Work Energy Power

JEE Main level, single correct, medium

148 of 183

Question 148 of 183

A body of mass 2kg2kg begins to move under the action of a time dependent force given by F⃗=(6ti^+6t2j^)N\vec{F}=\left(6t\hat{i}+6{t}^{2}\hat{j}\right)N. The power developed by the force at the time tt is given by:

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Answer: (D) (9t3+6t5)W\left(9{t}^{3}+6{t}^{5}\right)W

Given: F⃗=(6ti^+6t2j^)N\vec{F}=\left(6t\hat{i}+6{t}^{2}\hat{j}\right)N
Now, F⃗=ma⃗=(6ti^+6t2j^)\vec{F}=m\vec{a}=\left(6t\hat{i}+6{t}^{2}\hat{j}\right)
⇒a⃗=F⃗m=(3ti^+3t2j^)\Rightarrow \vec{a}=\dfrac{\vec{F}}{m}=\left(3t\hat{i}+3{t}^{2}\hat{j}\right)
Hence, velocity v⃗=∫0ta⃗dt=3t22i^+t3j^\vec{v}=\int _{0}^{t}\vec{a}dt=\dfrac{3{t}^{2}}{2}\hat{i}+{t}^{3}\hat{j}
Required power, P=F⃗.v⃗=(9t3+6t5)WP=\vec{F}.\vec{v}=\left(9{t}^{3}+6{t}^{5}\right)W

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