A body of mass 1 kg begins to move under the | vedclass

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(A) Given: Mass $m = 1 \ kg$,Force $\vec{F} = t \hat{i} + 2t^2 \hat{j}$.Using Newton's second law,$\vec{F} = m \vec{a}$,so $\vec{a} = \frac{\vec{F}}{m

Vedclass · Accepted Answer

$337.5$

Vedclass · Answer

(A) Given: Mass $m = 1 \ kg$,Force $\vec{F} = t \hat{i} + 2t^2 \hat{j}$.Using Newton's second law,$\vec{F} = m \vec{a}$,so $\vec{a} = \frac{\vec{F}}{m} = t \hat{i} + 2t^2 \hat{j}$.Velocity $\vec{v} = \int \vec{a} \ dt = \int (t \hat{i} + 2t^2 \hat{j}) \ dt = \frac{t^2}{2} \hat{i} + \frac{2t^3}{3} \hat{j}$ (assuming initial velocity is zero).Power $P = \vec{F} \cdot \vec{v} = (t \hat{i} + 2t^2 \hat{j}) \cdot (\frac{t^2}{2} \hat{i} + \frac{2t^3}{3} \hat{j})$.$P = \frac{t^3}{2} + \frac{4t^5}{3}$.At $t = 3 \ s$:$P = \frac{3^3}{2} + \frac{4(3^5)}{3} = \frac{27}{2} + 4(3^4) = 13.5 + 4(81) = 13.5 + 324 = 337.5 \ W$.

$A$ body of mass $1 \ kg$ begins to move under the action of a time-dependent force $\vec{F} = (t \hat{i} + 2t^2 \hat{j}) \ N$,where $\hat{i}$ and $\hat{j}$ are unit vectors along $x$ and $y$ axes. The power developed by the above force at time $t = 3 \ s$ will be: (in $W$)

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