Question

A bead of mass $\frac{1}{2}kg$ 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 $F=5N$ as shown. The speed of bead as it reaches the point $B$ is [Take $g=10m{s}^{-2}$]

• A. $14.14m{s}^{-1}$
• B. $7.07m{s}^{-1}$
• C. $4m{s}^{-1}$
• D. $25m{s}^{-1}$

Answer 1

Answer: (A)

$14.14m{s}^{-1}$

$W_{All}=\frac{1}{2}m{v}^2$

$\therefore W_F+W_{mg}+W_M=\frac{1}{2}m{v}^2$

$\therefore (5\times 5)+(\frac{1}{2}\times 10\times 5)+0=\frac{1}{2}\times \frac{1}{2}\times v^2$

$\therefore v=14.14m/s$