Physics - W.E.P. Question with Solution | TestHub

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 $\begin{array}{rl}\Delta K& =0\\ & =W_F+W_{Mg}+W_{\text{tension }}\end{array}$ [symbols have their usual meanings] $\begin{array}{rl}{W}_F& =F\times l\sin 45^{\circ },\\ W_{Mg}& =M_g\left(l-l\cos 45^{\circ }\right),W_{\text{tension }}=0\\ \therefore F& =Mg(\sqrt{2}-1)\end{array}$