A 0.580 kg glider on an air track is attached to the end of an ideal spring with

Question

A 0.580 kg glider on an air track is attached to the end of an ideal spring with force constant 450 N/m; it undergoes simple harmonic motion with an amplitude of 4.80×10-2 m.

Part A -
Compute the maximum speed of the glider.
vmax =1.34 m/s
Correct


Part B -
Compute the speed of the glider when it is at x = -1.30×10-2 m.
v =1.29 m/s
Correct


Part C -
Compute the magnitude of the maximum acceleration of the glider.
a_max = m/s^2

Part D -
Compute the acceleration of the glider at x = -1.30×10-2 m.
a = m/s^2

Part E -
Compute the total mechanical energy of the glider at any point in its motion.
Etotal = J

Explanation / Answer

Mass m = 0.580 kg
Force constant k =450 N/m
Amplitude A = 4.80×10 -2 m.
Angular frequency = [ k / m]

                               = 27.85 rad / s
The maximum speed of the glider v max = A

                                                            = 1.34 m/s

(B )The speed of the glider when it is at x = -1.30×10 -2 m is v = A 2 - x 2 ]

                                                                                           v =1.29 m/s
(C ).The magnitude of the maximum acceleration of the glider a max = A 2

                                                                                                       = 37.22 m/s 2

(D ).The acceleration of the glider at x = -1.30×10 -2 m is a = x 2
                                                                                     a = 10.08 m/s 2

(E).The total mechanical energy of the glider at any point in its motion E = ( 1/2) m 2 A 2

                                                                                                            = 0.52 J

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