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A man throws the bricks to a height of 12 m where they reach with a speed of \({12}{m}/{s}\). If he throws the bricks such that they just reach that height, what percentage of energy will be saved? \({\left({g}={9.8}{m}/{s}^{{{2}}}\right)}\)
(A) \({9.5}\%\)
(B) \({19}\%\)
(C) \({38}\%\)
(D) \({76}\%\)

Answer

Answer: C
In first Case, \({W}_{{{1}}}=\frac{{1}}{{2}}{m}{\left({v}_{{{1}}}\right)}^{{{2}}}+{m}{g}{h}\)
\(=\frac{{1}}{{2}}{m}{\left({12}\right)}^{{{2}}}+{m}\times{10}\times{12}\)
=72m + 120 m
and in second case, \({W}_{{{2}}}\) = mgh
=120 m
The percentage of energy saved
\(=\frac{{{192}{m}-{120}{m}}}{{{192}{m}}}\times{100}={38}\%\)
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