Answer:
(a) the mechanical energy of the system, U = 0.1078 J
(b) the maximum speed of the object, Vmax = 0.657 m/s
(c) the maximum acceleration of the object, a_max = 15.4 m/s²
Explanation:
Given;
Amplitude of the spring, A = 2.8 cm = 0.028 m
Spring constant, K = 275 N/m
Mass of object, m = 0.5 kg
(a) the mechanical energy of the system
This is the potential energy of the system, U = ¹/₂KA²
U = ¹/₂ (275)(0.028)²
U = 0.1078 J
(b) the maximum speed of the object

(c) the maximum acceleration of the object
