A 0.580 kg glider on an air track is attached to the end of an ideal spring with
ID: 1450279 • Letter: A
Question
A 0.580 kg glider on an air track is attached to the end of an ideal spring with force constant 455 N/m ; it undergoes simple harmonic motion with an amplitude of 4.40×102 m .
1-
Calculate the maximum speed of the glider.
Express your answer to three significant figures.
2-
Calculate the speed of the glider when it is at x = 1.30×102 m .
Express your answer to three significant figures.
3-
Calculate the magnitude of the maximum acceleration of the glider.
Express your answer to three significant figures.
4-
Calculate the acceleration of the glider at x = 1.30×102 m .
Express your answer to three significant figures.
5-
Calculate the total mechanical energy of the glider at any point in its motion.
Express your answer to three significant figures.
Explanation / Answer
1) angular frequency , w = sqrt(k/m)
= sqrt(455/0.580)
= 28 rad/sec
=> maximum speed of the glider = A * w
= 4.40 * 102 * 28
= 1.232 m/sec
2) speed of the glider = 28 * sqrt(0.0442 - 0.0132)
= 1.177 m/sec
3) magnitude of the maximum acceleration = 0.044 * 28 * 28
= 34.496 m/sec2
4) acceleration of the glider = - 28 * 28 * 0.013
= - 10.192 m/sec2
5) total mechanical energy of the glider = 1/2 * 455 * 0.044 * 0.044
= 0.4404 J
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