Rope sliding on double ramp, how much time would it take before all of the rope is on the right side?

Question

A 1kg rope of length 1.5m slides down a frictionless double ramp in the shape of an equilateral triangle, as shown in the image below. At time t=0, the rope is at rest with of its length 1m on the right side of the ramp.

How much time elapses before all of the rope is on the right side?

Note: Assume g=10

Answer 1

1kg - 1.5m
? - 1m

Mass of 1m of rope = 1*1/1.5 = 2/3 Kg
Mass of 0.5m of rope = 1- (2/3) = 1/3 Kg

Force on 0.5m Rope = (1/3)*10 = 10/3 N (downward)
Force on 1m Rope = (2/3)*10 = 20/3 N (downward)

Force on 0.5m Rope along the surface = 10/3*cos(30) = 10/3*0.866 = 2.887
Force on 1m Rope along the surface = 20/3*cos(30) = 20/3*0.866 = 5.773

Net force on the rope at any instant
= (5.773 + (10/3)*cos (30)*t) - (2.887 - (10/3)*cos (30)*t)
= (5.773 + 2.887*t) - (2.887 - 2.887*t)
= 2.887 + 5.773*t
Therefore net acceleration on the Rope
= ((2.887 + 5.773*t) N)/(1 Kg) = (2.887 + 5.773*t) ms^(-2)

Distance = (1/2)*acceleration*time^(2)
or D = 0.5*a*t^(2)
Required D is 0.5m of the rope on the left that has to come to the right, therefore time taken for this to happen is

0.5 = 0.5*(2.887 + 5.773*t)*t^(2)
1 = 2.887*t^(2) + 5.773*t^(3)
=> 5.773*t^(3) + 2.887*t^(2) - 1 =0
Solving the cubic equation gives you t = 0.43126s

time = 0.43126 s.