A bullet of mass 20 g leaves a riffle at an initial speed 100 m/s and strikes a target at the same level with speed 50 m/s. The amount of work done by the resistance of air will be

Options

A. 100 J
B. 25 J
C. 75 J
D. 50 J

Answer

To solve the problem, we will apply the work-energy theorem, which states that the work done by all forces acting on an object is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the bullet, \( m = 20 \, \text{g} = 20 \times 10^{-3} \, \text{kg} = 0.02 \, \text{kg} \)
- Initial speed of the bullet, \( v_i = 100 \, \text{m/s} \)
- Final speed of the bullet, \( v_f = 50 \, \text{m/s} \)

2. **Calculate Initial Kinetic Energy (\( KE_i \)):**
\[ KE_i = \frac{1}{2} m v_i^2 = \frac{1}{2} \times 0.02 \, \text{kg} \times (100 \, \text{m/s})^2 \]
\[ KE_i = 0.01 \times 10000 = 100 \, \text{J} \]

3. **Calculate Final Kinetic Energy (\( KE_f \)):**
\[ KE_f = \frac{1}{2} m v_f^2 = \frac{1}{2} \times 0.02 \, \text{kg} \times (50 \, \text{m/s})^2 \]
\[ KE_f = 0.01 \times 2500 = 25 \, \text{J} \]

4. **Calculate the Change in Kinetic Energy (\( \Delta KE \)):**
\[ \Delta KE = KE_f - KE_i = 25 \, \text{J} - 100 \, \text{J} = -75 \, \text{J} \]

5. **Apply the Work-Energy Theorem:**
According to the work-energy theorem:
\[ W = \Delta KE \]
Here, \( W \) is the work done by the resistance of air. Since the change in kinetic energy is negative, the work done by air resistance is:
\[ W = -75 \, \text{J} \]

6. **Find the Magnitude of Work Done by Air Resistance:**
The amount of work done by air resistance is:
\[ |W| = 75 \, \text{J} \]

### Final Answer:
The amount of work done by the resistance of air is **75 J**.

---