Question

A 5.00-g bullet is fired at a speed of 100 m/s into a fixed block of wood, it comes to rest after penetrating 6.00 cm into the wood. what is the average stopping force of the buller?

Answer 1

Mass Of The Bullet=5 g

In kg=5/1000=0.005 kg

Intial Velocity(u)Before striking the wood=100 m/s

Final Velocity (v)=0 m/s After Penetrating the wood.

Distance (s)=6cm=0.06 m

(- Accerlation) (a)=a

Using Motion of Equation

v²=u²+2as

0²=100²+2×a×0.06

0=10000+0.12a

0-10000=0.12a

-10000=0.12a

$a = \frac{ - 10000}{0.12} \\ \\ a = - 83333.33 \: m {s}^{2}$

We know Force=Mass ×Accerlation

Force=0.005×83333.33

=-416.66 N

Approx=417 N

$\boxed{\mathbf{Force=417 \:N}}$