Question

A body of mass 2kg rests on a rough inclined plane making an angle 30 degree with the horizontal. the coefficient of friction between the block and the plane is 0.7 the friction force on the block is​

Answer 1

Given,

Mass of the body, m=2kg

Angle of inclination between the block and the horizontal line is,ø=30°

coefficient of friction ,ų= 0.7

The force that tend to slide the body on the inclined plane is=mg sinø

=mg sin30°

=2×10×1/2(g=10m/s^2)

=10N

the limiting friction between the block and the hrizontal plane is

=ųmg cos30°

=0.7×2×10×0.865

=12.11N

So friction is equal to the applied force(•.•applied force <limiting friction)

So, frictional force is 10N

Answer 2

ANSWER:--------------

{ friction * Normal force }

      

 [ Ff           = Mu * N]

{Normal force N = m is the angle of inclination of plane.}

        [Ff = Mu * m g Cos = 0.7 * 2]

{ kg * 9.8m/s² * cos 30 = 11.88}

[Newtons}

{Suppose we calculate the component of weight along the inclined plane:}

  = m g Sin = 2 kg * 9.8m/s² * {sin

30 = 9.8 Newtons}

hope it helps:--

T!—!ANKS!!!