Now, it is given that the bullet is travelling with a velocity of $150\, m/s$.

Thus, when the bullet enters the block, its velocity = Initial velocity, $u = 150\, m/s$

Final velocity, $v = 0$ (since the bullet finally comes to rest)

Time taken to come to rest, $t= 0.03\, s$

According to the first equation of motion,$ v = u + at$

Acceleration of the bullet, $a$

$0 = 150 + (a \times  0.03 \,s)$

$a=\frac{-150}{0.03}=-5000 \,m / s ^{2}$

(Negative sign indicates that the velocity of the bullet is decreasing.)

According to the third equation of motion:

$v^2= u^2 + 2as$

$0 = (150)^2 + 2 ( – 5000) \,s$

$s=\frac{-(150)^{2}}{-2(5000)}=\frac{22500}{10000}=2.25\, m$

Hence, the distance of penetration of the bullet into the block is $2.25\, m$.

From Newton's second law of motion:

Force, $F =$ Mass $\times $ Acceleration

Mass of the bullet, $m = 10 \,g = 0.01\, kg$

Acceleration of the bullet, $a = 5000\, m/s^2$

$F = ma$

$= 0.01\times 5000 = 50\, N$