A particle is projected directly up along a rough plane of inclination α with velocity u. If after coming to rest the particle returns to the starting point with velocity v the coefficient of friction between the particle and the plane is

Engineering

Physics

Static and Kinetic Friction

Motion of Block on Inclined Surface

Straight Line Motion

Question

A particle is projected directly up along a rough plane of inclination α with velocity u. If after coming to rest the particle returns to the starting point with velocity v the coefficient of friction between the particle and the plane is

$\frac{u^{2}}{v^{2}} tanα$

$\frac{v^{2}}{u^{2}} tanα$

$\frac{u^{2} + v^{2}}{u^{2} - v^{2}} tanα$

$\frac{u^{2} - v^{2}}{u^{2} + v^{2}} tanα$

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

A particle moves up and down a rough inclined plane. The net acceleration differs due to friction direction. Up motion: a_up = g(sinα + μcosα). Down motion: a_down = g(sinα - μcosα).

Using v² = u² + 2as, for up motion: 0 = u² - 2a_ups. For down motion: v² = 0 + 2a_downs. Equating the distance s from both equations gives:

$\frac{u^{2}}{g (\sin α + μ \cos α)} = \frac{v^{2}}{g (\sin α - μ \cos α)}$

Solving this yields the coefficient of friction μ.

Final Answer: $\frac{u^{2} - v^{2}}{u^{2} + v^{2}} \tan α$