A body of mass 4 kg is placed on a plane at a point P haing coordinate (3, 4) m. Under the action of force , 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

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Solution

Q.
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 : [2025]

1 13 : 6

2 6 : 13

3 1 : 2

4 4 : 13

Ans.
(2)

$\vec{F} = (2 \hat{i} + 3 \hat{j}) N$ ; displacement $\vec{S} = (3 \hat{i} + 6 \hat{j}) m$

Acceleration, $\vec{a} = \frac{\vec{F}}{m} = (\frac{2 \hat{i} + 3 \hat{j}}{4}) m / s^{2}$

   $\vec{S} = \vec{u} t + \frac{1}{2} \vec{a} t^{2}$

$\Rightarrow 3 \hat{i} + 6 \hat{j} = 4 \vec{u} + \frac{1}{2} (\frac{2 \hat{i} + 3 \hat{j}}{4}) \cdot 16 = 4 \vec{u} + 4 \hat{i} + 6 \hat{j}$

or $\vec{u} = \frac{1}{4} (3 \hat{i} + 6 \hat{j} - 4 \hat{i} - 6 \hat{j}) \Rightarrow \vec{u} = - \frac{1}{4} \hat{i}$

Average Power $< P > = \vec{F} \cdot \frac{\vec{S}}{t}$

   $= \frac{(2 \hat{i} + 3 \hat{j}) \cdot (3 \hat{i} + 6 \hat{j})}{4} watt = 6 watt$

   $\vec{v}$ $at t = 4 \sec = (\frac{1}{2} \hat{i} + \frac{3}{4} \hat{j}) \times 4 = (2 \hat{i} + 3 \hat{j})$

   $P_{ins} = \vec{F} \cdot \vec{v} = (2 \hat{i} + 3 \hat{j}) (2 \hat{i} + 3 \hat{j}) = 13 W$

$\Rightarrow \frac{< P >}{P_{ins}} = \frac{6}{13}$

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