A light spring of length 10 cm is stretched | vedclass

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(D) Given: Natural length $L_0 = 10 \ cm$,extension $x_1 = 2 \ cm = 0.02 \ m$,mass $m = 20 \ g = 0.02 \ kg$,$g = 10 \ m/s^2$.At equilibrium,the spring

Vedclass · Accepted Answer

$8 	imes 10^{-3}$

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(D) Given: Natural length $L_0 = 10 \ cm$,extension $x_1 = 2 \ cm = 0.02 \ m$,mass $m = 20 \ g = 0.02 \ kg$,$g = 10 \ m/s^2$.At equilibrium,the spring force equals the gravitational force: $k x_1 = mg$.$k(0.02) = 0.02 	imes 10 \implies k = 10 \ N/m$.The total length becomes $14 \ cm$,so the new extension is $x_2 = 14 \ cm - 10 \ cm = 4 \ cm = 0.04 \ m$.The elastic potential energy stored in the spring is given by $U = \frac{1}{2} k x_2^2$.$U = \frac{1}{2} 	imes 10 	imes (0.04)^2 = 5 	imes 0.0016 = 0.008 \ J = 8 	imes 10^{-3} \ J$.

$A$ light spring of length $10 \ cm$ is stretched by $2 \ cm$ when a mass of $20 \ g$ is attached to its end. The mass is pulled until the total length of the spring becomes $14 \ cm$. What is the elastic potential energy stored in the spring (in Joules)?

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