By how much should the spring be compressed b | vedclass

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Question

(C) At point $P$ (which is at the same horizontal level as the center of the circular track),the centripetal force is provided by the normal force $N$

Vedclass · Accepted Answer

$\sqrt{\frac{3mgR}{k}}$

Vedclass · Answer

(C) At point $P$ (which is at the same horizontal level as the center of the circular track),the centripetal force is provided by the normal force $N$. Given that the centripetal force is $mg$,we have:$\frac{mv_P^2}{R} = mg$$	herefore v_P^2 = Rg$Using the law of conservation of energy between the lowest point $L$ and point $P$:$\frac{1}{2}mv_L^2 = \frac{1}{2}mv_P^2 + mgR$Substituting $v_P^2 = Rg$:$\frac{1}{2}mv_L^2 = \frac{1}{2}m(Rg) + mgR = \frac{3}{2}mgR$$v_L^2 = 3gR$The initial potential energy stored in the spring is converted into the kinetic energy of the block at point $L$:$\frac{1}{2}kx^2 = \frac{1}{2}mv_L^2$$\frac{1}{2}kx^2 = \frac{1}{2}m(3gR)$$kx^2 = 3mgR$$x = \sqrt{\frac{3mgR}{k}}$

By how much should the spring be compressed by the block so that the centripetal force at point $P$ is $mg$?

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$A$ $2 \ kg$ block slides on a horizontal surface at a speed of $4 \ m/s$ and strikes an uncompressed spring. The kinetic friction force is $15 \ N$ and the spring constant is $10,000 \ N/m$. By how many $cm$ will the spring be compressed?

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