A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of 45⁰ with the initial vertical direction is

 The situation is shown in fig. (9)

The constant force F required to take the body from position 1 to position 2 can be calculated by applying work-energy theorem. We assume that the body is taken slowly so that its speed doesn't change, i.e.$\mathrm{\Delta K}=0$

$$\begin{gathered} {\mathrm{W}}_{\mathrm{F}}+{\mathrm{W}}_{\mathrm{Mg}}+{\mathrm{W}}_{\text{tension }}=0 \\ \mathrm{Hero} {\mathrm{W}}_{\mathrm{F}}=\mathrm{F}\times \mathrm{lsin}{45}^{\circ } \\ {\mathrm{W}}_{\mathrm{Mg}}=\mathrm{Mg}\left(\mathrm{l}-\mathrm{lcos}{45}^{\circ }\right) \\ {\mathrm{W}}_{\text{tension }}=0 \\ \mathrm{F}\times \mathrm{lsin}{45}^{\circ }+\mathrm{Mg}\left(\mathrm{l}-\mathrm{lcos}{45}^{\circ }\right)=0 \\ \mathrm{or} \mathrm{F}=\mathrm{Mg}(\sqrt{2}-1) \end{gathered}$$