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 find the amount of work done by the resistance of air on the bullet, we can use the work-energy theorem, which states that the work done by the net forces acting on an object is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the bullet, \( m = 20 \, \text{g} = 0.02 \, \text{kg} \) (conversion from grams to kilograms)
- Initial speed of the bullet, \( u = 100 \, \text{m/s} \)
- Final speed of the bullet, \( v = 50 \, \text{m/s} \)

2. **Calculate the initial kinetic energy (KE_initial):**
\[ KE_{\text{initial}} = \frac{1}{2} m u^2 = \frac{1}{2} \times 0.02 \, \text{kg} \times (100 \, \text{m/s})^2 \]
\[ KE_{\text{initial}} = \frac{1}{2} \times 0.02 \times 10000 = 100 \, \text{J} \]

3. **Calculate the final kinetic energy (KE_final):**
\[ KE_{\text{final}} = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.02 \, \text{kg} \times (50 \, \text{m/s})^2 \]
\[ KE_{\text{final}} = \frac{1}{2} \times 0.02 \times 2500 = 25 \, \text{J} \]

4. **Calculate the change in kinetic energy (ΔKE):**
\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = 25 \, \text{J} - 100 \, \text{J} = -75 \, \text{J} \]

5. **Determine the work done by the resistance of air:**
According to the work-energy theorem, the work done \( W \) by the resistance of air is equal to the change in kinetic energy:
\[ W = \Delta KE = -75 \, \text{J} \]

The negative sign indicates that the work done by the resistance of air is in the opposite direction to the motion of the bullet.

### Final Answer:
The amount of work done by the resistance of air is \( W = -75 \, \text{J} \).

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