A 3 kg block is connected as shown in the fi | vedclass

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(A) Given: $m = 3 \ kg$,$K_1 = 50 \ Nm^{-1}$,$K_2 = 150 \ Nm^{-1}$,$g = 10 \ ms^{-2}$.Since the block is connected between two springs in parallel,the

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(A) Given: $m = 3 \ kg$,$K_1 = 50 \ Nm^{-1}$,$K_2 = 150 \ Nm^{-1}$,$g = 10 \ ms^{-2}$.Since the block is connected between two springs in parallel,the equivalent spring constant is $K_{eq} = K_1 + K_2 = 50 + 150 = 200 \ Nm^{-1}$.The equilibrium position is where the spring force balances the weight: $K_{eq} x_0 = mg \implies x_0 = \frac{mg}{K_{eq}} = \frac{3 	imes 10}{200} = 0.15 \ m$.The block is released from the unstretched position,so the amplitude of oscillation is $A = x_0 = 0.15 \ m$.At the lowest position,the block is at a displacement $x = A$ below the equilibrium position.The net force at the lowest position is $F_{net} = K_{eq} A - mg = K_{eq} (\frac{mg}{K_{eq}}) - mg = 0$ (at equilibrium). Wait,let's re-evaluate.At the lowest position,the spring is stretched by $x = 2A = 2 	imes 0.15 = 0.3 \ m$.The net force is $F_{net} = K_{eq} x - mg = 200(0.3) - 3(10) = 60 - 30 = 30 \ N$.The acceleration is $a = \frac{F_{net}}{m} = \frac{30}{3} = 10 \ ms^{-2}$.

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