Question

A body of mass 30 kg kept on a rough horizontal surface of coefficient of friction is 0.3 then frictional force between block and the surface is

Answer 1

Answer:

mass = 30kg

find normal reaction ( normal opposing force)

normal reaction = 30 x 10 = 300N

frictional force between block and the surface is

= coefficient of friction x normal reaction

=0.3 x 300

=90N

hope this will help you

Answer 2

Given

mass (m) = 30 kg

coefficient of friction = 0.3

Normal reaction = 30 × 10

= 300 N

frictional force between block and the surface = coefficient of friction × normal reaction

= 0.3 × 300 = 90 N

KILLER ✌