A 3 kg object has initial velocity `(6hati-2hatj)ms^(-1)`. What will be the total work done (in joule) on the object if its velocity changes to `(8hati+4hatj)ms^(-1)`?

Answer

To find the total work done on the object when its velocity changes, 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 Initial and Final Velocities:**
- Initial velocity \( \mathbf{V_i} = 6 \hat{i} - 2 \hat{j} \, \text{m/s} \)
- Final velocity \( \mathbf{V_f} = 8 \hat{i} + 4 \hat{j} \, \text{m/s} \)

2. **Calculate the Magnitude of the Initial Velocity:**
\[ |\mathbf{V_i}| = \sqrt{(6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} \, \text{m/s} \]

3. **Calculate the Magnitude of the Final Velocity:**
\[ |\mathbf{V_f}| = \sqrt{(8)^2 + (4)^2} = \sqrt{64 + 16} = \sqrt{80} \, \text{m/s} \]

4. **Calculate the Initial Kinetic Energy:**
The formula for kinetic energy is given by:
\[ KE = \frac{1}{2} m v^2 \]
For the initial kinetic energy \( KE_i \):
\[ KE_i = \frac{1}{2} \times 3 \, \text{kg} \times (|\mathbf{V_i}|)^2 = \frac{1}{2} \times 3 \times 40 = 60 \, \text{J} \]

5. **Calculate the Final Kinetic Energy:**
For the final kinetic energy \( KE_f \):
\[ KE_f = \frac{1}{2} \times 3 \, \text{kg} \times (|\mathbf{V_f}|)^2 = \frac{1}{2} \times 3 \times 80 = 120 \, \text{J} \]

6. **Calculate the Change in Kinetic Energy:**
\[ \Delta KE = KE_f - KE_i = 120 \, \text{J} - 60 \, \text{J} = 60 \, \text{J} \]

7. **Conclusion:**
The total work done on the object is equal to the change in kinetic energy:
\[ W = \Delta KE = 60 \, \text{J} \]

### Final Answer:
The total work done on the object is \( 60 \, \text{J} \).