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)

We are told that the total mechanical energy (the sum of kinetic and potential energies) of the pendulum bob is 0.02 J.

At the equilibrium (lowest) position the pendulum’s height is zero, so its gravitational potential energy is zero. Therefore all of the mechanical energy appears as kinetic energy:

$E\_k\; =\; 0.02 J$

The kinetic energy of a mass m moving with speed v is given by

$E\_k\; =\; \backslash frac\{1\}\{2\}\; m\; v^2$

We are given the bob’s mass as 20 g, which is

$m\; =\; 20 \backslash text\{g\}\; =\; 0.020 \backslash text\{kg\}$

Solving the kinetic‑energy equation for the speed v:

$v\; =\; \backslash sqrt\{\backslash frac\{2E\_k\}\{m\}\}$

Substituting the known values:

$v\; =\; \backslash sqrt\{\backslash frac\{2\; \backslash times\; 0.02\backslash \; \backslash text\{J\}\}\{0.020\backslash \; \backslash text\{kg\}\}\}\; =\; \backslash sqrt\{\backslash frac\{0.04\}\{0.020\}\}\; =\; \backslash sqrt\{2\}\; \backslash approx\; 1.41\backslash \; \backslash text\{m/s\}$

Thus the speed of the pendulum bob at the equilibrium position is approximately 1.41 m/s, which matches the provided correct option.