SATHEE: Kinematics Question 742

Kinematics Question 742

Question: If a particle of mass $m$ is moving in a horizontal circle of radius $r$ with a centripetal force $(- k / r^{2})$ , the total energy it’s [EAMCET (Med.) 1995; AMU (Engg.) 2001]

Options:

A) $- \frac{k}{2 r}$

B) $- \frac{k}{r}$

C) $- \frac{2 k}{r}$

D) $- \frac{4 k}{r}$

Show Answer

Answer:

Correct Answer: A

Solution:

$\frac{m v^{2}}{r} = \frac{k}{r^{2}}$

therefore $m v^{2} = \frac{k}{r}$

K.E.= $\frac{1}{2} m v^{2} = \frac{k}{2 r}$

P.E. $= \int F d r$

$= \int \frac{k}{r^{2}} d r = - \frac{k}{r}$

Total energy = K.E. + P.E. $= \frac{k}{2 r} - \frac{k}{r} = - \frac{k}{2 r}$

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Kinematics Question 742

Question: If a particle of mass m is moving in a horizontal circle of radius r with a centripetal force (−k/r2) , the total energy it’s [EAMCET (Med.) 1995; AMU (Engg.) 2001]

Options:

Answer:

Solution:

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Question: If a particle of mass $m$ is moving in a horizontal circle of radius $r$ with a centripetal force $(- k / r^{2})$ , the total energy it’s [EAMCET (Med.) 1995; AMU (Engg.) 2001]