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A car is running with a velocity 72kmph on a gravel Road, is stopped after travelling a distance of 30mts after disengaging it`s engine. The coefficient of friction between the road and the tyres is??
We know that change in kinetic energy =total work done.Here change in K.E=1/2×m×20^2-1/2×m×0^2=200m. (Initial speed=72km/hr or 20m/s. And final is 0m/s)Work=force×displacementTherefore, 200m=force×30. (Displacement=30m)Force=200m/30Coff. Of friction ×mg=200m/30Coff. Of friction=200/30×gCoff.=200/300=2/3=0.6
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