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(A) Let the block compress the spring by $x$ meters before coming to rest.According to the work-energy theorem,the initial kinetic energy of the block

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$5.5$

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(A) Let the block compress the spring by $x$ meters before coming to rest.According to the work-energy theorem,the initial kinetic energy of the block is equal to the sum of the potential energy stored in the spring and the work done against friction.Initial Kinetic Energy $(K_i)$ = $\frac{1}{2} m v^2 = \frac{1}{2} 	imes 2 	imes (4)^2 = 16 \ J$.Work done against friction $(W_f)$ = $f_k 	imes x = 15x$.Potential energy of the spring $(U_s)$ = $\frac{1}{2} k x^2 = \frac{1}{2} 	imes 10,000 	imes x^2 = 5,000x^2$.Applying the energy balance: $K_i = U_s + W_f$.$16 = 5,000x^2 + 15x$.$5,000x^2 + 15x - 16 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{-15 + \sqrt{15^2 - 4(5,000)(-16)}}{2 	imes 5,000} = \frac{-15 + \sqrt{225 + 320,000}}{10,000} = \frac{-15 + \sqrt{320,225}}{10,000} \approx \frac{-15 + 565.88}{10,000} \approx 0.055 \ m$.$x = 5.5 \ cm$.

$A$ $2 \ kg$ block slides on a horizontal floor with a speed of $4 \ m/s$. It strikes an uncompressed spring and compresses it until the block is motionless. The kinetic friction force is $15 \ N$ and the spring constant is $10,000 \ N/m$. The spring compresses by ............. $cm$.

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