A 0.250 kg air-track glider is attached to each end of the track by two coil spr
ID: 1580587 • Letter: A
Question
A 0.250 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.900 N to displace the glider to a new equilibrium position, x= 0.090 m.
Find the effective spring constant of the system.
1.00×101 N/m
The glider is now released from rest at x= 0.090 m. Find the maximum x-acceleration of the glider.
3.60 m/s^2
Find the x-coordinate of the glider at time t= 0.350T, where T is the period of the oscillation.
? - this is where I need help
Find the kinetic energy of the glider at x=0.00 m.
4.05×10-2 J
Explanation / Answer
1.
F = kx
k = F/x
= (0.900 N) / (0.090 m)
= 10 N/m = 1 x 101 N/m
2. a = (k/m)x
Maximum acceleration occurs when x is greatest; in this case, x_max = 0.090 m.
a = (10 N/m)(0.090 m) / (0.250 kg)
= 3.6 m/s^2
3.
x(t) = Acos(t + ), with ^2 = k/m
A = amplitude = x_max = 0.090 m
is a constant; to find it, set t = 0:
t = 0, x(t) = 0.090 m
0.090 m = (0.090 m)(cos())
cos() = 1
= 0
T = 2 (m/k)
0.350T = 0.350*2 (m/k)
t = 0.350*2 (m/k)
t = 0.350*2 (m/k) * (k/m)
= 0.70
x(0.350T) = (0.090 m)cos(0.70 + 0)
= -0.052900672 m
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