Question

A 40g bullet traveling at 450 m/s passed through a board and comes out traveling 300m/s. The board is 7.6cm thick. What is the force of friction applied by the wood to the bullet?

Answer 1

f = 29605.2 [N]

Step-by-step explanation:

To solve this problem we must use the following equation of kinematics, then use Newton's second law.

`v_(f)^(2) =v_(o)^(2) -2*a*x`

where:

Vf = final velocity = 300 [m/s]

Vo = initial velocity = 450 [m/s]

a = acceleration [m/s²]

x = distance = 7.6 [cm] = 0.076 [m]

Now replacing and clearing a

2*a*0.076 = (450² - 300²)

a = 740131.57 [m/s²]

Now using Newton's second law which tells us that the force on a body is equal to the product of mass by acceleration.

f = m*a

where:

f = friction force [N]

m = mass = 40 [g] = 0.04 [kg]

f = 0.04*740131.57

f = 29605.2 [N]