Explanation:
(a) Hooke's law:
F = kx
7.50 N = k (0.0300 m)
k = 250 N/m
(b) Angular frequency:
ω = √(k/m)
ω = √((250 N/m) / (0.500 kg))
ω = 22.4 rad/s
Frequency:
f = ω / (2π)
f = 3.56 cycles/s
Period:
T = 1/f
T = 0.281 s
(c) EE = ½ kx²
EE = ½ (250 N/m) (0.0500 m)²
EE = 0.313 J
(d) A = 0.0500 m
(e) vmax = Aω
vmax = (0.0500 m) (22.4 rad/s)
vmax = 1.12 m/s
amax = Aω²
amax = (0.0500 m) (22.4 rad/s)²
amax = 25.0 m/s²
(f) x = A cos(ωt)
x = (0.0500 m) cos(22.4 rad/s × 0.500 s)
x = 0.00919 m
(g) v = dx/dt = -Aω sin(ωt)
v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)
v = -1.10 m/s
a = dv/dt = -Aω² cos(ωt)
a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)
a = -4.59 m/s²
Answer:
5.5 x 10^5 N/C
Explanation:
t = 0.001 s
Δp = - 8.8 x 10^-17 kg m /s
Force is equal to the rate of change of momentum.
F = Δp / Δt
F = (8.8 x 10^-17) / 0.001 = 8.8 x 10^-14 N
q = 1.6 x 10^-19 C
Electric field, E = F / q = (8.8 x 10^-14) / (1.6 x 10^-19)
E = 5.5 x 10^5 N/C