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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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When a body is thrown vertically upwards with a velocity 50 m/s, the percentage of its initial kinetic energy converted to potential energy after 4s is $(g=10\;m/s^2)$

\[(a)\;96\;\% \quad (b)\;50\;\% \quad (c)\;24\;\% \quad (d)\;4\;\%\]
Can you answer this question?
 
 

1 Answer

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Vertical height reached after $4s$ is
$h=ut -\large\frac{1}{2}$$gt^2$
$\quad=50 \times 4 -\large\frac{1}{2}$$\times 10 \times 16 =120 m$
Percentage of energy converted
$\quad=\large\frac{mgh}{\Large\frac{1}{2}mv^2} $$\times 100$
$\quad=\large\frac{2 \times 10 \times 120}{50 \times 50}$$ \times 100$
$\quad=96 \;\%$
Hence a is the correct answer.
answered Jul 22, 2013 by meena.p
edited Jun 5, 2014 by lmohan717
 

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