When a 1.0,kg mass hangs attached to a spring | vedclass

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(C) First,calculate the force constant $(k)$ of the spring using Hooke's Law: $F = kx$,where $F = mg$.Given: $m = 1.0\,kg$,$g = 10\,m/s^2$,and extensi

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$2.5$

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(C) First,calculate the force constant $(k)$ of the spring using Hooke's Law: $F = kx$,where $F = mg$.Given: $m = 1.0\,kg$,$g = 10\,m/s^2$,and extension $x = 2\,cm = 0.02\,m$.$k = \frac{mg}{x} = \frac{1.0 	imes 10}{0.02} = 500\,N/m$.Next,determine the total extension of the spring when its length becomes $60\,cm$. The original length is $50\,cm$,so the total extension $x_{total} = 60\,cm - 50\,cm = 10\,cm = 0.1\,m$.The elastic potential energy $(U)$ stored in the spring is given by $U = \frac{1}{2}kx^2$.$U = \frac{1}{2} 	imes 500 	imes (0.1)^2 = 250 	imes 0.01 = 2.5\,J$.

When a $1.0\,kg$ mass hangs attached to a spring of length $50\,cm$,the spring stretches by $2\,cm$. The mass is pulled down until the length of the spring becomes $60\,cm$. What is the amount of elastic energy stored in the spring in this condition,if $g = 10\,m/s^2$? (in $Joule$)

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