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 F→=8i^+2j^N. Then the rate at which the kinetic energy of the particle is increasing at point B is

Displacement d→=r→B−r→A=3−2i^+5−1j^ m⇒d→=i^+4j^mWork done on the particle, W=F→.d→=8i^+2j^.i^+4j^⇒ W=16 JouleBy work - energy theorem, 12mV2=W⇒12.2.V2=16⇒V=4 m/sSince, velocity vector must lie along the force vector, we can write, V→=V.V^=V.F^=4688i^+2j^m/s∴ Rate of increase of kinetic energy =F→.V→=8i^+2j^.4688i^+2j^J/s=468 Watt

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 F→=8i^+2j^N. Then the rate at which the kinetic energy of the particle is increasing at point B is

Displacement d→=r→B−r→A=3−2i^+5−1j^ m⇒d→=i^+4j^mWork done on the particle, W=F→.d→=8i^+2j^.i^+4j^⇒ W=16 JouleBy work - energy theorem, 12mV2=W⇒12.2.V2=16⇒V=4 m/sSince, velocity vector must lie along the force vector, we can write, V→=V.V^=V.F^=4688i^+2j^m/s∴ Rate of increase of kinetic energy =F→.V→=8i^+2j^.4688i^+2j^J/s=468 Watt

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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

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$2 \sqrt{74} W$

$4 \sqrt{52} W$

$2 \sqrt{54} W$

$4 \sqrt{68} W$

answer is D.

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Detailed 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 $= \vec{F} . \vec{V} = (8 \hat{i} + 2 \hat{j}) . [\frac{4}{\sqrt{68}} (8 \hat{i} + 2 \hat{j})] J / s = 4 \sqrt{68} Watt$