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(B) Given: Mass $m = 4 \ kg$,Force $\overrightarrow{F} = (2 \hat{i} + 3 \hat{j}) \ N$,Displacement $\overrightarrow{d} = \overrightarrow{OQ} - \overri

Vedclass · Accepted Answer

$6: 13$

Vedclass · Answer

(B) Given: Mass $m = 4 \ kg$,Force $\overrightarrow{F} = (2 \hat{i} + 3 \hat{j}) \ N$,Displacement $\overrightarrow{d} = \overrightarrow{OQ} - \overrightarrow{OP} = (6-3) \hat{i} + (10-4) \hat{j} = (3 \hat{i} + 6 \hat{j}) \ m$,Time $t = 4 \ s$.Work done $W = \overrightarrow{F} \cdot \overrightarrow{d} = (2 \hat{i} + 3 \hat{j}) \cdot (3 \hat{i} + 6 \hat{j}) = (2 	imes 3) + (3 	imes 6) = 6 + 18 = 24 \ J$.Average power $\langle P angle = \frac{W}{t} = \frac{24}{4} = 6 \ W$.Acceleration $\overrightarrow{a} = \frac{\overrightarrow{F}}{m} = \frac{2 \hat{i} + 3 \hat{j}}{4} = (0.5 \hat{i} + 0.75 \hat{j}) \ m/s^2$.Velocity at $t = 4 \ s$ is $\overrightarrow{v} = \overrightarrow{u} + \overrightarrow{a}t = 0 + (0.5 \hat{i} + 0.75 \hat{j}) 	imes 4 = (2 \hat{i} + 3 \hat{j}) \ m/s$.Instantaneous power $P_{ins} = \overrightarrow{F} \cdot \overrightarrow{v} = (2 \hat{i} + 3 \hat{j}) \cdot (2 \hat{i} + 3 \hat{j}) = 4 + 9 = 13 \ W$.The ratio of average power to instantaneous power is $\frac{\langle P angle}{P_{ins}} = \frac{6}{13}$.

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