Question

The bob of a pendulum is released from a horizontal position $A$ as shown in the figure. If the length of the pendulum is $1.5m$ , what is the speed with which the bob arrives at the lowermost point $B$ , given that it dissipated $5\%$ of its initial energy against air resistance?

• A. $5m{s}^{-1}$
• B. $5.5m{s}^{-1}$
• C. $5.3m{s}^{-1}$
• D. $4.4m{s}^{-1}$

Answer 1

Answer: (C)

At the point $A$ the energy of the pendulum is entirely potential energy. At point B, the energy of the pendulum is entirely kinetic energy. It means that as the bob pendulum lowers from A to B, the potential energy is converted into kinetic energy. Thus, at $B$ , KE = PE. But $5\%$ of the potential energy is dissipated against air resistance.

$KE$ at $B=95\%ofPEatA$ .... (i)
$\therefore$ From equation (i), $\frac{1}{2}m{v}^2=\frac{95}{100}mgh$
$v^2=2\times \frac{95}{100}gh=2\times \frac{95}{100}\times 9.8\times 1.5$
$v=\sqrt{\frac{19\times 9.8\times 1.5}{10}}=\sqrt{27.93}=5.285m{s}^{-1}$
$v=5.3m{s}^{-1}$