{v}^{{{2}}}=\frac{{{1}}}{{{2}}}\times{1000}{\left({10}\right)}^{{{2}}}\)
\(={0.5}\times{10}^{{{5}}}{J}\)
Mechanical energy \(={U}+{K}\)
\(={1.96}\times{10}^{{{6}}}+{0.5}\times{10}^{{{5}}}={2.01}\times{10}^{{{6}}}{J}\)
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