The potential energy of the iron ball at its maximum displacement is equal to the work done in lifting it to that height against the force of gravity. The maximum displacement of the ball is when the thread makes a 30° angle with the vertical, and the height of the ball above its equilibrium position is given by:
h = 6m * sin(30°) = 3m
The potential energy of the ball at this displacement is:
PE = mgh
PE = 3kg * 9.81 m/s^2 * 3m
PE = 88.29 J
At the highest point of its oscillation, all of the potential energy is converted into kinetic energy, which is given by:
KE = 1/2 * m * v^2
where v is the velocity of the ball at this point. Since the total mechanical energy (potential energy + kinetic energy) is conserved, the kinetic energy at the maximum displacement is equal to the potential energy:
KE = PE
1/2 * m * v^2 = 88.29 J
Solving for v, we get:
v = sqrt(2 * PE / m)
v = sqrt(2 * 88.29 J / 3kg)
v = 6.17 m/s
Therefore, the maximum potential energy of the iron ball is 88.29 J, the maximum kinetic energy is also 88.29 J, and its velocity while passing through the mean position is 6.17 m/s.