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
![V_{max} =\omega*A= \sqrt{\frac{K}{M} } *A\\\\V_{max} = \sqrt{\frac{275}{0.5} } *0.028\\\\V_{max} = 0.657 \ m/s](https://tex.z-dn.net/?f=V_%7Bmax%7D%20%3D%5Comega%2AA%3D%20%20%5Csqrt%7B%5Cfrac%7BK%7D%7BM%7D%20%7D%20%2AA%5C%5C%5C%5CV_%7Bmax%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B275%7D%7B0.5%7D%20%7D%20%2A0.028%5C%5C%5C%5CV_%7Bmax%7D%20%3D%200.657%20%5C%20m%2Fs)
(c) the maximum acceleration of the object
![a_{max} = \frac{KA}{M} \\\\a_{max} = \frac{275*0.028}{0.5}\\\\a_{max} = 15.4 \ m/s^2](https://tex.z-dn.net/?f=a_%7Bmax%7D%20%3D%20%5Cfrac%7BKA%7D%7BM%7D%20%5C%5C%5C%5Ca_%7Bmax%7D%20%3D%20%5Cfrac%7B275%2A0.028%7D%7B0.5%7D%5C%5C%5C%5Ca_%7Bmax%7D%20%3D%2015.4%20%5C%20m%2Fs%5E2)