Question

The bob of a simple pendulum of length $200~\text{cm}$ is released from horizontal position.
If $10%$ of its initial energy is lost to air resistance, then the speed of bob at the mean position is
(Acceleration due to gravity $= 10~\text{m/s}^2$)

• A. $6~\text{m/s}$
• B. $3~\text{m/s}$
• C. $12~\text{m/s}$
• D. $2~\text{m/s}$

Hint:

If energy is lost, subtract it from total potential energy before applying conservation laws.

Answer 1

The Correct Option is A

Height of bob from mean position: $h = 2~\text{m}$
Initial potential energy: $PE = mgh = 2mg$
$10%$ loss $\Rightarrow$ energy converted to KE = $0.9 \cdot 2mg = 1.8mg$
Set this equal to kinetic energy at mean: $\dfrac{1}{2}mv^2 = 1.8mg$
$v^2 = 3.6g = 36 \Rightarrow v = 6~\text{m/s}$