A 1.450 kg air-track glider is attached to each end of the track by two coil spr

Question

A 1.450 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= 0.250 m.

Find the x-coordinate of the glider at time t= 0.650T, where T is the period of the oscillation.

Find the kinetic energy of the glider at x=0.00 m.

A 1.450 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= 0.250 m. Find the x-coordinate of the glider at time t= 0.650T, where T is the period of the oscillation. Find the kinetic energy of the glider at x=0.00 m.

Explanation / Answer

c) Find the x-coordinate of the glider at time t= 0.510T, where T is the period of the oscillation Position (x) = Amplitude * cos (2 * p * frequency * time) Amplitude is the maximum distance that the object moves away from its equilibrium position = 0.250 Period = 2 * p * (m/k)^0.5 = Frequency =1/period = 1/(2p) * (k/m)^0.5 = 1/(2p) * (3?/1.450)^0.5 Position (x) = 0.250 * cos (2 * p * [1/(2p) * (3?/1.450)^0.5] * 0.510) d) Find the kinetic energy of the glider at x=0.00 m. The work done as the glider was moved = Average force * distance This work is converted into kinetic energy when the block is released. The maximum kinetic energy occurs when the glider has moved 0.250 meter back to position x = 0 As the glider is moved 0.250 meter, the average force = ½ * (0 + 0.500) Work = 0.350 * 0.250 = Kinetic energy

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