JEE Main · PhysicsMedium

$A particle moves under the influence of a force F = - k r, where k is a constant and r is the position vector. What is the potential energy of the particle at a distance r from the origin?$

A. $\frac{1}{2} k r^{2}$
B. $- \frac{1}{2} k r^{2}$
C. $k r^{2}$
D. $- k r^{2}$

Show correct answer & step-by-step solution

Correct answer: A — $\frac{1}{2} k r^{2}$

Solution

$The potential energy U (r) is given by U (r) = - \int F \cdot d r .$
$U (r) = - \int (- k r) d r .$
$U (r) = \int k r d r .$
$U (r) = \frac{1}{2} k r^{2} .$
$Hence the answer is (A).$

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