A block of mass 2 kg placed on an inclined plane just slides down when its angle of inclination is 45°. Calculate the(i) coefficient of friction between the surfaces of inclined plane and block(ii) acceleration of the block when angle of inclination of the plane is 30°(iii) magnitude of frictional force when angle of inclination is 30° ?(g = 10 ms^-2)
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Since it starts sliding down when angle is 45 , means the downward component of weight i.e., mgsin45 equals the frictional force i.e., (miu) mgcos45 so on equating the two , we get miu(coefficient of friction) to be tan45=1.
its obvious that block didn't slide till the angle of inclination doesn't get 45 so at 30 no motion no acceleration will be there.
the value of frictional force at 30 would be equal to the inclined component of the weight i.e.,mgsin30 , actually the frictional force is self adjusting force it can have the maximum value of mgsin45 as we calculated but it doesn't act wholly until needed here only force equal to mgsin30 is required to keep the block at rest so no need to act fully.
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