A block of mass 'm' (as shown in the figure) | vedclass

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(D) According to the work-energy theorem,the work done by the spring force is equal to the change in kinetic energy of the block.$W_{	ext{net}} = K_f

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(D) According to the work-energy theorem,the work done by the spring force is equal to the change in kinetic energy of the block.$W_{	ext{net}} = K_f - K_i$Given that the initial kinetic energy is $K_i = E = \frac{1}{2}mv^2$.When the speed is halved,the final speed is $v' = \frac{v}{2}$.The final kinetic energy is $K_f = \frac{1}{2}m(\frac{v}{2})^2 = \frac{1}{4}(\frac{1}{2}mv^2) = \frac{E}{4}$.The work done by the spring in compressing it by a distance $x = 25\,cm = 0.25\,m = \frac{1}{4}\,m$ is $W = -\frac{1}{2}Kx^2$.Applying the work-energy theorem:$-\frac{1}{2}Kx^2 = K_f - K_i$$-\frac{1}{2}K(\frac{1}{4})^2 = \frac{E}{4} - E$$-\frac{1}{2}K(\frac{1}{16}) = -\frac{3E}{4}$$\frac{K}{32} = \frac{3E}{4}$$K = \frac{3E 	imes 32}{4} = 24E$Comparing this with $nE$,we get $n = 24$.

$A$ block of mass '$m$' (as shown in the figure) moving with kinetic energy $E$ compresses a spring through a distance $25\,cm$ when its speed is halved. The value of the spring constant of the used spring will be $nE\;N\,m^{-1}$ for $n=$

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