A 0.650 kg air-track glider is attached to each end of the track by two coil spr

Question

A 0.650 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.700 N to displace the glider to a new equilibrium position, x= 0.210 m.

Find the effective spring constant of the system.
3.33 N/m

The glider is now released from rest at x= 0.210 m. Find the maximum x-acceleration of the glider.
1.08 m/s^2

Need help with last two.

(A)Find the x-coordinate of the glider at time t= 0.350T, where T is the period of the oscillation.

(B)Find the kinetic energy of the glider at x=0.00 m

Explanation / Answer

1. 3.57 N/m
F = kx
k = F/x
= (0.750 N) / (0.210 m)
= 3.57 N/m

2. 1.15 m/s^2
a = (k/m)x
Maximum acceleration occurs when x is greatest; in this case, x_max = 0.210 m.
a = (3.57 N/m)(0.210 m) / (0.650 kg)
= 1.154 m/s^2

Round to two sig figs.

3. 0.21 m
x(t) = Acos(?t + ?), with ?^2 = k/m
A = amplitude = x_max = 0.210 m
? is a constant; to find it, set t = 0:
t = 0, x(t) = 0.210 m
0.210 m = (0.210 m)(cos(?))
cos(?) = 1
? = 0
T = 2?? (m/k)
0.350 T = 0.350 2?? (m/k)
t = 0.350*2?? (m/k)
?t = 0.350*2?? (m/k) * ? (k/m)
= 0.700?
x(0.350 T) = (0.210 m) cos(0.700? + 0)
= 0.2098 m
Round to two sig figs

4. 0.37 J
The velocity is greatest at the equilibrium; thus, KE is greatest when x = 0.00 m. By conservation of energy:
Max PE = Max KE
Max KE = (1/2)kA^2, where A = amplitude = x_max
= (1/2)(3.57 N/m)(0.210 m)^2
= 0.375 J

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