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

This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
$1.41 \text{ m/s}$

Pendulum Energy and Speed Calculation

The total mechanical energy ($E$) of a simple pendulum is the sum of its kinetic energy ($KE$) and potential energy ($PE$): $E = KE + PE$

We are given that the total energy is $E = 0.02 \text{ J}$.

At the equilibrium position, the potential energy ($PE$) is minimum, typically considered zero. Therefore, at this point, all the energy is kinetic energy ($KE_{max}$). $E = KE_{max} + 0$ $KE_{max} = E = 0.02 \text{ J}$

Calculating Speed at Equilibrium

The formula for kinetic energy is $KE = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the speed.

Given:

  • Total Energy, $E = 0.02 \text{ J}$
  • Mass, $m = 20 \text{ g} = 0.02 \text{ kg}$ (converted to kilograms)

At equilibrium, $KE_{max} = \frac{1}{2}mv^2$. We set this equal to the total energy:

$0.02 \text{ J} = \frac{1}{2} \times (0.02 \text{ kg}) \times v^2$

Now, solve for $v$:

$0.02 = 0.01 \times v^2$

$v^2 = \frac{0.02}{0.01}$

$v^2 = 2$

$v = \sqrt{2} \text{ m/s}$

Calculating the square root:

$v \approx 1.414 \text{ m/s}$

Rounding to two decimal places, the speed is approximately $1.41 \text{ m/s}$.

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