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Two bodies of weight 10N and 1.5N are connected to the two ends of a light inextensible String, passing over a smooth pulley. The weight 10N is placed on a rough horizontal surface while the weight of 1.5N is hanging vertically in air. Initially the friction between the weight 10N and the table is just sufficient to prevent motion. If an additional weight of 0.5N is added to the weight 1.5N, determine 

(i) The acceleration of the two weight. 

(ii) Tension in the string after adding additional weight of 0.5N to the weight 1.5N

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Initially when W1 = 1.5N, then the body is in equilibrium. i.e. both in rest or a = 0,

Then consider block W1

RV = 0; T = W1 = 1.5N ...(1)

Consider block W2

RV = 0; R = W2 = ...(2)

Fr – T = 0; Fr = T = 1.5N ...(3)

But, Fr = µR = µW2; µW2 = 1.5; µ X 10 = 1.5, µ = 0.15 ...(4)

Now when Weight W1 = 2.0N, body moves down Now the tension on both side be T1

Consider block W1W1 – T1 = ma2 – T1 = (2/g)a ...(5)

Consider block W2

T1 – Fr = ma

T1 – µW2 = (10/g)a 

T1 – 1.5 = (10/g)a

Solve the equation (5) and (6) for T1 and a, we get

T1 = 1.916N, a = 0.408m/sec2

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