Question

A block of mass √2 kg is kept on a rough horizontal surface (mu = 0.5). A horizontal force of 5 N is applied on the block. The resultant of normal reaction and the frictional force acting on the object is​

Answer 1

N=mg−Fsin37∘N=20−53Ffmax=μN=0.5(20−53F)=(10−103F)

It must be equal to Fcos37∘

F×54=10−103F1011F=10F=11100N