A block of mass 2 kg is attached to one end | vedclass

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(B) Given:Natural length of spring $L = 2 \ m$Mass of block $m = 2 \ kg$Spring constant $k = 200 \ N/m$Initial compression $x_i = L - 1 = 2 - 1 = 1 \

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$10[1-(2-x)^2]^{1/2} \ m/s$

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(B) Given:Natural length of spring $L = 2 \ m$Mass of block $m = 2 \ kg$Spring constant $k = 200 \ N/m$Initial compression $x_i = L - 1 = 2 - 1 = 1 \ m$Final compression at distance $x$ from the wall is $x_f = L - x = (2 - x) \ m$Using the principle of conservation of mechanical energy:$K_i + U_i = K_f + U_f$Since the block is released from rest,$K_i = 0$.$0 + \frac{1}{2} k x_i^2 = \frac{1}{2} m v^2 + \frac{1}{2} k x_f^2$$\frac{1}{2} m v^2 = \frac{1}{2} k (x_i^2 - x_f^2)$$m v^2 = k (x_i^2 - x_f^2)$Substituting the values:$2 	imes v^2 = 200 	imes [1^2 - (2 - x)^2]$$v^2 = 100 	imes [1 - (2 - x)^2]$$v = 10 \sqrt{1 - (2 - x)^2} \ m/s$$v = 10 [1 - (2 - x)^2]^{1/2} \ m/s$

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