Question

An 8.0-g bullet with a speed of 800 m/s is shot into a wooden block and penetrates 20 cm before stopping. What is the average force of the wood on the bullet? Assume the block does not move.

a) 320,000 N
b) 40,000 N
c) 4,000 N
d) 32,000 N

Answer 1

To find the average force exerted by the wood on the bullet, we can use the equation Favg = m * a, where Favg is the average force, m is the mass of the bullet, and a is the acceleration. Using the given information, the average force of the wood on the bullet is approximately 32,000 N.

Step-by-step explanation:

To find the average force exerted by the wood on the bullet, we can use the equation:

Favg = m * a

where Favg is the average force, m is the mass of the bullet, and a is the acceleration.

First, we need to calculate the acceleration using the formula:

a = ∆v / ∆t

where ∆v is the change in velocity and ∆t is the change in time.

Given that the bullet stops completely, its final velocity is 0 m/s. The initial velocity can be found using the equation:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled. Rearranging the equation, we get:

u = sqrt(v^2 - 2as)

Using the given values, we can calculate the acceleration and the average force exerted by the wood on the bullet.

The average force of the wood on the bullet is approximately 32,000 N.