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 as 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 mass and initial/final velocities:**
- Mass (m) = 3 kg
- Initial velocity (V1) = \(6 \hat{i} - 2 \hat{j}\) m/s
- Final velocity (V2) = \(8 \hat{i} + 4 \hat{j}\) m/s

2. **Calculate the initial kinetic energy (KE1):**
\[ KE_1 = \frac{1}{2} m V_1^2 \]
First, find \(V_1^2\):
\[ V_1^2 = (6^2 + (-2)^2) = 36 + 4 = 40 \, \text{m}^2/\text{s}^2 \]
Now calculate \(KE_1\):
\[ KE_1 = \frac{1}{2} \times 3 \times 40 = \frac{120}{2} = 60 \, \text{J} \]

3. **Calculate the final kinetic energy (KE2):**
\[ KE_2 = \frac{1}{2} m V_2^2 \]
First, find \(V_2^2\):
\[ V_2^2 = (8^2 + 4^2) = 64 + 16 = 80 \, \text{m}^2/\text{s}^2 \]
Now calculate \(KE_2\):
\[ KE_2 = \frac{1}{2} \times 3 \times 80 = \frac{240}{2} = 120 \, \text{J} \]

4. **Calculate the change in kinetic energy (ΔKE):**
\[ \Delta KE = KE_2 - KE_1 = 120 \, \text{J} - 60 \, \text{J} = 60 \, \text{J} \]

5. **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 Joules**.