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)  \[Mg\left( \sqrt{2}+1 \right)\]  

B)  \[Mg\sqrt{2}\]

C)  \[\frac{Mg}{\sqrt{2}}\]                                

D)  \[Mg\left( \sqrt{2}-1 \right)\]

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