Question

A bead of mass$\frac{1}{2}kg$ which is at rest, starts to move from end A of a smooth fixed quarter ring of radius $5m$ in vertical plane under the action of a constant horizontal force $F=5N$ as shown. Then the speed of the bead as it reaches the end B is

• A. $14.14m/s$
• B. $7.07m/s$
• C. $5m/s$
• D. $25m/s$

Answer 1

The correct option is A $14.14m/s$

$W_{force}+W_{gravity}=\frac{1}{2}m{v}^{2}F.R+mgR+0=\frac{1}{2}m{v}^{2}25+25=\frac{1}{2}\times \frac{1}{2}\times v^{2}v=14.14m/s$