A 2 kg block of wood rests on a long table top. A 5 g bullet moving horizontally with a speed of `150 ms^(-1)` is shot into the block and sticks to it. The block then slides 2.7m along the table top and comes to a stop. The force of friction between the block and the table is

Options

A. 0.052 N
B. 3.63 N
C. 2.50 N
D. 1.04 N

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- Mass of the block (M) = 2 kg
- Mass of the bullet (m) = 5 g = 0.005 kg (convert grams to kilograms)
- Initial speed of the bullet (v₁) = 150 m/s
- Distance traveled by the block after the bullet sticks (s) = 2.7 m

### Step 2: Use conservation of momentum to find the final velocity (v_f) after the bullet sticks to the block
According to the law of conservation of momentum:
\[ \text{Initial momentum} = \text{Final momentum} \]
\[ m \cdot v_1 + M \cdot 0 = (M + m) \cdot v_f \]
Substituting the values:
\[ 0.005 \cdot 150 + 2 \cdot 0 = (2 + 0.005) \cdot v_f \]
Calculating the left side:
\[ 0.75 = 2.005 \cdot v_f \]
Now, solving for \(v_f\):
\[ v_f = \frac{0.75}{2.005} \approx 0.374 \text{ m/s} \]

### Step 3: Calculate the deceleration (a) using the equation of motion
We will use the equation:
\[ v^2 = u^2 + 2as \]
Where:
- \(v\) = final velocity = 0 (the block comes to a stop)
- \(u\) = initial velocity = \(v_f\) = 0.374 m/s
- \(s\) = distance = 2.7 m
- \(a\) = acceleration (deceleration in this case)

Rearranging the equation gives:
\[ 0 = (0.374)^2 + 2a(2.7) \]
\[ 0.139876 = -5.4a \]
Now solving for \(a\):
\[ a = -\frac{0.139876}{5.4} \approx -0.0259 \text{ m/s}^2 \]

### Step 4: Calculate the frictional force (F_f)
The frictional force can be calculated using Newton's second law:
\[ F_f = (M + m) \cdot a \]
Substituting the values:
\[ F_f = (2 + 0.005) \cdot (-0.0259) \]
\[ F_f = 2.005 \cdot (-0.0259) \approx -0.0519 \text{ N} \]
The negative sign indicates that the frictional force acts in the opposite direction of motion.

### Step 5: Present the final answer
The magnitude of the frictional force is approximately:
\[ F_f \approx 0.052 \text{ N} \]