First slide

Power

Question

A particle of mass 2 kg is at rest at point A (2m, 1m). It is displaced to point B (3m, 5m) under the action of a constant force $\vec{F} = (8 \hat{i} + 2 \hat{j}) N$ . Then the rate at which the kinetic energy of the particle is increasing at point B is

Difficult

A

$2 \sqrt{74} W$
B

$4 \sqrt{52} W$
C

$2 \sqrt{54} W$
D

$4 \sqrt{68} W$

Solution

Displacement $\vec{d} = {\vec{r}}_{B} - {\vec{r}}_{A} = (3 - 2) \hat{i} + (5 - 1) \hat{j} m$
$\Rightarrow \vec{d} = (\hat{i} + 4 \hat{j}) m$
Work done on the particle, $W = \vec{F} . \vec{d} = (8 \hat{i} + 2 \hat{j}) . (\hat{i} + 4 \hat{j})$
$\Rightarrow W = 16 Joule$
By work - energy theorem, $\frac{1}{2} m V^{2} = W \Rightarrow \frac{1}{2} .2. V^{2} = 16 \Rightarrow V = 4 m / s$
Since, velocity vector must lie along the force vector, we can write, $\vec{V} = V . \hat{V} = V . \hat{F} = \frac{4}{\sqrt{68}} (8 \hat{i} + 2 \hat{j}) m / s$
$∴$ Rate of increase of kinetic energy $= \vec{F} . \vec{V} = (8 \hat{i} + 2 \hat{j}) . [\frac{4}{\sqrt{68}} (8 \hat{i} + 2 \hat{j})] J / s = 4 \sqrt{68} Watt$

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