Frank has a mass of 30 kg. He is sitting on a swing which is 2.3 m long. His dad pulls him up to a height of 1.2 m above his lowest point and releases him. b. What is his velocity at the bottom of his path?

His velocity at the bottom of his path is approximately 4.85 m/s

Explanation:

From the question, we have;

The mass of Frank, m = 30 kg

The length of the swing which he is sitting on, l = 2.3 m

The height above the lowest point to which the dad pulls him, h = 1.2 m

Potential energy, P.E. = m·g·h

Where;

g = The acceleration due to gravity ≈ 9.8 m/s²

The maximum potential energy Frank gains, P.E. is goven as follows;

P.E. = 30 kg × 9.81 m/s² × 1.2 m = 353.16 J

Therefore, by the principle of conservation of energy, we have;

The maximum kinetic energy at the bottom of the swing, K.E. = P.E. =  353.16 J

K.E. = 1/2·m·v²

Where;

v = The velocity at the bottom of his path

∴ K.E. = 1/2 × 30 kg × v² = 353.16 J

v² = 353.16 J/(1/2 × 30 kg)

v² = 23.544 m²/s²

v = √(23.544 m²/s²) ≈ 4.85 m/s

His velocity at the bottom of his path, v ≈ 4.85 m/s