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 lower most point $B$, given that it dissipated $5$% of its initial energy against air resistance?

• A. $5m/s$
• B. $5.5m/s$
• C. $5.3m/s$
• D. $4.4m/s$

Answer 1

The correct option is C $5.3m/s$

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 equation (i), $\frac{1}{2}m{v}^2=\frac{95}{100}mgh$
$v^2=2\times \frac{95}{100}gh=2\times \frac{95}{100}\times 9.8\times 1.5$
$v=\sqrt{\frac{19\times 9.8\times 1.5}{10}}=\sqrt{27.93}=5.285{ms}^{-1}$
$v=5.3{ms}^{-1}$.