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 × 0.03 s)

(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

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 x Acceleration

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

Acceleration of the bullet, a = 5000 m/s^{2 }

F = ma = 0.01×5000 = 50 N