Question

A bullet with a mass of 12 g moving horizontally strikes a fixed block of wood and penetrates a distance of 5.2 cm. The speed of the bullet just before the collision is 640 m/s. What is the average force that the wood exerted on the bullet

Answer 1

Answer:

F = 47261.5 N = 47.26 KN

Explanation:

First, we will calculate the deceleration of the bullet by using the third equation of motion:

$2as = v_f^2-v_i^2$

where,

a = deceleration = ?

s = distace covered = 5.2 cm = 0.052 m

vf = final speed = 0 m/s

vi = initial speed = 640 m/s

Therefore,

$2(a)(0.052\ m) = (0\ m/s)^2-(640\ m/s)^2\\\\a = -\frac{409600\ m^2/s^2}{0.104\ m}\\\\a=-3938461.5\ m/s^2$

negative sign shows deceleration.

Now for the force, we will use Newton's Second Law of Motion:

F = ma

where,

F = Force = ?

m = mass of bullet = 12 g = 0.012 kg

Therefore,

F = (0.012 kg)(3938461.5 m/s²)

F = 47261.5 N = 47.26 KN