Work Energy Power

JEE Main level, numerical answer, medium

154 of 183

Question 154 of 183

A body of mass 1kg1kg begins to move under the action of a time dependent force F⃗=(ti^+3t2j^)N\vec{F}=\left(t\hat{i}+3{t}^{2}\hat{j}\right)N, where i^\hat{i} and j^\hat{j} are the unit vectors along xx and yy axis. The power developed by above force, at the time t=2st=2s, will be _______WW.

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Answer: 100

Given, F⃗=(ti^+3t2j^)N\vec{F}=\left(t\hat{i}+3{t}^{2}\hat{j}\right)N and m=1kgm=1kg
Now, we can write from Newton's second law, F⃗=ma⃗=mdv⃗dt\vec{F}=m\vec{a}=m\dfrac{d\vec{v}}{dt}
Then, dv⃗=(ti^+3t2j^)dtd\vec{v}=\left(t\hat{i}+3{t}^{2}\hat{j}\right)dt
Integrating the above,
∫0vdv⃗=∫0t(ti^+3t2j^)dt⇒v⃗=t22i^+t3j^{\int }_{0}^{v}d\vec{v}={\int }_{0}^{t}\left(t\hat{i}+3{t}^{2}\hat{j}\right)dt \Rightarrow \vec{v}=\dfrac{{t}^{2}}{2}\hat{i}+{t}^{3}\hat{j}
Now, power developed by the force is P=F⃗⋅v⃗=(ti^+3t2j^)⋅(t22i^+t3j^)P=\vec{F}\cdot \vec{v}=\left(t\hat{i}+3{t}^{2}\hat{j}\right)\cdot \left(\dfrac{{t}^{2}}{2}\hat{i}+{t}^{3}\hat{j}\right)
=(t32+3t5)=\left(\dfrac{{t}^{3}}{2}+3{t}^{5}\right)
At t=2st=2s, power is P=(232+3(2)5)=4+(3×32)=100WP=\left(\dfrac{{2}^{3}}{2}+3{\left(2\right)}^{5}\right)=4+\left(3\times 32\right)=100W

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