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

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Aicharya · Accepted Answer

$\mathrm{k}=8 	imes 10^3 \mathrm{Nm}^{-1}$ From figure, $T=40 \mathrm{~N} \Rightarrow$ Force in spring, $\mathrm{F}=4 \mathrm{~T}=160 \mathrm{~N}$ $	herefore$ Extension in the spring, $x=\frac{F}{K}=\frac{160}{8 	imes 10^3}=2 \mathrm{~cm}$

A 4kg mass is suspended as shown in figure. All pulleys are frictionless and spring constant K is $8 \times…

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