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 ms^(-1)`
B. `7.07 ms^(-1)`
C. `5 ms^(-1)`
D. `25 ms^(-1)`

Answer

A

Explanation

(a) From work-energy theorrem
`W_(F)+W_(mg)=(1)/(2)mv^(2)`
`rArr " " F.R+mgR=(1)/(2)mv^(2)`
`rArr " " 5xx5+(1)/(2)xx10xx5=(1)/(2)xx(1)/(2)xxv^(2)`
`rArr " " v=sqrt(200)=14.14" ms"^(-1)`