Question

A block of mass 2 kg rests on a plane inclined at 30° with the horizontal. The coefficient of friction between the block and the surface is 0.7. What will be the frictional force acting on the block?

Answer 1

frictional force at a given angle = umgcos a

u= 0.7

g =10 ( approx)

m=2 kg

so frictional force= 0.7*10*2* cos 30

after simplification we get

7*1.732=12.1243