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. (a) `14.14ms^-1`
B. (b) `7.07ms^-1`
C. (c) `5ms^-1`
D. (d) `25ms^-1`

Answer

A

Explanation

Applying the work-energy theorem, we get
`1/2xxmv^2-0=FxxR+mgxxR`
`1/2xx1/2xxv^2=5xx5+1/2xx10xx5=50`
`v=sqrt(200)=14.14ms^-1`