A 4 text{ kg} mass is suspended as shown in t | vedclass

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(B) Given: Mass $m = 4 	ext{ kg}$,Spring constant $K = 8 	imes 10^3 	ext{ Nm}^{-1}$,$g = 10 	ext{ ms}^{-2}$.From the free body diagram of the lowe

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$2 	ext{ cm}$

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(B) Given: Mass $m = 4 	ext{ kg}$,Spring constant $K = 8 	imes 10^3 	ext{ Nm}^{-1}$,$g = 10 	ext{ ms}^{-2}$.From the free body diagram of the lower pulley,the tension $T$ in the string supporting the $4 	ext{ kg}$ mass is $T = mg = 4 	imes 10 = 40 	ext{ N}$.The lower pulley is supported by two segments of the string,each with tension $T$. Thus,the force exerted by the lower pulley on the upper pulley is $2T = 2 	imes 40 = 80 	ext{ N}$.The upper pulley is supported by the spring and the string fixed to the ground. The total downward force on the upper pulley is the sum of the tension from the string fixed to the ground $(2T)$ and the force from the lower pulley $(2T)$.Therefore,the total force $F$ in the spring is $F = 2T + 2T = 4T = 4 	imes 40 = 160 	ext{ N}$.Using Hooke's Law,the extension $x$ in the spring is $x = \frac{F}{K} = \frac{160}{8 	imes 10^3} = 20 	imes 10^{-3} 	ext{ m} = 2 	imes 10^{-2} 	ext{ m} = 2 	ext{ cm}$.

$A$ $4 \text{ kg}$ mass is suspended as shown in the figure. All pulleys are frictionless and the spring constant $K$ is $8 \times 10^3 \text{ Nm}^{-1}$. The extension in the spring is $\left(g=10 \text{ ms}^{-2}\right)$.

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