Question

The bob of a pendulum was released from a horizontal position. The length of the pendulum is 10 m. If it dissipates 10% of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is: [Use, \( g : 10 \, \text{m/s}^2 \)]

• A. \( 6\sqrt{5} \, \text{ms}^{-1} \)
• B. \( 5\sqrt{6} \, \text{ms}^{-1} \)
• C. \( 5\sqrt{5} \, \text{ms}^{-1} \)
• D. \( 2\sqrt{5} \, \text{ms}^{-1} \)

Answer 1

The Correct Option is A

To determine the pendulum bob's speed at its lowest point, accounting for energy loss due to air resistance, we first establish the initial potential energy. When released from a horizontal position, the bob possesses potential energy calculated as:

\[ E_{\text{initial}} = m \cdot g \cdot h \]

where:

• \(m\) represents the bob's mass (this variable will cancel out).
• \(g = 10 \, \text{m/s}^2\) is the acceleration due to gravity.
• \(h = 10 \, \text{m}\) is the initial height.

Consequently, the initial energy is:

\[ E_{\text{initial}} = m \cdot 10 \cdot 10 = 100m \, \text{J} \]

The problem states that 10% of this initial energy is dissipated. This means 90% is converted into kinetic energy at the lowest point. Therefore, the energy at the lowest point is:

\[ E_{\text{kinetic}} = 0.9 \times 100m = 90m \, \text{J} \]

The kinetic energy at the lowest point is also defined by:

\[ E_{\text{kinetic}} = \frac{1}{2} m v^2 \]

Equating the available kinetic energy with this formula yields:

\[ \frac{1}{2} m v^2 = 90m \]

Upon canceling the mass (\(m\)) from both sides:

\[ \frac{1}{2} v^2 = 90 \]

Multiplying both sides by 2 to solve for \(v^2\):

\[ v^2 = 180 \]

Taking the square root of both sides gives the speed:

\[ v = \sqrt{180} = \sqrt{36 \cdot 5} = 6\sqrt{5} \, \text{ms}^{-1} \]

The bob's speed upon reaching the lowest point is thus \(6\sqrt{5} \, \text{ms}^{-1}\).

\( 6\sqrt{5} \, \text{ms}^{-1} \)