A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of \[{{45}^{o}}\] with the initial vertical direction is
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] \[{{W}_{F}}=F\times l\sin {{45}^{o}},\] \[{{W}_{Mg}}={{M}_{g}}(1-lcos{{45}^{o}}),\] \[{{W}_{tension}}=0\] \[\therefore \] \[F=Mg(\sqrt{2}-1)\]