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A bead of mass 1/2 kg starts from rest f...

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)`]

A

`14.14ms^-1`

B

`7.07ms^-1`

C

`5ms^-1`

D

`25ms^-1`

Text Solution

Verified by Experts

The correct Answer is:
A
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`
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