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A particle moves on a rough horizontal ground with some initial velocity v .if 3/4 of it's kinetic energy is lost due to friction in time t then coefficient of friction between particle and the ground is. (a) v/2gt (b) v/4gt (c) 3v/4gt (d) v/gt

A particle moves on a rough horizontal ground with some initial velocity v .if 3/4 of it's kinetic energy is lost due to friction in time t then coefficient of friction between particle and the ground is.                                 (a) v/2gt (b) v/4gt (c) 3v/4gt (d) v/gt
 

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1 Answers

Saurav
224 Points
5 years ago
3/4 of ke lost therefore 1/4 of ke left
Ke=1/2mv square  
As mass does not changed final velocity=half of initial velocity as ke is one fourth now
Acceleration due to frixn = v/2t
Frixn force is kN 
Where k is frixn coefficient
kmg is frixn force therefore it is accelaratacc hence k=v/2tg

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