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

Options

A. `(Mg)/(sqrt(2))`
B. `Mg(sqrt(2) -1)`
C. `Mg(sqrt(2) + 1)`
D. `Mg sqrt(2)`

Answer

B

Explanation

According to work energy theorem "work done by all forces is equal to change in kinetic energy"
`W_("total") = Delta K`
`W_(F) = W_("gravity") = (k_("final") -k_("initial")) = (0-0) =0`

Work done by `F, W_(F) = F xx l sin 45^@ = (Fl)/(sqrt(2))`
Work done by gravity,
`W_("gravity") = -Mg(l-l cos45^@)=Mgl (1-(1)/(sqrt(2)))`
`:. (Fl)/(sqrt(2)) +(-(Mgl(sqrt(2) -1))/(sqrt(2))) = 0`
or `F = Mg (sqrt(2) -1)`.