A block of mass 100 g is lying on an inclined plane of angle 30°. The frictional force on thisblock …….N.

A block of mass 100 g is lying on an inclined plane of angle 30°. The

1.	A block of mass 100 g is lying on an inclined plane of angle 30°. The frictional force on thisblock …….N.
Answer» `4.9xx10^(2)` `4.9xx10^(-1)` `4.9xx10^(-1)` `4.9 xx10^(1)` Solution :Weight of the blockW=mg Thecomponentof thisweightparallelto thesurfaceof aslopeW sin `theta` GIVESTHE frictionforce F = mgsin `theta` `=mgsin 30^(@)=0.1 xx 9.8xx (1)/(2) = 0.49 N` `f= 4.9 xx 10^(-1) N`