As the object on inclined plane is at rest net force acting on the object is zero. Force applied on object by earth is Mg. Resultant of Normal reaction and friction force will cancel gravitational force. So the force applied by inclined plane on block is Mg in vertically  upward direction.

Friction force action on the object is f=μ.N= μ.mg

Retardation of the block is  a=μ.g = 0.5× 10 =5

Using equation of motion. Stopping distance = u^2/2a=(10 × 10)/( 2* 5)=10 m

Using equation of motion

v= u + at

⇒ 0= 6 + a ×10

⇒ a= -0.6 m/sec^2

Maximum Friction force is μN = μ mg so, a= μ.g

⇒ -0.6 = - μ. 10

⇒μ = 0.06 

Since \(3/4^{\text{th}}\) of kinetic energy is lost, the remaining kinetic energy is \(1/4^{\text{th}}\), meaning the final velocity \(v = v_0/2\). Using \(v = v_0 - at_0\) where \(a = \mu g\), we get \(v_0/2 = v_0 - \mu gt_0 ⇒ \mu = \frac{v_0}{2gt_0}\).

Retardation \(a = g(\sin\theta + \mu\cos\theta) = 10(\sin 30^\circ + \frac{1}{\sqrt{3}}\cos 30^\circ) = 10(0.5 + 0.5) = 10\text{ m/s}^2\). Using \(v^2 = u^2 - 2as ⇒ 0 = 1^2 - 2(10)s\), we get \(s = 0.05\text{ m} = 5\text{ cm}\).

The hanging part of length \( \frac{2L}{5} \) exerts a pulling force of \( \frac{2}{5} Mg \). The part on the table of length \( \frac{3L}{5} \) experiences a maximum friction force of \( \mu \frac{3}{5} Mg \). Equating the forces at the verge of sliding gives \( \mu = \frac{2}{3} \).