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 \[45{}^\circ \] 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: C

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 doesnt change, then                 \[\Delta K=0\]                 \[={{W}_{F}}+{{W}_{mg}}+{{W}_{tension}}\] [symbols have their usual meanings]                 \[{{W}_{F}}=F\times l\sin {{45}^{o}}\]                 \[{{W}_{mg}}={{M}_{g}}(l-l\cos {{45}^{o}}),\,\,{{W}_{tension}}=0\] \[\therefore \]  \[F=Mg(\sqrt{2}-1)\]