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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
3/4 of ke lost therefore 1/4 of ke leftKe=1/2mv square As mass does not changed final velocity=half of initial velocity as ke is one fourth nowAcceleration due to frixn = v/2tFrixn force is kN Where k is frixn coefficientkmg is frixn force therefore it is accelaratacc hence k=v/2tg
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