• question_answer 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 $45{}^\circ$ with the initial vertical direction is A)  $Mg(\sqrt{2}+1)$         B)  $Mg\sqrt{2}$ C) $\frac{Mg}{\sqrt{2}}$                                  D) $Mg(\sqrt{2}-1)$

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                     $={{W}_{F}}+{{W}_{Mg}}+{{W}_{tension}}$ [symbols have their usual meanings] $={{W}_{F}}=F\times l\,\sin \,{{45}^{o}},$ ${{W}_{ & Mg}}={{M}_{g}}(l-l\,\cos \,{{45}^{o}}),$${{W}_{tension}}=0$ $\therefore$  $F=Mg(\sqrt{2}-1)$