A 20-g bullet with an initial velocity of 3.0 $\times$ 10⁴ cm/s penetrates 6.0 cm into a stationary wall.The average force exerted on the bullet by the wall in bringing it to rest is approximately A)1.5 $\times$ 10⁴ N B)2.3 $\times$ 10³ N C)9.8 $\times$ 10^-3 N D)2.0 $\times$ 10² N E)4.7 $\times$ 10^-2 N

Question

Quizplus · Accepted Answer

The average force exerted on the bullet can be calculated using the work-energy principle. The work done by the force is equal to the change in kinetic energy of the bullet. The initial kinetic energy of the bullet is given by $ \frac{1}{2}mv^2 $, where $ m = 0.02 $ kg (converting 20 g to kg) and $ v = 300 $ m/s (converting 30000 cm/s to m/s). The final kinetic energy is 0 (since the bullet comes to rest). The work done by the force is also equal to the force times the distance over which it acts ( $ F \times d $ ), where $ d = 0.06 $ m (converting 6 cm to m). Solving for $ F $ gives us the average force.

A 20-G Bullet with an Initial Velocity of 3 ×\times× 104 Cm/s Penetrates 6

A 20-G Bullet with an Initial Velocity of 3 $\times$ 10⁴ Cm/s Penetrates 6