Final Answer:
The maximum velocity of the 0.0100 kg mass object, when it bounces 3.00 cm above and below its equilibrium position, is (c) (0.45 m/s). Therefore the correct option is c). (0.45 m/s).
Step-by-step explanation:
To determine the maximum velocity of the object, we can use the conservation of mechanical energy, considering both kinetic and potential energy. At the equilibrium position, all the initial potential energy is converted to kinetic energy. The potential energy at the maximum displacement (3.00 cm above the equilibrium position) is then equal to the kinetic energy at the equilibrium position.
The potential energy (PE) at the maximum displacement is given by (PE = mgh), where (m) is the mass, (g) is the acceleration due to gravity, and (h) is the height. The kinetic energy (KE) at the equilibrium position is given by KE = 1/2mv², where (v) is the velocity.
Setting these two equal, we get mgh = 1/2mv². Solving for (v), we find v = √2gh. Substituting the known values m = 0.0100kg, g = 9.8 m/s², and h = 0.0300 m, we get v = √2 × 9.8 × 0.0300 ≈ 0.45 m/s.
Therefore, the maximum velocity of the object is (0.45 m/s), and the correct answer is (c) (0.45 m/s).Therefore the correct option is c). (0.45 m/s).