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Question

(D) The block reaches its maximum speed when its acceleration becomes zero.At this point,the spring force balances the applied force $F$,so $kx = F$,w

Vedclass · Accepted Answer

$\frac{F}{\sqrt{mk}}$

Vedclass · Answer

(D) The block reaches its maximum speed when its acceleration becomes zero.At this point,the spring force balances the applied force $F$,so $kx = F$,which gives $x = \frac{F}{k}$.Applying the work-energy theorem,the total work done on the block equals the change in its kinetic energy:$W_{	ext{spring}} + W_F = \Delta K.E.$$-\frac{1}{2}kx^2 + Fx = \frac{1}{2}mv_{\max}^2$Substituting $x = \frac{F}{k}$ into the equation:$-\frac{1}{2}k\left(\frac{F}{k}ight)^2 + F\left(\frac{F}{k}ight) = \frac{1}{2}mv_{\max}^2$$-\frac{F^2}{2k} + \frac{F^2}{k} = \frac{1}{2}mv_{\max}^2$$\frac{F^2}{2k} = \frac{1}{2}mv_{\max}^2$$v_{\max}^2 = \frac{F^2}{mk}$$v_{\max} = \frac{F}{\sqrt{mk}}$

Column-$I$	Column-$II$
$(1)$ Potential Energy $=$ Kinetic Energy	$(a)$ At height $\frac{2h}{3}$
$(2)$ Potential Energy $= 2 \times$ Kinetic Energy	$(b)$ Constant at every point
	$(c)$ At height $\frac{h}{2}$

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