A block of mass m = 5 text{ kg} is released f | vedclass

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(C) Let the block be released from height $h = L \sin(30^{\circ}) = 10 	imes 0.5 = 5 	ext{ m}$.Using the Work-Energy Theorem: $W_{	ext{gravity}} +

Vedclass · Accepted Answer

$1 	ext{ m}$

Vedclass · Answer

(C) Let the block be released from height $h = L \sin(30^{\circ}) = 10 	imes 0.5 = 5 	ext{ m}$.Using the Work-Energy Theorem: $W_{	ext{gravity}} + W_{	ext{friction}} + W_{	ext{spring}} = \Delta KE = 0$.Work done by gravity: $W_g = mgh = 5 	imes 10 	imes 5 = 250 	ext{ J}$.Work done by friction on the horizontal surface: $W_f = -\mu mg(d + x) = -0.5 	imes 5 	imes 10 	imes (2 + x) = -25(2 + x) = -50 - 25x$.Work done by the spring: $W_s = -\frac{1}{2} kx^2 = -\frac{1}{2} 	imes 100 	imes x^2 = -50x^2$.Equating the total work to zero: $250 - 50 - 25x - 50x^2 = 0$.$200 - 25x - 50x^2 = 0$.Dividing by $25$: $8 - x - 2x^2 = 0$,or $2x^2 + x - 8 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 - 4(2)(-8)}}{4} = \frac{-1 \pm \sqrt{65}}{4}$.Since $x > 0$,$x = \frac{-1 + 8.06}{4} \approx 1.76 	ext{ m}$.Re-evaluating the provided options,if we assume the friction acts only on the $2 	ext{ m}$ part and the spring compression is $x$,the equation is $250 - (0.5 	imes 5 	imes 10 	imes 2) - (0.5 	imes 5 	imes 10 	imes x) - 50x^2 = 0$.$250 - 50 - 25x - 50x^2 = 0 \Rightarrow 200 - 25x - 50x^2 = 0 \Rightarrow 2x^2 + x - 8 = 0$.Given the options,if $x=1$,$2(1)^2 + 1 - 8 = -5 
eq 0$. If $x=2$,$2(4) + 2 - 8 = 2 
eq 0$. There is a discrepancy in the provided options vs the physics model. Assuming the intended answer based on standard problem types is $x=1$ or $x=2$ depending on friction parameters,but mathematically $x \approx 1.76$.

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