Answer:
a. 145 N/m b. 1.29 Hz c. 1.62 m/s d. 0 m e. 13.2 m/s² f. ± 0.2 m g. 2.9 J h. 0.54 m/s i. 4.39 m/s²
Explanation:
a. The force constant of the spring
The spring force F = kx and k = F/x where k is the spring constant. F = 29.0 N and x = 0.200 m
k = 29.0 N/0.200 m = 145 N/m
b. The frequency of oscillations, f
f = 1/2π√(k/m) m = mass = 2.20 kg
f = 1/2π√(145 N/m/2.20 kg) = 1.29 Hz
c. maximum speed of the object
The maximum elastic potential energy of the spring = maximum kinetic energy of the object
1/2kx² = 1/2mv²
v = (√k/m)x where v is the maximum speed of the object
v = (√145/2.2)0.2 = 1.62 m/s
d Where does the maximum speed occur?
The maximum speed occurs at 0 m
e. The maximum acceleration
a = kx/m = 145 × 0.2/2.2 = 13.2 m/s²
f. The maximum acceleration occurs at x = ± 0.2 m
g. The total energy of the system is the maximum elestic potential energy of the system
E = 1/2kx² = 1/2 × 145 × 0.2² = 2.9 J
h. When x = x₀/3
1/2k(x₀/3)² = 1/2mv²
kx₀²/9 = mv²
v = 1/3(√k/m)x₀ = 1/3(√145/2.2)0.2 = 0.54 m/s
i When x = x₀/3
a = kx₀/3m = 145 × 0.2/(2.2 × 3)= 4.39 m/s²
