Question
A body of mass $$1\,kg$$ begins to move under the action of a time dependent force $$F = \left( {2t\,\hat i + 3{t^2}\hat j} \right)N,$$ where $${\hat i}$$ and $${\hat j}$$ are unit vectors along $$X$$ and $$Y$$ axes. What power will be developed by the force at the time $$\left( t \right)$$ ?
A.
$$\left( {2{t^2} + 4{t^4}} \right)W$$
B.
$$\left( {2{t^3} + 3{t^4}} \right)W$$
C.
$$\left( {2{t^3} + 3{t^5}} \right)W$$
D.
$$\left( {2t + 3{t^3}} \right)W$$
Answer :
$$\left( {2{t^3} + 3{t^5}} \right)W$$
Solution :
According to question, a body of mass $$1\,kg$$ begins to move under the action of time dependent force,
$$F = \left( {2t\hat i + 3{t^2}\hat j} \right)N$$
where $${\hat i}$$ and $${\hat j}$$ are unit vectors along $$X$$ and $$Y$$-axes.
$$\eqalign{
& \because F = ma \cr
& \Rightarrow a = \frac{F}{m} \cr
& \Rightarrow a = \frac{{\left( {2t\hat i + 3{t^2}\hat j} \right)}}{1}\,\,\,\left( {\because m = 1\,kg} \right) \cr
& \Rightarrow a = \left( {2t\hat i + 3{t^2}\hat j} \right)m/{s^2} \cr
& \because {\text{acceleration,}}\,\,a = \frac{{dv}}{{dt}} \cr
& \Rightarrow dv = adt\,......\left( {\text{i}} \right) \cr} $$
Integrating both sides, we get
$$\eqalign{
& \int {dv} = \int a \,dt = \int {\left( {2t\hat i + 3{t^2}\hat j} \right)dt} \cr
& v = {t^2}\hat i + {t^3}\hat j \cr} $$
$$\because $$ Power developed by the force at the time $$t$$ will be given as
$$\eqalign{
& P = F \cdot v = \left( {2t\hat i + 3{t^2}\hat j} \right).\left( {{t^2}\hat i + {t^3}\hat j} \right) \cr
& = \left( {2t.{t^2} + 3{t^2}.{t^3}} \right) \cr
& P = \left( {2{t^3} + 3{t^5}} \right)W \cr} $$