A bead of mass ½ kg starts form rest from A to move in a vertical plane along a smooth fixed quarter ring of radius 5 m, under the action of a constant horizontal force F= 5N as shown in the figure. The speed of bead as it reaches point B is
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a
14.14 ms-1
b
7.07ms-1
c
5m-1
d
25m-1
answer is A.
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Detailed Solution
Applying the work energy theorem we getchange in KE =Work done by horizontal force and gravitational force⇒12×mv2−0=F×R+mg×Rsubstitute the given values12×12×v2=5×5+12×10×5⇒v2=4x50 v=200=14.14mS−1