A) \[Mg(\sqrt{2}+1)\]
B) \[Mg\sqrt{2}\]
C) \[\frac{Mg}{\sqrt{2}}\]
D) \[Mg(\sqrt{2}-1)\]
Correct Answer: D
Solution :
[d] 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 |
\[\Delta K=\]Change in kinetic energy\[=0\] |
\[={{W}_{F}}+{{W}_{Mg}}+{{W}_{tension}}\] .(i) |
(symbols have their usual meanings) |
\[{{W}_{F}}=F\times l\,\sin {{45}^{o}},\] |
\[{{W}_{Mg}}={{M}_{g}}(l-l\cos {{45}^{o}}),{{W}_{tension}}=0\] |
Putting these values in Eq. (i), we get |
\[F=Mg(\sqrt{2}-1)\] |
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