Question

A bead of mass 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 = 5 N as shown in Fig. 8.263. The speed of bead as it reaches point B is

• A. 14.14$m{s}^{-1}$
• B. 7.07$m{s}^{-1}$
• C. 5$m{s}^{-1}$
• D. 25$m{s}^{-1}$

Answer 1

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

According to work energy theorem.

Work done by all forces is equal to change in kinetic energy.

$+\frac{1}{2}\times 10\times 5+5\times 5=\frac{1}{2}\times \frac{1}{2}\times V^2$

$\Rightarrow V^2=200$

$\Rightarrow V=10\sqrt{2}$

$\Rightarrow V=14.14$ m/s