A 0.250 kg air-track glider is attached to each end of the track by two coil spr

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= 0.150 m Eind the effective spring constant of the system. 6.00 N/m are correct. Your receipt no. is 160-7839Previous Tries The glider is now released from rest at x= 0.150 m. Find the maximum x-acceleration of the glider. 3.60 m/s 2 Your are poerne tis 160-2472 Previous Tries Your receipt no. is 160-2472eious ms Find the x-coordinate of the glider at time t- 0.410T, where T is the period of the oscillation. Calculate the angular velocity and the period, then the time at which you must find the position. Submit Answer Incorrect. Tries 1/12 Previous Tries Find the kinetic energy of the glider at x-0.00 m. Submit Answer Tries 0/12

Explanation / Answer

3. x(t) = Acos(t + ), with ^2 = k/m
A = amplitude = x_max = 0.150 m
is a constant; to find it, set t = 0:
t = 0, x(t) = 0.150 m
0.150 m = (0.150 m)(cos())
cos() = 1
= 0
T = 2 (m/k)
0.410T = 0.410*2 (m/k)
t = 0.410*2 (m/k)
t = 0.410*2 (m/k) * (k/m)
= 0.820
x(0.410T) = (0.150 m)cos(0.820 + 0)
= -0.15 m

4. 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)(6 N/m)(0.150 m)^2
= 0.0675 J

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