Question

A bead of mass $\frac{1}{2}\text{kg}$ starts from rest from point $\text{A}$ and moves in a vertical plane along a smooth fixed quarter ring of radius $5\text{m}$, under the action of a constant horizontal force $F=5\text{N}$ as shown in the figure. The speed of the bead as it reaches the point $\text{B}$ is (Take $g=10{\text{m/s}}^2)$

• A. $14.14\text{m/s}$
  
• B. $7.07\text{m/s}$
• C. $5\text{m/s}$
• D. $25\text{m/s}$

Answer 1

The correct option is A $14.14\text{m/s}$


From work-energy theorem,

$W_{net}=\Delta KE$

$W_F+W_{mg}+W_N=\frac{1}{2}m{v}^2-0$

$F\left(R\right)+mgR+0=\frac{1}{2}m{v}^2$

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

$\therefore$ $v=\sqrt{200}=14.14\text{m/s}$

Why this Question? Key concept : Work-Energy Theorem The theorem states that the total work done on a system equals the change in its kinetic energy $W_{net}=\Delta KE$