Question

A block of mass 2 kg starts moving when the angle of inclination of the inclined plane is 60o. if the coefficient of kinetic friction is 0.6, the frictional force is

Answer 1

HELLO DEAR,


Contact force is the resultant of Frictional force and normal reaction


The frictional force can be calculated = μᴋN

Where N = mgcosΦ
And, μᴋ = coefficient of kinetic friction
NOW,

Given that:-

g = 10m/s²
M = 2kg
μᴋ = 0.6
Φ = 60°


N = μk × mgcosΦ

⇒0.6 × 2 × 10 × cos60

⇒1.2 × 10 × 1/2
∴ [ cos60° = 1/2]

⇒12 × 1/2

⇒6 Newton



I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

heya..!!!


The formula of frictional force = μᴋN

Where,

N = mgcos∅
And,
μᴋ = coefficient of kinetic friction
NOW,

Given that:-

g = 10m/s²
M = 2kg
μᴋ = 0.6
∅ = 60°


N = μk × mgcos8

= 0.6 × 2 × 10 × cos60

= 6 × 2 × 1/2
∴ [ cos60° = 1/2]

= 6 × 1

= 6 Newton



I HOPE ITS HELP YOU,