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 calculated by using work-energy theorem. Let us assume that body is taken slowly so that its speed doesn?t change, then\[\Delta K=0\] \[={{W}_{F}}+{{W}_{Mg}}+{{W}_{tension}}\] [symbols have their usual meanings] \[{{W}_{F}}=F\times l\sin {{45}^{o}}\] \[{{W}_{Mg}}={{M}_{g}}(l-l\cos {{45}^{o}})\] \[{{W}_{tension}}=0\] \[\therefore \] \[F=Mg(\sqrt{2}-1)\]You need to login to perform this action.
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