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In the situation shown in Fig. all contact surfaces are smooth. The force constant of the spring is K. Two forces F are applied as shown. The maximum elongation produced in the spring is how many times of F/K (initially the spring is relaxed)

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answer is 2.

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Detailed Solution

(2)
Maximum elongation is given by
$x_{\max} = \frac{2 [F_{1} m_{2} + F_{2} m_{1}]}{K (m_{1} + m_{2})}$
Here $F_{1} = F_{2} = F; m_{1} = m a n d m_{2} = M$

Put the values and solve to get $x_{\max} = 2 F / K$

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