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) \( 6:13 \).

Answer 2

Step 1: Given information.
Mass of the body, \( m = 4 \, \text{kg} \)
Initial position, \( P(3, 4) \)
Final position, \( Q(6, 10) \)
Force, \( \mathbf{F} = (2\hat{i} + 3\hat{j}) \, \text{N} \)
Time of motion, \( t = 4 \, \text{s} \).

Step 2: Find the displacement vector.
\[ \overrightarrow{PQ} = (6 - 3)\hat{i} + (10 - 4)\hat{j} = 3\hat{i} + 6\hat{j} \] So, displacement \( \mathbf{s} = 3\hat{i} + 6\hat{j} \).

Step 3: Calculate the work done by the force.
\[ W = \mathbf{F} \cdot \mathbf{s} = (2\hat{i} + 3\hat{j}) \cdot (3\hat{i} + 6\hat{j}) = 2(3) + 3(6) = 6 + 18 = 24 \, \text{J}. \]

Step 4: Calculate average power.
\[ P_{\text{avg}} = \frac{W}{t} = \frac{24}{4} = 6 \, \text{W}. \]

Step 5: Calculate acceleration.
The force and mass are related by Newton’s second law:
\[ \mathbf{F} = m\mathbf{a} \Rightarrow \mathbf{a} = \frac{\mathbf{F}}{m} = \frac{1}{4}(2\hat{i} + 3\hat{j}) = 0.5\hat{i} + 0.75\hat{j}. \]
Step 6: Find initial velocity.
We know displacement \( \mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2 \).
\[ 3\hat{i} + 6\hat{j} = \mathbf{u}(4) + \frac{1}{2}(0.5\hat{i} + 0.75\hat{j})(16) \] \[ 3\hat{i} + 6\hat{j} = 4\mathbf{u} + (4\hat{i} + 6\hat{j}) \] \[ 4\mathbf{u} = (3 - 4)\hat{i} + (6 - 6)\hat{j} = -\hat{i}. \] \[ \mathbf{u} = -\frac{1}{4}\hat{i}. \]

Step 7: Find velocity at \( t = 4 \, \text{s} \).
\[ \mathbf{v} = \mathbf{u} + \mathbf{a}t = \left(-\frac{1}{4}\hat{i}\right) + (0.5\hat{i} + 0.75\hat{j}) \times 4 \] \[ \mathbf{v} = -0.25\hat{i} + (2\hat{i} + 3\hat{j}) = 1.75\hat{i} + 3\hat{j}. \]

Step 8: Instantaneous power at the end of 4 s.
\[ P_{\text{inst}} = \mathbf{F} \cdot \mathbf{v} = (2\hat{i} + 3\hat{j}) \cdot (1.75\hat{i} + 3\hat{j}) = 2(1.75) + 3(3) = 3.5 + 9 = 12.5 \, \text{W}. \]

Step 9: Ratio of average power to instantaneous power.
\[ P_{\text{avg}} : P_{\text{inst}} = 6 : 12.5 = 12 : 25 = 6 : 13. \]

Final Answer:
\[ \boxed{6 : 13} \]