Question

A bead of mass $\frac{1}{2}$ kg starts from rest 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. 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

The correct option is A $14.14m{s}^{-1}$

Total work-done by bead is the sum of work done by force F and gravity

$W_F=F.s=5\times 5=25J$, $W_g=mgh=\frac{1}{2}10\times 5=5\times 5=25J$

$W_{net}=25+25=50J$
By work energy theorem,
$W_{net}=\Delta KE$

$⟹\frac{1}{2}m{v}^2=50$

$⟹v^2=50\times 2\times 2=200$
$v=\sqrt{200}=14.14m/s$