Question

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° with the initial vertical direction is :

• A. $Mg\left(\sqrt{2}+1\right)$
• B. $Mg\sqrt{2}$
• C. $\frac{Mg}{\sqrt{2}}$
• D. $Mg \left(\sqrt{2}-1\right)$

Answer 1

Answer: (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 is taken slowly so that its speed doesn’t change, then $\Delta$K = 0
$=W_{F}+W_{M g}+W_{tension}$
[symbols have their usual meanings]
$W_{F}=F \times l \, sin \,45^{\circ}, $
$W_{Mg}=Mg \left(l-l \cos 45^{\circ}\right), w_{tension}=0$
$\therefore \, \quad F=Mg \left(\sqrt{2}-1\right)$