Calculating the Average Resistive Force Exerted by a Wooden Block on a Bullet

In this article, we will delve into the physics behind the interaction between a bullet and a wooden block. We will use various equations and methods to find the average resistive force exerted by the block on the bullet as it stops. This scenario involves a detailed analysis of the bullet's motion, its deceleration, and the forces acting upon it.

Given Data and Initial Conditions

The bullet has a mass of 0.04 kg and is initially moving with a velocity of 90 m/s. It enters a heavy wooden block and comes to a complete stop after traversing a distance of 6 cm (0.06 meters).

Method 1: Using Work-Energy Theorem

Let's start with a method that utilizes the work-energy theorem.

The initial kinetic energy of the bullet is given by:

(KE_i frac{1}{2}mv^2 frac{1}{2} times 0.04 ,text{kg} times (90 ,text{m/s})^2 1620 , text{J})

Since the bullet comes to a complete stop, its final kinetic energy (KE_f 0). The work done by the resistive force is equal to the change in kinetic energy:

(W Delta KE KE_f - KE_i -1620 , text{J})

The work done is also given by:

(W F_{res} times d)

Where (F_{res}) is the resistive force and (d) is the distance over which the force acts. Substituting the known values:

(-1620 , text{J} F_{res} times 0.06 , text{m})

Solving for (F_{res}) gives:

(F_{res} frac{-1620 , text{J}}{0.06 , text{m}} -27000 , text{N})

The negative sign indicates that the force is in the opposite direction to the motion of the bullet, confirming that it is a resistive force. Thus, the average resistive force exerted by the block on the bullet is:

(F_{res} 2700 , text{N})

Method 2: Using Kinematic Equations and Newton's Second Law

We can also solve this problem using kinematic equations and Newton's second law.

First, we find the deceleration (a) using the kinematic equation:

(v^2 u^2 2as)

Since the final velocity (v 0), the equation simplifies to:

(0 90^2 2a times 0.06)

Rearranging for (a):

(2a times 0.06 -8100)

(a frac{-8100}{0.12} -67500 , text{m/s}^2)

Using Newton's second law, (F ma), we find the resistive force:

(F 0.04 , text{kg} times 67500 , text{m/s}^2 2700 , text{N})

Method 3: Using the Third Kinematic Equation

Another approach is to use the third kinematic equation:

(v^2 - u^2 2as)

As before, the final velocity (v 0 , text{m/s}). The initial velocity (u 90 , text{m/s}) and the distance (s 0.06 , text{m}). Substituting these values, we get:

(0 - 90^2 2a times 0.06)

(-8100 2a times 0.06)

(2a -8100/0.12 -67500 , text{m/s}^2)

Using Newton's second law, (F ma), we find the resistive force:

(F 0.04 , text{kg} times 67500 , text{m/s}^2 2700 , text{N})

Conclusion

Using different methods, we consistently find that the average resistive force exerted by the wooden block on the bullet is 2700 Newtons. This demonstrates the reliability of the physical principles involved in solving such problems.

Keywords

Average resistive force, bullet physics, resistance force calculation