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A particle of mass 0.5 kg is moving in xy plane with acceleration $\vec{a} = (3 \hat{i} + 2 t \hat{j}) m / s^{2} .$ At time t, its velocity is given by $\vec{V} = (2 t \hat{i} + 3 t^{2} \hat{j}) m / s .$ Then rate of work done on the particle at t = 1 sec is

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Correct option is A

(1)Force acting on the particle, F→=ma→=0.53i^+2tj^N∴ Power, P=F→.V→=0.53i^+2tj^ . 2ti^+3tj^=12 6t+6t3 wattP=3t+t3 wattAt t=1 sec, P=31+1watt=6 watt.

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A particle of mass 0.5 kg is moving in xy plane with acceleration a→=3i^+2tj^m/s2. At time t, its velocity is given by V→=2ti^+3t2j^m/s. Then rate of work done on the particle at t = 1 sec is

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Ready to Test Your Skills?

free study material

A particle of mass 0.5 kg is moving in xy plane with acceleration $\vec{a} = (3 \hat{i} + 2 t \hat{j}) m / s^{2} .$ At time t, its velocity is given by $\vec{V} = (2 t \hat{i} + 3 t^{2} \hat{j}) m / s .$ Then rate of work done on the particle at t = 1 sec is