A mass of M ~kg is suspended by a weightless string. The horizontal...

Here, the constant horizontal force required to take the body from position 1 to position 2 can be calculated by using work-energy theorem. Let us assume that body is taken slowly so that its speed doesn't change, then $$\begin{aligned} \Delta K & =0 \\ & =W_F+W_{M g}+W_{\text{tension }} \end{aligned}$$ [symbols have their usual meanings] $$\begin{aligned} W_F & =F \times l \sin 45^{\circ}, \\ W_{M g} & =M_g\left(l-l \cos 45^{\circ}\right), W_{\text{tension }}=0 \\ \therefore \quad F & =M g(\sqrt{2}-1) \end{aligned}$$