A mass of 2 kg, initially at a height of 1.2 | vedclass

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(C) Let the compression of the spring be $x$ (in meters). According to the law of conservation of mechanical energy, the total potential energy lost b

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$50$

Vedclass · Answer

(C) Let the compression of the spring be $x$ (in meters). According to the law of conservation of mechanical energy, the total potential energy lost by the mass equals the potential energy gained by the spring. The total height fallen by the mass is $(h + x)$, where $h = 1.2 \ m$. So, $mg(h + x) = \frac{1}{2} kx^2$. Substituting the given values: $2 	imes 10 	imes (1.2 + x) = \frac{1}{2} 	imes (2 	imes 10^4) 	imes x^2$. $20(1.2 + x) = 10^4 x^2$. $24 + 20x = 10000x^2$. $10000x^2 - 20x - 24 = 0$. Dividing by $4$: $2500x^2 - 5x - 6 = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{5 \pm \sqrt{(-5)^2 - 4(2500)(-6)}}{2(2500)} = \frac{5 \pm \sqrt{25 + 60000}}{5000} = \frac{5 \pm \sqrt{60025}}{5000} = \frac{5 \pm 245}{5000}$. Since $x > 0$, $x = \frac{250}{5000} = 0.05 \ m$. Converting to $mm$: $0.05 \ m = 50 \ mm$.

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