Question

A mass of $M$ kg is suspended by a weightless string. The horizontal force required to displace it until the string makes an angle of ${45}^o$ with the initial vertical is :

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

Answer 1

Answer: (A)

$Mg(\sqrt{2}-1)$

Here, the constant horizontal force required to take the body from position 1 to position 2 can be calculated by using work energy theorem. It is given that body is taken slowly so that its speed doesn't change, then
$\Delta K.E=0$
$=W_F+W_{Mg}+W_{tension}$
[symbols have their usual meanings]
$W_F=F\times lsin{45}^0,$
$W_{Mg}=-Mg(l-lcos{45}^0)$

and $W_{tension}=0$ ($\because$ tension is always perpendicular to the motion of block)

$F\times lsin{45}^o-Mg(l-lcos{45}^o)+0=0$
$F=Mg(\sqrt{2}-1)$