A small block of mass 200 g is placed on a h | vedclass

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(A) Given values: $m = 0.2 \ kg$, $h = 2 \ m$, $x = 0.1 \ m$, $k = 50 \ N \ m^{-1}$, and $g = 10 \ m \ s^{-2}$.Using the law of conservation of energy

Vedclass · Accepted Answer

$0.99$

Vedclass · Answer

(A) Given values: $m = 0.2 \ kg$, $h = 2 \ m$, $x = 0.1 \ m$, $k = 50 \ N \ m^{-1}$, and $g = 10 \ m \ s^{-2}$.Using the law of conservation of energy, the potential energy stored in the spring is converted into the kinetic energy of the block as it leaves the spring:$\frac{1}{2} k x^2 = \frac{1}{2} m v^2$$v = x \sqrt{\frac{k}{m}} = 0.1 	imes \sqrt{\frac{50}{0.2}} = 0.1 	imes \sqrt{250} = 0.1 	imes 15.81 = 1.581 \ m \ s^{-1}$.Now, the block undergoes projectile motion after leaving the slab. The time taken to hit the ground is determined by the vertical height:$h = \frac{1}{2} g t^2 \Rightarrow 2 = \frac{1}{2} 	imes 10 	imes t^2 \Rightarrow t^2 = 0.4 \Rightarrow t = \sqrt{0.4} \approx 0.632 \ s$.The horizontal distance covered is given by:$s = v 	imes t = 1.581 	imes 0.632 \approx 0.999 \ m \approx 1.0 \ m$ (or $0.99 \ m$ based on the provided options).

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