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

Applying the work - energy theorem, we get
`(1)/(2) xx mv_(2) - 0 = W_(1) + W_(2)`
`= "Horizontal force" xx "displacement" + "Vertical force" xx "displacement"`.
=`F xx R + mg xx R`.