Question

A body of mass 4 kg is placed on a plane at a point P having coordinate $(3,4)$ m. Under the action of force $\vec{F} = (2\hat{i} + 3\hat{j})$ 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 of:

• A. $6 : 13$
• B. $4 : 3$
• C. $1 : 2$
• D. $6 : 13$

Hint:

For constant force motion starting from rest, verify displacement using $\vec{s} = \frac{1}{2}\vec{a}t^2$ to confirm consistency.

Answer 1

The Correct Option is A

Concept:

Average power $= \dfrac{\text{Work done}}{\text{Time}}$
Instantaneous power $= \vec{F} \cdot \vec{v}$
Work done by a constant force $= \vec{F} \cdot \vec{s}$
Step 1: Displacement of the body: \[ \vec{s} = (6 - 3)\hat{i} + (10 - 4)\hat{j} = 3\hat{i} + 6\hat{j} \]
Step 2: Work done by the force: \[ W = \vec{F} \cdot \vec{s} = (2\hat{i} + 3\hat{j}) \cdot (3\hat{i} + 6\hat{j}) \] \[ W = 6 + 18 = 24\,\text{J} \]
Step 3: Average power: \[ P_{\text{avg}} = \frac{W}{t} = \frac{24}{4} = 6\,\text{W} \]
Step 4: Acceleration of the body: \[ \vec{a} = \frac{\vec{F}}{m} = \frac{1}{4}(2\hat{i} + 3\hat{j}) = 0.5\hat{i} + 0.75\hat{j} \]
Step 5: Velocity at the end of 4 sec (initial velocity zero): \[ \vec{v} = \vec{a}t = 4(0.5\hat{i} + 0.75\hat{j}) = 2\hat{i} + 3\hat{j} \]
Step 6: Instantaneous power at $t = 4$ sec: \[ P_{\text{inst}} = \vec{F} \cdot \vec{v} = (2\hat{i} + 3\hat{j}) \cdot (2\hat{i} + 3\hat{j}) \] \[ P_{\text{inst}} = 4 + 9 = 13\,\text{W} \]
Step 7: Required ratio: \[ P_{\text{avg}} : P_{\text{inst}} = 6 : 13 \]