Question

a block of mass 2kg starts moving when the angle of inclination of the inclined plane is 60 degree. If the coefficient of kinetic friction is 0.6, the frictional force is

Answer 1

The mass of 2 kg is placed above an inclined plane with 60° as the angle of inclination.

So, the normal reaction force on the body will be opposite to the force experienced by the body due to gravity. At the given angle, to balance the gravitational pull, normal reaction force has to be m*g, or, 2*cos 60°*g.

Considering g = 10 m/s²,
Normal Reaction (N) = $2 * cos 60$°$* 10$ = $2 * \frac{1}{2} * 10$ = 10 m/s².

Given value of coefficient of kinetic friction = 0.6

By the formula frictional force = coefficient of friction * N,
Calculated frictional force on the body = 10 * 0.6 N = 6 N.

Hence, ANSWER is 6 newton.

Answer 2

Hello.

Good Question and Keep Progressing.



Given Conditions---

    Mass(m) = 2 kg.

   Coefficient of Kinetic Friction (μk) = 0.6

     Angle(θ) = 60°

  Using the Formula,

     Frictional Force = μk × N
       
           N =  mg × Cos θ 

  Taking g = 10 m/s.
           N = 2 × 10 × Cos 60°
               = 20 × 1/2                      [Since Cos 60° = 1/2]
               = 10 newton


Thus, Frictional Force = μk × N
                                  = 0.6 × 10
                                  = 60 N 


I anticipate it will help u.

Have a Marvelous Day.