The P.E. of a certain spring when stretched f | vedclass

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(D) The initial potential energy of the spring is given by the formula $U = \frac{1}{2} k x^2$.Given $U_1 = 10\, J$ for $x_1 = 0.3\, m$,we have $10 =

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$12.5$

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(D) The initial potential energy of the spring is given by the formula $U = \frac{1}{2} k x^2$.Given $U_1 = 10\, J$ for $x_1 = 0.3\, m$,we have $10 = \frac{1}{2} k (0.3)^2$.Solving for the spring constant $k$: $k = \frac{20}{0.09} = \frac{2000}{9}\, N/m$.The spring is stretched an additional $0.15\, m$,so the new total extension is $x_2 = 0.3 + 0.15 = 0.45\, m$.The final potential energy is $U_2 = \frac{1}{2} k x_2^2 = \frac{1}{2} 	imes \frac{2000}{9} 	imes (0.45)^2$.$U_2 = \frac{1000}{9} 	imes 0.2025 = 1000 	imes 0.0225 = 22.5\, J$.The work done to stretch the spring further is the change in potential energy: $W = U_2 - U_1 = 22.5\, J - 10\, J = 12.5\, J$.

The $P.E.$ of a certain spring when stretched from natural length through a distance $0.3\, m$ is $10\, J$. The amount of work in joule that must be done on this spring to stretch it through an additional distance $0.15\, m$ will be ................ $J$.

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