A force `3hat i+4 hat j - 5 hatk` N is acting on a particle. If the velocity of particle at an instant is `2hati + 2 hat j + 2 hatk` m/s , find the instantaneous power developed by the force (in watts).

Explanation

To find the instantaneous power developed by the force acting on the particle, we can use the formula for power in terms of force and velocity:

\[ P = \mathbf{F} \cdot \mathbf{v} \]

Where:
- \( P \) is the instantaneous power,
- \( \mathbf{F} \) is the force vector,
- \( \mathbf{v} \) is the velocity vector,
- \( \cdot \) denotes the dot product.

### Step 1: Identify the force and velocity vectors
Given:
- Force vector \( \mathbf{F} = 3\hat{i} + 4\hat{j} - 5\hat{k} \) N
- Velocity vector \( \mathbf{v} = 2\hat{i} + 2\hat{j} + 2\hat{k} \) m/s

### Step 2: Calculate the dot product \( \mathbf{F} \cdot \mathbf{v} \)
The dot product of two vectors \( \mathbf{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \) and \( \mathbf{B} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k} \) is given by:

\[ \mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 + a_3b_3 \]

Applying this to our force and velocity vectors:

\[ \mathbf{F} \cdot \mathbf{v} = (3)(2) + (4)(2) + (-5)(2) \]

### Step 3: Perform the calculations
Calculating each term:
- \( 3 \times 2 = 6 \)
- \( 4 \times 2 = 8 \)
- \( -5 \times 2 = -10 \)

Now, summing these results:

\[ \mathbf{F} \cdot \mathbf{v} = 6 + 8 - 10 \]

\[ \mathbf{F} \cdot \mathbf{v} = 14 - 10 = 4 \]

### Step 4: Conclusion
The instantaneous power \( P \) is equal to the dot product we calculated:

\[ P = 4 \text{ watts} \]

### Final Answer
The instantaneous power developed by the force is **4 watts**.

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