The glider is now released from rest at x= 0.110 m. Find the maximum x-accelerat

Question

The glider is now released from rest at x= 0.110 m. Find the maximum x-acceleration of the glider.
2.57 m/s^2

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

4.95

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A 0.350 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= 0.110 m. Find the effective spring constant of the system. 8.18 N/m The glider is now released from rest at x= 0.110 m. Find the maximum x-acceleration of the glider. 2.57 m/s^2 Find the x-coordinate of the glider at time t= 0.650T, where T is the period of the oscillation. 4.95 times 10-2 J

Explanation / Answer

A .350 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= 0.110 m.

Force = -k * (distance stretched)
0.900 = -k *0.110
k =8.18 N/m

As the glider is moved 0.110 m, the force increases from 0 N to 0.900 N
When the glider is released, the force decrease from 0.700 N to 0 N as it moves back to the original position. Then increases from 0 N to 0.900 N, in the opposite direction, as the glider slides to the farthest position.

Max acceleration = max force

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