Question

Find the average retarding force when a 5 g bullet, travelling at 500 m/s, penetrates a 

block of wood a distance of 5 cm before it stops.

Answer 1

Let's first find the deceleration of the bullet:

$$v^2=v_0^2+2ad,$$ $$a=\dfrac{v^2-v_0^2}{2d}$$ $$a=\dfrac{0-(500\ \dfrac{m}{s})^2}{2\cdot5\cdot10^{-2}\ m}=-2.5\cdot10^{6}\ \dfrac{m}{s^2}.$$

The sign minus means that the bullet decelerates.

Finally, we can find the average retarding force from the Newton's Second Law of Motion:

$$F_{ret,avg}=ma=5\cdot10^{-3}\ kg\cdot(-2.5\cdot10^{6}\ \dfrac{m}{s^2})=-12500\ N.$$

The sign minus means that the average retarding force directed in the opposite direction to the motion of the bullet.

Answer:

$F_{ret,avg}=12500\ N,$ in the opposite direction to the motion of the bullet.