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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A $30\; g$ bullet initially travelling at $120 \;m/s$ penetrates $12\; cm$, into a wooden block. The average resistance exerted by block is

\[(a)\;2850\;N \quad (b)\;2200\; N \quad (c)\;2000\;N \quad (d)\;1800\; N\]

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Work done = Kinetic energy lost
$F . s= m \large \frac{v^2-v^2}{2}$
$F= m \large \frac{v^2-v^2}{2s}$
$\quad=\large\frac{30 \times 10^{-3} \times (120)^2}{2 \times 12 \times 10^{-2}}$
$\quad=1800 N$
Hence d is the correct answer. 


answered Jul 15, 2013 by meena.p
edited Feb 10, 2014 by meena.p

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