A block of mass 5 kg is placed on a rough inclined plane. The inclination of the plane is gradually increased until the block just begins to slide down. The sin of angle made by the plane with horizontal is (3/5) when the block start sliding. The coefficient of friction between the block and the plane is ( g = 10 m s- 2 )

Question

A block of mass 5 kg is placed on a rough inclined plane. The inclination of the plane is gradually increased until the block just begins to slide down. The sin of angle made by the plane with horizontal is (3/5) when the block start sliding. The coefficient of friction between the block and the plane is ( g = 10 m s- 2 ) (A) (3/5) (B) (3/4) (C) (4/5) (D) (2/3)

Tardigrade Admin · Accepted Answer

The block will just start sliding when the angle θ made by the inclined plane with the horizontal will be equal to the angle of repose. θ=θ text repose = tan -1(μ) ⇒ tanθ =μ μ =(3/4)

$\frac{3}{5}$

$\frac{3}{4}$

$\frac{4}{5}$

$\frac{2}{3}$

The block will just start sliding when the angle $θ$ made by the inclined plane with the horizontal will be equal to the angle of repose.
$θ = θ_{repose} = tan^{- 1} (μ)$
$\Rightarrow t an θ = μ$
$μ = \frac{3}{4}$

53​

43​

54​

32​

The block will just start sliding when the angle θ made by the inclined plane with the horizontal will be equal to the angle of repose. θ=θrepose ​=tan−1(μ) ⇒tanθ=μ μ=43​

$\frac{3}{5}$

$\frac{3}{4}$

$\frac{4}{5}$

$\frac{2}{3}$

The block will just start sliding when the angle $θ$ made by the inclined plane with the horizontal will be equal to the angle of repose.
$θ = θ_{repose} = tan^{- 1} (μ)$
$\Rightarrow t an θ = μ$
$μ = \frac{3}{4}$