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

The problem states that the sum of the kinetic energy (KE) and potential energy (PE) of a simple pendulum bob is 0.02 J, and we are to find the speed of the pendulum bob at its equilibrium position. At the equilibrium position of a simple pendulum, the entire energy is in the form of kinetic energy because the potential energy is zero.   

1. The total energy of the pendulum is given as 0.02 J: \(E = KE + PE = 0.02\ \text{J}\).
2. At the equilibrium position, PE = 0, therefore: \(KE = 0.02 \ \text{J}\).
3. The formula for kinetic energy (KE) is: \(\text{KE} = \frac{1}{2}mv^2\), where \(m\) is the mass and \(v\) is the velocity.
4. Substitute the given values: \(\frac{1}{2} \times 20\ \text{g} \times v^2 = 0.02\ \text{J}\).
5. Convert the mass from grams to kilograms: \(20\ \text{g} = 0.02\ \text{kg}\).
6. Plug the values into the equation: \(\frac{1}{2} \times 0.02 \times v^2 = 0.02\).
7. Solve for \(v^2\): \(v^2 = \frac{0.02 \times 2}{0.02}\).
8. Calculate and take the square root to find \(v\): \(v^2 = 2\) → \(v = \sqrt{2} \approx 1.41\ \text{m/s}\).

Thus, the speed of the simple pendulum bob at the equilibrium position is approximately 1.41 m/s.