Post Utme Osustech 2024 Physics Question 38

Practice objective / multiple choice question 38 from the 2024 Post Utme Osustech Physics examination.

Post Utme Osustech 2024 Physics Objective / Multiple Choice
Question 38 POST_UTME_OSUSTECH • 2024 • PHYSICS • objective

A particle of mass $m$ is moving in a circular orbit of radius $r$ under the influence of a central force given by $F(r) = -\frac{k}{r^2}$. If the particle's angular momentum is $L = 2mvr$, what is the total energy of the particle?

Answer Options

A. E = \frac{1}{2}mv^2 + \frac{k}{r}
Correct Answer B. E = \frac{1}{2}mv^2 - \frac{k}{r}
C. E = \frac{1}{2}mv^2 + \frac{k}{r^2}
D. E = \frac{1}{2}mv^2 - \frac{k}{r^2}
Correct Answer
B
Correct Option:
E = \frac{1}{2}mv^2 - \frac{k}{r}

About This Question

This is Post Utme Osustech 2024 Physics Question 38. It is one of the objective questions from the 2024 Post Utme Osustech Physics examination.

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