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 \[\Delta K=\]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 \[F=Mg(\sqrt{2}-1)\]You need to login to perform this action.
You will be redirected in
3 sec