Initially when W_{1} = 1.5N, then the body is in equilibrium. i.e. both in rest or a = 0,

Then consider block W_{1}

R_{V} = 0; T = W_{1} = 1.5N ...(1)

Consider block W_{2}

R_{V} = 0; R = W_{2} = ...(2)

F_{r} – T = 0; F_{r} = T = 1.5N ...(3)

But, F_{r} = µR = µW_{2;} µW_{2} = 1.5; µ X 10 = 1.5, µ = 0.15 ...(4)

Now when Weight W_{1 }= 2.0N, body moves down Now the tension on both side be T_{1}

Consider block W_{1}W_{1} – T_{1} = ma_{2} – T_{1} = (2/g)a ...(5)

Consider block W_{2}

T_{1} – F_{r} = ma

T_{1} – µW_{2} = (10/g)a

T_{1 }– 1.5 = (10/g)a

Solve the equation (5) and (6) for T_{1} and a, we get

T_{1} = 1.916N, a = 0.408m/sec^{2}