A bullet of mass `0.05` kg moving with a speed of `80m//s` enters a wooden block and is stopped after a distance of 0.40 m The average resistive force exerted by the block on the bullet is:

Options

A. 300 N
B. 20 N
C. 400 N
D. 40 N

Explanation

To find the average resistive force exerted by the wooden block on the bullet, we can follow these steps:

### Step 1: Calculate the initial kinetic energy of the bullet.
The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
where:
- \( m = 0.05 \, \text{kg} \) (mass of the bullet)
- \( v = 80 \, \text{m/s} \) (initial speed of the bullet)

Substituting the values:
\[ KE = \frac{1}{2} \times 0.05 \, \text{kg} \times (80 \, \text{m/s})^2 \]
\[ KE = \frac{1}{2} \times 0.05 \times 6400 \]
\[ KE = 0.025 \times 6400 = 160 \, \text{J} \]

### Step 2: Use the work-energy principle to find the average resistive force.
According to the work-energy principle, the work done by the resistive force is equal to the change in kinetic energy. Since the bullet comes to a stop, its final kinetic energy is 0. Therefore, the work done by the resistive force (W) is equal to the initial kinetic energy:
\[ W = F \cdot d \]
where:
- \( F \) is the average resistive force
- \( d = 0.40 \, \text{m} \) (distance over which the bullet is stopped)

Setting the work done equal to the initial kinetic energy:
\[ F \cdot d = KE \]
\[ F \cdot 0.40 = 160 \]

### Step 3: Solve for the average resistive force \( F \).
Rearranging the equation to solve for \( F \):
\[ F = \frac{KE}{d} \]
Substituting the values:
\[ F = \frac{160}{0.40} \]
\[ F = 400 \, \text{N} \]

### Conclusion:
The average resistive force exerted by the block on the bullet is **400 N**.

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