Question

The sum of kinetic energy and potential energy of a simple pendulum bob is 0.02 J. The speed of the simple pendulum bob at equilibrium position is approximately: (Consider mass of the bob = 20 g)

• A. 14.1 m/s
• B. 1.41 m/s
• C. 2.0 m/s
• D. 0.2 m/s

Hint:

In conservation of energy problems, always convert mass to kilograms (SI units) immediately to avoid decimal errors in your final result.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
According to the law of conservation of mechanical energy, the total energy (KE + PE) remains constant. At the equilibrium position (lowest point), the potential energy is zero (reference level), so all the energy is converted into kinetic energy.

Step 2: Key Formula or Approach:
1. Total Energy ($E$) = Max Kinetic Energy 2. $E = \frac{1}{2} mv^2$

Step 3: Detailed Explanation:
Given: $E = 0.02$ J, $m = 20$ g = 0.02 kg. (Note: Assuming the "eV" in the user prompt was a typo for "J" given the typical scale of such physics problems, as 0.02 eV would result in an extremely small, non-listed velocity). 1. Set up the energy equation: \[ 0.02 = \frac{1}{2} \times 0.02 \times v^2 \] 2. Simplify: \[ 0.02 = 0.01 \times v^2 \] \[ v^2 = \frac{0.02}{0.01} = 2 \] 3. Calculate $v$: \[ v = \sqrt{2} \approx 1.41 \text{ m/s} \]

Step 4: Final Answer:
The speed at the equilibrium position is approximately 1.41 m/s.