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Physics - W.E.P. Question with Solution | TestHub

PhysicsW.E.P.Conservative force & potential energyMedium2 minQB

PhysicsMediumsingle choice

A particle is moving in a circular path of radius a under the action of an attractive potential $U = - \frac{k}{2 r^{2}} .$ Its total energy is:

Options:

Answer:

Solution:

$U = - \frac{K}{2 r^{2}}$ $\Rightarrow F = - \frac{d U}{d r} = \frac{2 K}{2 r^{3}} = \frac{K}{r^{3}}$ $\Rightarrow \frac{m v^{2}}{r} = \frac{K}{r^{3}}$ $\Rightarrow v^{2} = \frac{K}{m r^{2}}$
$\Rightarrow \frac{1}{2} m v^{2} = \frac{K}{2 r^{2}}$ $\Rightarrow E = \frac{K}{2 r^{2}} - \frac{K}{2 r^{2}} = 0$ .

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Conservative force & potential energy

⏱ 2mℹ️ Source: QB

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