A force `vecF=(3veci+4hatj)N` acts on a 2 kg movable object that moves from an initial position `vecd_(1)=(-3hatij-2hatj)m` to a final position `vecd_(f)=(5hati+4hatj)` m is 6 s. The average power (in watt) delivered by the force during the interval is equal to:

Explanation

To solve the problem of finding the average power delivered by the force acting on the object, we can follow these steps:

### Step 1: Identify the given quantities
- Force vector: \(\vec{F} = 3\hat{i} + 4\hat{j} \, \text{N}\)
- Initial position: \(\vec{d}_i = -3\hat{i} - 2\hat{j} \, \text{m}\)
- Final position: \(\vec{d}_f = 5\hat{i} + 4\hat{j} \, \text{m}\)
- Time interval: \(t = 6 \, \text{s}\)

### Step 2: Calculate the displacement vector
The displacement vector \(\vec{d}\) is given by the difference between the final and initial positions:
\[ \vec{d} = \vec{d}_f - \vec{d}_i \]
Substituting the values:
\[ \vec{d} = (5\hat{i} + 4\hat{j}) - (-3\hat{i} - 2\hat{j}) \]
\[ = 5\hat{i} + 4\hat{j} + 3\hat{i} + 2\hat{j} \]
\[ = (5 + 3)\hat{i} + (4 + 2)\hat{j} \]
\[ = 8\hat{i} + 6\hat{j} \, \text{m} \]

### Step 3: Calculate the work done by the force
The work done \(W\) by the force is given by the dot product of the force vector and the displacement vector:
\[ W = \vec{F} \cdot \vec{d} \]
Calculating the dot product:
\[ W = (3\hat{i} + 4\hat{j}) \cdot (8\hat{i} + 6\hat{j}) \]
\[ = (3 \times 8) + (4 \times 6) \]
\[ = 24 + 24 \]
\[ = 48 \, \text{J} \]

### Step 4: Calculate the average power
The average power \(P\) is given by the work done divided by the time interval:
\[ P = \frac{W}{t} \]
Substituting the values:
\[ P = \frac{48 \, \text{J}}{6 \, \text{s}} \]
\[ = 8 \, \text{W} \]

### Final Answer
The average power delivered by the force during the interval is \(8 \, \text{W}\).

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