Question

A body of mass \(4\) kg is placed at a point \(P\) having coordinates \( (3,4) \) m. Under the action of force \( \mathbf{F} = (2\hat{i} + 3\hat{j}) \) N, it moves to a new point \(Q\) having coordinates \( (6,10) \) m in \(4\) sec. The average power and instantaneous power at the end of \(4\) sec are in the ratio:

• A. \( 1:2 \)
• B. \( 6:13 \)
• C. \( 4:3 \)
• D. \( 13:6 \)

Hint:

HERE WE HAVE TO USE Average power and Instantaneous power formula

Answer 1

The Correct Option is B

The displacement vector: \[ \mathbf{d} = (6-3)\hat{i} + (10-4)\hat{j} = 3\hat{i} + 6\hat{j} \] Work done: \[ W = \mathbf{F} \cdot \mathbf{d} = (2\hat{i} + 3\hat{j}) \cdot (3\hat{i} + 6\hat{j}) \] \[ = (2 \times 3) + (3 \times 6) = 6 + 18 = 24 { J} \] Average power: \[ P_{{avg}} = \frac{W}{t} = \frac{24}{4} = 6 { W} \] Instantaneous power: \[ P_{{inst}} = \mathbf{F} \cdot \mathbf{v} = 13 { W} \] \[ \frac{P_{{avg}}}{P_{{inst}}} = \frac{6}{13} \] Thus, the correct answer is (2) \( 6:13 \).