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(D) Let the length of the string be $\ell$. When the mass is displaced by an angle $	heta = 45^\circ$,the vertical height gained by the mass is $h =

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$Mg(\sqrt{2} - 1)$

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(D) Let the length of the string be $\ell$. When the mass is displaced by an angle $	heta = 45^\circ$,the vertical height gained by the mass is $h = \ell - \ell \cos 45^\circ = \ell(1 - \frac{1}{\sqrt{2}})$.The horizontal displacement of the mass is $x = \ell \sin 45^\circ = \frac{\ell}{\sqrt{2}}$.Using the work-energy theorem,the work done by the external horizontal force $F$ must equal the change in potential energy of the mass (assuming it starts and ends at rest,so $\Delta K = 0$):$W_F + W_g = 0$$F \cdot x - Mg \cdot h = 0$$F \cdot (\frac{\ell}{\sqrt{2}}) = Mg \cdot \ell(1 - \frac{1}{\sqrt{2}})$$F = Mg \cdot \sqrt{2}(1 - \frac{1}{\sqrt{2}})$$F = Mg(\sqrt{2} - 1)$

$A$ mass of $M \ kg$ is suspended by a weightless string of length $\ell$. The horizontal force required to displace it until the string makes an angle of $45^\circ$ with the initial vertical direction is:

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