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 $\text{1}.\text{5 m}$ , what is the speed with which the bob arrives at the lower most point B, given that it dissipated 5% of its initial energy against air resistance?

• A. $\text{5 m}/\text{s}$
• B. 5.5 m/s
• C. $\text{5}.\text{3 m}/\text{s}$
• D. $\text{4}.\text{4 m}/\text{s}$

Answer 1

Answer: (C)

At the point A, the energy of the pendulum is entirely PE. At point B, the energy of the pendulum is entirely KE. It means that as the bob pendulum lowers from A to B, PE is converted into KE. Thus, at B, KE=PE. But 5% of the PE is dissipated against air resistance.
KE at B = 95% of PE at A ... (i) $\therefore$ From Eq. (i), $\frac{1}{2}m{v}^2=\frac{95}{100}mgh$ $v^2=2\times \frac{95}{100}gh=2\times \frac{95}{100}\times 9.8\times 1.5$ $v=\sqrt{\frac{19\times 9.8\times 1.5}{10}}=\sqrt{27.93}=5.285\text{m}{\text{s}}^{\text{-1}}$ $v=5.3\text{m}{\text{s}}^{\text{-1}}$