In the setup shown,a 200, N block is supporte | vedclass

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(C) Let $T_1$ be the tension in the right string and $F_s$ be the spring force in the left string.At point $O$,the forces are in equilibrium.Resolving

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$3000$

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(C) Let $T_1$ be the tension in the right string and $F_s$ be the spring force in the left string.At point $O$,the forces are in equilibrium.Resolving forces horizontally: $F_s \cos 37^\circ = T_1 \cos 53^\circ$.Since $\cos 37^\circ = 4/5$ and $\cos 53^\circ = 3/5$,we have $F_s (4/5) = T_1 (3/5) \implies 4F_s = 3T_1 \implies T_1 = \frac{4}{3}F_s$.Resolving forces vertically: $F_s \sin 37^\circ + T_1 \sin 53^\circ = 200$.Substituting $T_1$: $F_s (3/5) + (\frac{4}{3}F_s) (4/5) = 200$.$\frac{3}{5}F_s + \frac{16}{15}F_s = 200 \implies \frac{9+16}{15}F_s = 200 \implies \frac{25}{15}F_s = 200$.$F_s = 200 	imes \frac{15}{25} = 200 	imes 0.6 = 120\, N$.Using Hooke's Law $F_s = kx$,where $x = 4\, cm = 0.04\, m$.$120 = k 	imes 0.04 \implies k = \frac{120}{0.04} = 3000\, N/m$.

In the setup shown,a $200\, N$ block is supported in equilibrium with the help of strings and a spring. At point $O$,the strings are knotted. The extension in the spring is $4\, cm$. The force constant of the spring is closest to ............ $N/m$ $[g = 10\, m/s^2]$.

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