Question 12 Marks
The bob of a simple pendulum is imparted a velocity $5 m s ^{-1}$ when it is at its mean position. To what maximum vertical height will it rise on reaching to its extreme position if $60 \%$ of its energy is lost in overcome friction of air?
Answer
View full question & answer→Potential energy at the extreme position $=40 \%$ of kinetic energy at the resting position.
$\begin{aligned}
& m g h=\frac{40}{100} \times\left(\frac{1}{2} m v^2\right) \\
& \frac{40}{100} \times\left(\frac{1}{2} m v^2\right)=m g h
\end{aligned}$
$0.4 \times 0.5 \times m \times v ^2= mgh$
$0.2 \times v^2=10 \times h$
$0.2 \times 5 \times 5=10 h$
$H=5 / 10=0.5 m$
$\begin{aligned}
& m g h=\frac{40}{100} \times\left(\frac{1}{2} m v^2\right) \\
& \frac{40}{100} \times\left(\frac{1}{2} m v^2\right)=m g h
\end{aligned}$
$0.4 \times 0.5 \times m \times v ^2= mgh$
$0.2 \times v^2=10 \times h$
$0.2 \times 5 \times 5=10 h$
$H=5 / 10=0.5 m$