Question

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

Answer 1

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

Answer 2

14.14 m/s. Is it right