A bead of mass 12kg  starts from rest from A to move in a vertical plane along a smooth fixed quarter ring of radius 5m, under a force F=5 N.  The speed of bead is

Given mass of bead m=12kg Radius of ring R=5 mForce F=5 NThe speed of bead v=?According work-energy theorem.Work done = change in kinetic energyW=ΔKE=12mv2−12mu2                  (∵u=0)⇒F×R+mgR=12mv2−0                 (∵g=0)⇒12×12v2=5×5+12×10×5⇒14v2=25+25⇒v2=50×4=200⇒v=200=102=14.14 m/s