Question

The coefficient of friction of an inclined plane is 1/root 3 . If it is inclined at angle 30 degree with the horizontol , what will be the downward acceleration of the block placed on the inclined plane

Answer 1

0 m/s²

Step-by-step explanation:

Draw a free body diagram (see attached).

Sum of the forces normal to the incline:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces parallel to the incline:

∑F = ma

mg sin θ − f = ma

mg sin θ − Nμ = ma

Substituting:

mg sin θ − (mg cos θ) μ = ma

g sin θ − g μ cos θ = a

a = g (sin θ − μ cos θ)

Given that g = 9.8 m/s², θ = 30°, and μ = 1/√3:

a = (9.8 m/s²) (sin 30° − 1/√3 cos 30°)

a = (9.8 m/s²) (1/2 − 1/2)

a = 0 m/s²

The acceleration down the incline is 0 m/s².