• # question_answer A mass of M kg is suspended by a weightless string. The minimum force that is required to displace it until the string makes an angle of ${{45}^{o}}$ 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 calculate by using work-energy theorem. Let us assume that body is taken slowly so, that its speed does not change, then $\Delta K=0$ ${{W}_{F}}+{{W}_{Mg}}\,+{{W}_{tenstion}}=\Delta KE$ (Symbols have their usual meaning) ${{W}_{F}}\,=F\times l\sin {{45}^{0}}$ ${{W}_{Mg}}\,=Mg(l-\cos {{45}^{0}}),\,\,{{W}_{tension}}\,=0$ $\therefore \,\,F=Mg\,(\sqrt{2}-1)$