A bead of mass 12 kg starts from rest from “A” to move in a vertical plane along a smooth fixed quarter ring of radius 5m, under the action of a constant horizontal force F = 5 N as shown. The speed of bead as it reaches point “B” is
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a
14.14 m/s
b
7.07 m/s
c
5 m/s
d
25 m/s
answer is A.
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Detailed Solution
Applying the work - energy theorem, we get12 x mv2-0=W1+W2= Horizontal force × displacement + Vertical force × displacement.= F x R + mg x R