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.**