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

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Question

(A) Given: Mass $m = 0.18 \ kg$,Spring constant $k = 2 \ N/m$,Coefficient of friction $\mu = 0.1$,Displacement $x = 0.06 \ m$,Initial velocity $V = N/

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Given: Mass $m = 0.18 \ kg$,Spring constant $k = 2 \ N/m$,Coefficient of friction $\mu = 0.1$,Displacement $x = 0.06 \ m$,Initial velocity $V = N/10 \ m/s$.Using the Work-Energy Theorem,the work done by all forces equals the change in kinetic energy:$W_{	ext{friction}} + W_{	ext{spring}} = K_f - K_i$$-(\mu mg)x - \frac{1}{2}kx^2 = 0 - \frac{1}{2}mV^2$Substituting the values:$-(0.1 	imes 0.18 	imes 10 	imes 0.06) - \frac{1}{2} 	imes 2 	imes (0.06)^2 = -\frac{1}{2} 	imes 0.18 	imes V^2$$-(0.0108) - (0.0036) = -0.09 V^2$$-0.0144 = -0.09 V^2$$V^2 = \frac{0.0144}{0.09} = 0.16$$V = \sqrt{0.16} = 0.4 \ m/s$Given $V = N/10$,so $0.4 = N/10 \implies N = 4$.

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