A particle of mass 0.020 kg moves along a curve with velocity `5.0 hat(i)+18 hat(k) m//s`. After some time, the velocity changes to `9.0 hat(i) + 22 hat(j)` m/s due to action of a single force. Find the work done on the particle during this interval of time.

Answer

To find the work done on the particle during the interval of time, we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Identify the mass and velocities**:
- Mass of the particle, \( m = 0.020 \, \text{kg} \)
- Initial velocity, \( \mathbf{v_i} = 5 \hat{i} + 18 \hat{k} \, \text{m/s} \)
- Final velocity, \( \mathbf{v_f} = 9 \hat{i} + 22 \hat{j} \, \text{m/s} \)

2. **Calculate the initial kinetic energy (\( KE_i \))**:
- The formula for kinetic energy is given by:
\[ KE = \frac{1}{2} m v^2 \]
- First, find the magnitude of the initial velocity:
\[ v_i = \sqrt{(5)^2 + (18)^2} = \sqrt{25 + 324} = \sqrt{349} \]
- Now calculate the initial kinetic energy:
\[ KE_i = \frac{1}{2} \times 0.020 \times (349) = 0.010 \times 349 = 3.49 \, \text{J} \]

3. **Calculate the final kinetic energy (\( KE_f \))**:
- Find the magnitude of the final velocity:
\[ v_f = \sqrt{(9)^2 + (22)^2} = \sqrt{81 + 484} = \sqrt{565} \]
- Now calculate the final kinetic energy:
\[ KE_f = \frac{1}{2} \times 0.020 \times (565) = 0.010 \times 565 = 5.65 \, \text{J} \]

4. **Calculate the work done (\( W \))**:
- The work done is the change in kinetic energy:
\[ W = KE_f - KE_i = 5.65 \, \text{J} - 3.49 \, \text{J} = 2.16 \, \text{J} \]

### Final Answer:
The work done on the particle during this interval of time is \( \boxed{2.16 \, \text{J}} \).