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If elongation of spring is 4m then $F_{min}$ for equilibrium ${(g =10 m s}^{2})$

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120 N

60 N

200 N

80 N

answer is D.

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Detailed Solution

$2 F_{\min} + f_{l} = K x$

$F_{\min} = \frac{50 (4) - 0.4 (100)}{2} = 80 N$

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