A `30gm` bullet initially at `120m//s` penetrates `12cm` into a wooden block. The average block. The average resistance exerted by the wooden block is.

Options

A. `2850N`
B. `2200N`
C. `2000N`
D. `1800N`

Explanation

To solve the problem of finding the average resistance exerted by the wooden block on the bullet, we can follow these steps:

### Step 1: Convert Units
Convert the mass of the bullet from grams to kilograms and the distance from centimeters to meters.
- Mass of the bullet: \(30 \, \text{g} = 0.03 \, \text{kg}\)
- Distance penetrated: \(12 \, \text{cm} = 0.12 \, \text{m}\)

### Step 2: Identify Initial and Final Velocities
Identify the initial and final velocities of the bullet.
- Initial velocity, \(u = 120 \, \text{m/s}\)
- Final velocity, \(v = 0 \, \text{m/s}\) (since the bullet comes to a stop)

### Step 3: Use the Equation of Motion
Use the equation of motion to find the acceleration (or deceleration) of the bullet as it penetrates the block:
\[ v^2 = u^2 + 2as \]
Where:
- \(v\) = final velocity
- \(u\) = initial velocity
- \(a\) = acceleration
- \(s\) = distance penetrated

Substituting the known values:
\[ 0 = (120)^2 + 2a(0.12) \]
\[ 0 = 14400 + 0.24a \]
\[ 0.24a = -14400 \]
\[ a = \frac{-14400}{0.24} = -60000 \, \text{m/s}^2 \]

### Step 4: Calculate the Average Resistance Force
Now, use Newton's second law to find the average resistance force exerted by the wooden block:
\[ F = ma \]
Where:
- \(F\) = force (resistance)
- \(m\) = mass of the bullet
- \(a\) = acceleration (deceleration in this case)

Substituting the values:
\[ F = 0.03 \, \text{kg} \times (-60000 \, \text{m/s}^2) \]
\[ F = -1800 \, \text{N} \]

The negative sign indicates that the force is acting in the opposite direction to the motion of the bullet, which is expected for a resistance force.

### Final Answer
The average resistance exerted by the wooden block is \(1800 \, \text{N}\).

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