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(D) The work done to compress the spring is stored as elastic potential energy $(U_s = \frac{1}{2} k x^2)$.Given the force $F = kx = 5 \,N$ and compre

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$2 \,m$

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(D) The work done to compress the spring is stored as elastic potential energy $(U_s = \frac{1}{2} k x^2)$.Given the force $F = kx = 5 \,N$ and compression $x = 0.2 \,m$, the elastic potential energy is $U_s = \frac{1}{2} Fx = \frac{1}{2} 	imes 5 \,N 	imes 0.2 \,m = 0.5 \,J$.When the spring is released, this energy is converted into gravitational potential energy $(U_g = mgh)$ at the maximum height $h$.Equating the energies: $mgh = \frac{1}{2} Fx$.Substituting the values: $m = 25 \,g = 0.025 \,kg$, $g = 10 \,m/s^2$, $F = 5 \,N$, and $x = 0.2 \,m$.$0.025 \,kg 	imes 10 \,m/s^2 	imes h = 0.5 \,J$.$0.25 	imes h = 0.5$.$h = \frac{0.5}{0.25} = 2 \,m$.

$A$ sphere of mass $25 \,g$ is placed on a vertical spring. The spring is compressed by $0.2 \,m$ using a force of $5 \,N$. When the spring is released, the mass reaches a maximum height of (take $g = 10 \,m/s^2$):

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