Question

A bead of mass `1/2 kg` starts from rest from A to move in a vertical place along a smooth fixed quarter ring of radius `5 m`, under the action of a constant horizontal force `f=5 N` as shown. The speed of bead as it reaches the point (B) is [Take `g=10 ms^(-2)`]

A. (a) `14.14 ms^(1)`
B. (b) `7.07 ms^(-1)`
C. (c) `4 ms^(-1)`
D. (d) `25 ms^(-1)`

Answer 1

Correct Answer - A
`W_("All") =1/2mv^(2)`
`:. W_(F) + W_(mg) + W_(N) =1/2mv^(1)`
`:. (5 xx 5) + (1/2 xx 10 xx 5) + 0 = 1/2 xx 1/2 xx v^(2)`
`:. v=14.14 m//s`.