A body of mass 1kg begins to move under the action of a time...

Given, \overset{\to}{F} = \left(t\overset{^}{i} + 3{t}^{2}\overset{^}{j}\right)N andm = 1kg$</p><p>Now, we can write from Newton's second law,$\overset{\to}{F} = m\overset{\to}{a} = m\dfrac{d \overset{\to}{v}}{d t}$</p><p>Then,$d \overset{\to}{v} = \left(t\overset{^}{i} + 3{t}^{2}\overset{^}{j}\right) d t$</p><p>Integrating the above, </p><p>${∫}_{0}^{v} d \overset{\to}{v} = {∫}_{0}^{t}\left(t\overset{^}{i} + 3{t}^{2}\overset{^}{j}\right) d t ⇒ \overset{\to}{v} = \dfrac{{t}^{2}}{2}\overset{^}{i} + {t}^{3}\overset{^}{j}$</p><p>Now, power developed by the force is$P = \overset{\to}{F} \cdot \overset{\to}{v} = \left(t\overset{^}{i} + 3{t}^{2}\overset{^}{j}\right) \cdot \left(\dfrac{{t}^{2}}{2}\overset{^}{i} + {t}^{3}\overset{^}{j}\right)$</p><p>$= \left(\dfrac{{t}^{3}}{2} + 3{t}^{5}\right)$</p><p>At$t = 2s$, power is$P = \left(\dfrac{{2}^{3}}{2} + 3{\left(2\right)}^{5}\right) = 4 + \left(3 \times 32\right) = 100W$