A spring-mass system vibrates such that the m | vedclass

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(B) According to the Work-Energy Theorem,the total work done on the system is equal to the change in kinetic energy.$W_{total} = \Delta KE$Since the m

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$\frac{2 \mu mg}{K}$

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(B) According to the Work-Energy Theorem,the total work done on the system is equal to the change in kinetic energy.$W_{total} = \Delta KE$Since the mass starts from rest at $x = -a$ and stops momentarily at $x = b$,the change in kinetic energy is $\Delta KE = 0 - 0 = 0$.The work done by the spring force is $W_s = \int_{-a}^{b} -Kx \, dx = -K \left[ \frac{x^2}{2} ight]_{-a}^{b} = -\frac{K}{2}(b^2 - a^2) = \frac{K}{2}(a^2 - b^2)$.The work done by friction is $W_f = -\mu mg(a + b)$ because the total distance traveled is $(a + b)$ and friction acts opposite to the displacement.Setting $W_{total} = W_s + W_f = 0$:$\frac{K}{2}(a^2 - b^2) - \mu mg(a + b) = 0$$\frac{K}{2}(a - b)(a + b) = \mu mg(a + b)$Since $a + b 
eq 0$,we can divide both sides by $(a + b)$:$\frac{K}{2}(a - b) = \mu mg$$(a - b) = \frac{2 \mu mg}{K}$Thus,the reduction in amplitude is $\frac{2 \mu mg}{K}$.

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