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,the

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).

Column-$I$	Column-$II$
$(1)$ Work done is zero	$(a)$ By gravitational force (when body moves down)
$(2)$ Work done is positive	$(b)$ Against gravitational force (when body moves up)
$(3)$ Work done is negative	$(c)$ By centripetal force

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Match the items in Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ Work done is zero $(a)$ By gravitational force (when body moves down)

$(2)$ Work done is positive $(b)$ Against gravitational force (when body moves up)

$(3)$ Work done is negative $(c)$ By centripetal force

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