A) \[Mg(\sqrt{2}+1)\]
B) \[Mg\sqrt{2}\]
C) \[\frac{Mg}{\sqrt{2}}\]
D) \[Mg(\sqrt{2}-1)\]
Correct Answer: D
Solution :
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) \[{{W}_{F}}\,=F\times l\sin {{45}^{0}}\] \[{{W}_{Mg}}\,=Mg(l-\cos {{45}^{0}}),\,\,{{W}_{tension}}\,=0\] \[\therefore \,\,F=Mg\,(\sqrt{2}-1)\]
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