A small glider is placed against a compressed spring at the bottom of an air tra

Question

A small glider is placed against a compressed spring at the bottom of an air track that slopes upward at an angle of 41.0 above the horizontal. The glider has mass 9.00×102 kg . The spring has 680 N/m and negligible mass. When the spring is released, the glider travels a maximum distance of 1.30 m along the air track before sliding back down. Before reaching this maximum distance, the glider loses contact with the spring.

Part A

What distance was the spring originally compressed?

Part B

When the glider has traveled along the air track 0.300 m from its initial position against the compressed spring, is it still in contact with the spring?

Part C

What is the kinetic energy of the glider at this point?

Explanation / Answer

The energy of the compressed spring is converted into the potential energy of the glider and initial kinetic energy of the glide. As the glider slides up the track, its initial kinetic energy decreases to 0 joules as the its potential energy increases to the maximum.

So the PE of the spring = PE max of the glider


Potential energy of spring = ½ * K * d^2
k = 640
d = distance compressed
Potential energy of spring = ½ * 680 * d^2
PE max = mass * g * height
Height = total distance * sin

PE max = 0.0900 * 9.8 * (1.30 * sin 41°)

Potential energy of spring = PE max of glider

theta = 41 deg
M =9.00 * 10^-2 kg
k = 680 N/m
Vertical height travelled h = 1.30 m * sin(theta) = 1.30 m * sin(41.0 deg) = .853 m
By conservation of energy,
1/2 * k *d2 = Mgh
1/2 * 680 *d^2 = 9.00 * 10^-2 * 9.8 * .853
340 * d^2 = 0.752
d^2 = 0.69384/340 = 20.11 * 10^-4
x = sqrt(20.11 * 10^-4) = 4.48 * 10^-2 m
Ans: 4.48 * 10^-2 m

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