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Q.
If a bullet of mass $10 \,gm$ moving with velocity $150m/sec$ penetrates the wooden block upto $6 \,cm$. Then the average force imposed by the bullet on the block is
Laws of Motion
Solution:
Using $v ^{2}= u ^{2}+2 as$
$0 = (150)^2 + 2 \times a \times 0.06$
$ \frac{-22500}{2 \times 0.06} = a $
$\Rightarrow a = \frac{-22500}{0.12}$
(-ve sign indicates opposite direction)
$F = 10^2 \times \frac{22500}{0.12} = 1875\,N$