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(B) Initially,the spring is in equilibrium: $k l = Mg \Rightarrow k = \frac{5 	imes 10}{0.1} = 500 \ N/m$. Let $h$ be the maximum height reached by t

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$0.2$

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(B) Initially,the spring is in equilibrium: $k l = Mg \Rightarrow k = \frac{5 	imes 10}{0.1} = 500 \ N/m$. Let $h$ be the maximum height reached by the block from the equilibrium position. Using the law of conservation of energy,the total energy at the equilibrium position equals the total energy at the maximum height: $\frac{1}{2} M v^2 + \frac{1}{2} k l^2 = Mgh + \frac{1}{2} k (h - l)^2$. Substituting the values: $\frac{1}{2} 	imes 5 	imes (2)^2 + \frac{1}{2} 	imes 500 	imes (0.1)^2 = 5 	imes 10 	imes h + \frac{1}{2} 	imes 500 	imes (h - 0.1)^2$. $10 + 2.5 = 50h + 250(h^2 - 0.2h + 0.01)$. $12.5 = 50h + 250h^2 - 50h + 2.5$. $10 = 250h^2$. $h^2 = \frac{10}{250} = \frac{1}{25}$. $h = 0.2 \ m$.

$A$ block of mass $M = 5 \ kg$ is suspended from one end of a spring. The spring extends by $l = 0.1 \ m$ due to the mass of the block. The block is given an upward velocity of $v = 2 \ m/s$. What is the maximum height $h$ (in $m$) reached by the block? $(g = 10 \ m/s^2)$

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