A ball of mass 100 gm is projected upwards with velocity `(10 m//s)`. It returns back with `(6 m//s)`. Find work done by air resistance .

Explanation

To solve the problem of finding the work done by air resistance on a ball projected upwards, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the ball, \( m = 100 \, \text{g} = 0.1 \, \text{kg} \) (conversion from grams to kilograms).
- Initial velocity when projected upwards, \( v_i = 10 \, \text{m/s} \).
- Final velocity when returning downwards, \( v_f = 6 \, \text{m/s} \).

2. **Use the Work-Energy Theorem:**
- According to the work-energy theorem, the total work done on an object is equal to the change in its kinetic energy.
- The formula for kinetic energy (KE) is given by:
\[ KE = \frac{1}{2} m v^2 \]
- Therefore, the change in kinetic energy (\( \Delta KE \)) can be expressed as:
\[ \Delta KE = KE_f - KE_i = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \]

3. **Calculate the Initial and Final Kinetic Energies:**
- Calculate the final kinetic energy when the ball returns:
\[ KE_f = \frac{1}{2} m v_f^2 = \frac{1}{2} \times 0.1 \, \text{kg} \times (6 \, \text{m/s})^2 = \frac{1}{2} \times 0.1 \times 36 = 1.8 \, \text{J} \]
- Calculate the initial kinetic energy when the ball is projected:
\[ KE_i = \frac{1}{2} m v_i^2 = \frac{1}{2} \times 0.1 \, \text{kg} \times (10 \, \text{m/s})^2 = \frac{1}{2} \times 0.1 \times 100 = 5 \, \text{J} \]

4. **Calculate the Change in Kinetic Energy:**
- Now, substitute the values into the change in kinetic energy equation:
\[ \Delta KE = KE_f - KE_i = 1.8 \, \text{J} - 5 \, \text{J} = -3.2 \, \text{J} \]

5. **Determine the Work Done by Air Resistance:**
- The work done by air resistance (\( W_{\text{air}} \)) is equal to the change in kinetic energy:
\[ W_{\text{air}} = \Delta KE = -3.2 \, \text{J} \]

### Final Answer:
The work done by air resistance is \( -3.2 \, \text{J} \).

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