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

`(1)`

Explanation

Applying the work - energy theorem, we get
`1/2 xx mv^(2) - 0 = W_(1) + W_(2) `
`= Fxx R + mg xx R`