Question

A block of mass 2 kg rests on a rough inclined plane making an angle of 30 with the horizontal if 0.6 what is the frictional force on the block

Answer 1

use formula m.g.sin tita here m=2,tita=30,g=10(approximately)

Answer 2

Explanation:

Given that,

Mass of the block, m=2 kg

Angle of inclination between the block and the horizontal line is given as :

θ=30

0

Also,

Coefficient of friction, μ=0.65

The force that tends to slide the body on the inclined plane can be given by,

f=mgsinθ

[A]

F=2×9.8×0.5=9.8N