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

Options

A. `(Mg)/(sqrt2)`
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 kinetic energy”.
`:. W_("total") = Delta K`
`W_(F) + W_("gravity") + W_(T) = (K_("final") - K_("initial")) = (0-0) = 0`
Work done by `F , W_(F) = F xx l sin 45^@ = (F1)/(sqrt(2))`
Work done by gravity ,
`W_("gravity") = -Mg(1 - l cos 45^@) = - M gl(1 - 1/(sqrt2))`
Work done by tension = 0
`:. (F1)/(sqrt2) + (- (Mgl(sqrt(2) - 1))/(sqrt(2))) = 0 " or " F = Mg(sqrt(2) - 1)` .