A bead of mass `1/2 kg` starts from rest from A to move in a vertical place along a smooth fixed quarter ring of radius `5 m`, under the action of a constant horizontal force `f=5 N` as shown. The speed of bead as it reaches the point (B) is [Take `g=10 ms^(-2)`]

Options

A. 14.14 m/s
B. 7.07 m/s
C. 5 m/s
D. 25 m/s

Answer

A

Explanation

`1/2 xx mv^2 - 0 = F xx R + mg xx R , 1/2 xx 1/2 xx v^2 = 5 xx 5 + 1/2 xx (10 xx 5) = 50, v = sqrt(200) = 14.14 m//s`.