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

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(C) Given force: $\vec{F} = t \hat{i} + 3t^2 \hat{j} \ N$ and mass $m = 1 \ kg$.Using Newton's second law,$\vec{F} = m \vec{a} = m \frac{d\vec{v}}{dt}

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(C) Given force: $\vec{F} = t \hat{i} + 3t^2 \hat{j} \ N$ and mass $m = 1 \ kg$.Using Newton's second law,$\vec{F} = m \vec{a} = m \frac{d\vec{v}}{dt}$.Since $m = 1 \ kg$,$\frac{d\vec{v}}{dt} = t \hat{i} + 3t^2 \hat{j}$.Integrating with respect to time $t$ (assuming initial velocity is zero at $t=0$): $\vec{v} = \int (t \hat{i} + 3t^2 \hat{j}) dt = \frac{t^2}{2} \hat{i} + t^3 \hat{j} \ m/s$.Power $P$ is given by the dot product of force and velocity: $P = \vec{F} \cdot \vec{v}$.$P = (t \hat{i} + 3t^2 \hat{j}) \cdot (\frac{t^2}{2} \hat{i} + t^3 \hat{j}) = \frac{t^3}{2} + 3t^5$.At $t = 2 \ s$,$P = \frac{2^3}{2} + 3(2^5) = \frac{8}{2} + 3(32) = 4 + 96 = 100 \ W$.

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

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