Q. A mass of M kg suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of $45^{0\ }$with the initial vertical direction is
A) $Mg(\sqrt{}2\ + 1)$
B) $Mg\sqrt{2}$
C) $\frac{Mg}{\sqrt{2}}$
D) $Mg\left( \sqrt{2} - 1 \right)$
Explanation
(d) 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 taken slowly so that its speed doesn’t change, then $\mathrm{\Delta}K = 0$
$= W_{F} + W_{Mg} + W_{tension}$
(symbols have their usual meanings )
$W_{F} = F \times l\sin{45{^\circ}},$
$W_{Mg} = Mg(l - l\cos{45{^\circ}),W_{tension} = 0}$
$\therefore\ F = Mg(\sqrt{2} - 1)$
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Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.