(1) The effective spring constant of the system is 7.14 N/m.
(2) The maximum x-acceleration of the glider is 0.9 m/s².
(3) The x-coordinate of the glider at time t= 0.650T is 0.28 m.
(4) The kinetic energy of the glider at x=0.00 m is 0.0175 J.
The effective spring constant of the system is calculated as follows;
F = kx
where;
k = F/x
k = 0.5/0.07
k = 7.14 N/m
a = ω²x
where;
ω = √k/m
ω = √(7.14/0.55)
ω = 3.6 rad/s
a = (3.6)² x 0.07
a = 0.9 m/s²
T = 2πx/v
T = 2πx/(ωx)
T = 2π/ω
T = 2π/(3.6)
T = 1.75 seconds
t = 0.65T
t = 0.65 x 1.75
t = 1.14 seconds
x = vt
x = (ωx)t
x = (3.6 x 0.07) x 1.14
x = 0.28 m
apply the principle of conservation of energy
Kinetic energy = work done by the spring
Kinetic energy = average force x distance
Kinetic energy = ¹/₂(0.5 N) x 0.07 m
Kinetic energy = 0.0175 J
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The complete question is below:
A 0.550 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.500 N to displace the glider to a new equilibrium position, x= 0.070 m.
1. Find the effective spring constant of the system.
2. The glider is now released from rest at x= 0.070 m. Find the maximum x-acceleration of the glider.
3. Find the x-coordinate of the glider at time t= 0.650T, where T is the period of the oscillation.
4. Find the kinetic energy of the glider at x=0.00 m.
