Work Energy Power

JEE Main level, single correct, medium

125 of 183

Question 125 of 183

A particle is moving in a circle of radius rr under the action of a force F=αr2F=\alpha {r}^{2} which is directed towards centre of the circle. Total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy=0=0 for r=0r=0):

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Answer: (A) 56αr3\dfrac{5}{6}\alpha {r}^{3}

Force F=αr2F=\alpha {r}^{2} will act as centripetal force

⇒αr2=mv2r⇒mv2=αr3\Rightarrow \alpha {r}^{2}=\dfrac{m{v}^{2}}{r}\Rightarrow m{v}^{2}=\alpha {r}^{3}

∴∴ Kinetic energy K.E.=12mv2K.E.=\dfrac{1}{2}m{v}^{2}

⇒K.E.=αr32\Rightarrow K.E.=\dfrac{\alpha {r}^{3}}{2}

Also dU=−F⃗.dr⃗dU=-\vec{F}.d\vec{r}

∆U=−∫F⃗.dr⃗∆U=-\int \vec{F}.d\vec{r}

=−∫αr2(−dr)=-\int \alpha {r}^{2}\left(-dr\right)

=α∫0rr2dr=\alpha \int _{0}^{r}{r}^{2}dr

U−0=αr33U-0=\dfrac{\alpha {r}^{3}}{3}

⇒U=αr33\Rightarrow U=\dfrac{\alpha {r}^{3}}{3}

So total mechanical energy

E=U+KE=U+K

=αr33+αr32=\dfrac{\alpha {r}^{3}}{3}+\dfrac{\alpha {r}^{3}}{2}

E=5αr36E=\dfrac{5\alpha {r}^{3}}{6}

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