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 Topic:
This problem involves "Work, Energy, and Power." It uses the Principle of Conservation of Mechanical Energy. In a frictionless pendulum system, the total mechanical energy (the sum of kinetic and potential energy) remains constant throughout the motion.
Step 2: Key Formulas and Approach:

Total Mechanical Energy ($E_{total}$) = $K.E. + P.E. = \text{constant}$.
At equilibrium position (the lowest point), potential energy is minimum (taken as zero).
Thus, at equilibrium: $E_{total} = K.E._{max}$.
$K.E. = \frac{1}{2} m v^2$.

Step 3: Detailed Explanation:

Identify given values: Total Energy $E = 0.02 \text{ J}$. Mass $m = 20 \text{ g} = 0.02 \text{ kg}$ (conversion to SI units is essential).
Equate energies: At the equilibrium point, all the energy stored in the pendulum system is in the form of motion. \[ 0.02 = \frac{1}{2} \cdot m \cdot v^2 \]
Substitute mass: \[ 0.02 = \frac{1}{2} \cdot (0.02) \cdot v^2 \]
Solve for v: \[ 0.02 = 0.01 \cdot v^2 \] \[ v^2 = \frac{0.02}{0.01} = 2 \] \[ v = \sqrt{2} \]
Approximate: $\sqrt{2} \approx 1.414$.
This speed is the maximum speed the bob reaches during its entire swing.
Step 4: Final Answer:
The speed at the equilibrium position is approximately 1.41 m/s.