Question

The bob of a pendulum is released from a horizontal position. If the length of pendulum is $2m$, what is the speed with which the bob arrives at the lower most point. Assume that $10$% of its energy is dissipated against air resistance.
(Take $g=10m{s}^{-2})$.

• A. $4m{s}^{-1}$
• B. $6m{s}^{-1}$
• C. $8m{s}^{-1}$
• D. $10m{s}^{-1}$

Answer 1

The correct option is B $6m{s}^{-1}$

The situation is as shown in the figure.
Potential energy of the bob at $A=mgh$
Kinetic energy of the bob at $B=\frac{1}{2}m{v}^2$
As $10$% of the energy of the bob is dissipated against air resistance.
$\therefore \frac{1}{2}m{v}^2=\frac{90}{100}$ of $mgh$ or $v^2=2\times \frac{90}{100}\times g\times h$
$v=\sqrt{\frac{2\times 90\times g\times h}{100}}=\sqrt{\frac{2\times 90\times 10\times 2}{100}}=6m{s}^{-1}$.