A block of mass 0.18 kg is attached to a spr | vedclass

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(B) Given:Mass of block,$m = 0.18 \ kg$Spring constant,$k = 2 \ N/m$Coefficient of friction,$\mu = 0.1$Distance travelled,$x = 0.06 \ m$Acceleration d

Vedclass · Accepted Answer

$4$

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(B) Given:Mass of block,$m = 0.18 \ kg$Spring constant,$k = 2 \ N/m$Coefficient of friction,$\mu = 0.1$Distance travelled,$x = 0.06 \ m$Acceleration due to gravity,$g = 10 \ m/s^2$Initial velocity,$v = N/10 \ m/s$According to the work-energy theorem,the work done by all forces is equal to the change in kinetic energy.$W_{	ext{spring}} + W_{	ext{friction}} = \Delta K$$-\frac{1}{2} kx^2 - \mu mgx = 0 - \frac{1}{2} mv^2$Rearranging the terms:$\frac{1}{2} mv^2 = \frac{1}{2} kx^2 + \mu mgx$$v^2 = \frac{kx^2}{m} + 2\mu gx$Substituting the values:$v^2 = \frac{2 	imes (0.06)^2}{0.18} + 2 	imes 0.1 	imes 10 	imes 0.06$$v^2 = \frac{2 	imes 0.0036}{0.18} + 0.12$$v^2 = \frac{0.0072}{0.18} + 0.12$$v^2 = 0.04 + 0.12 = 0.16$$v = \sqrt{0.16} = 0.4 \ m/s$Given $v = N/10$,so $0.4 = N/10$,which implies $N = 4$.

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