question_answer

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] \[{{W}_{F}}=F\times l\sin {{45}^{o}},\] \[{{W}_{Mg}}={{M}_{g}}(1-lcos{{45}^{o}}),\] \[{{W}_{tension}}=0\] \[\therefore \] \[F=Mg(\sqrt{2}-1)\]