A bullet of mass 10 g travelling horizontally with a velocity of 150 m s...

Given, initial velocity, $u=150m{s}^{-1}$, final velocity, $v=0$ (because the bullet has finally come to rest), mass of the bullet, $m=10g=0.01kg$ and time taken to come to rest, t=0.03 s.
Let the acceleration of a bullet be a.
We employ the first equation of motion, v=u+at,
$\Rightarrow 0=150+a\times 0.03$
$\Rightarrow a=\frac{-150}{0.03}=-5000m{s}^{-2}$
(A negative sign indicates that the bullet's velocity is decreasing.)
Now we use third equation of motion:
$v^2=u^2+2$as,
$\Rightarrow 0={150}^2+2\times (-5000)s$
$\Rightarrow s=\frac{22500}{10000}=2.25m$
Thus, the distance of penetration of the bullet into the block is 2.25 m.
Newton's second law of motion states:
Force, F=m×a=0.01×5000=50 N
Therefore, the magnitude of force exerted by the wooden block on the bullet is 50 N.