# A bead of mass 1/2 kg starts from rest from A to move in a vertical plane along a smooth fixed quarter ring of radium 5 m, under the action of a constant horizontal force F = 5N . Then speed of bead as it reaches the point B is [Taken g = 10 ms-2] Options: 14.14 ms-1 7.07 ms-1 4 ms-1 25 ms -1

1
by RKS

2015-09-15T02:00:24+05:30

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Looks like the bead is moving down the altitude while sliding along the quarter ring.

Work done by the force = force * displacement horizontally
W = F * Radius = 5 N * 5 m = 25 J

K E = Work done by Force 5 N  +  change in PE,   due to conservation of energy

1/2 m v²  =  25 N + 1/2 kg * 10 m /s²  * 5 m = 50 N

v² = 200  =>     v = 10 √2  m/sec.  = 14.14 m/s

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