A 45.0 g object connected toa spring with a force constant of 40.0 N/m oscillate

Question

A 45.0 g object connected toa spring with a force constant of 40.0 N/m oscillates on a horizontal,frictionless surface with an amplitude of 5.00 cm. (a) Find the total energy of the system.
mJ

(b) Find the speed of the object when the positionis 1.25 cm.
m/s

(c) Find the kinetic energy when the position is 3.50 cm.
mJ

(d) Find the potential energy when the positionis 3.50 cm.
mJ (a) Find the total energy of the system.
mJ

(b) Find the speed of the object when the positionis 1.25 cm.
m/s

(c) Find the kinetic energy when the position is 3.50 cm.
mJ

(d) Find the potential energy when the positionis 3.50 cm.
mJ

Explanation / Answer

Hi, The Total energy of the simple harmonic system isalways constant and is given by                                                           E = (1/2) k A2   where 'k' is the springconstant and                                                                                               'A' is the amplitude.                                                Hence,E = 0.5 * 40 * (5 * 10-2)2                                                              = 0.05 J = 50 mJ. The speed of the object at a distance of 1.25 cm isgiven by                                                            v= (A2 - x2)                                                               =(k/m)  (A2 -x2)                                                               =(40/0.045) [(5 * 10-2)2 - (1.25* 10-2)2]                                                               =29.8 (0.0025 - 0.000156)                                                              = 1.44 m/sec. The K.E. when at a position of 3.5cm is given by                                                          K = (1/2) m v2                                                               =0.5 * 0.045 * [29.8 *  [(5 *10-2)2 - (3.5 *10-2)2]]2                                                              = 0.5 * 0.045 * 29.8 * (0.0025 - 0.000156)                                                              = 0.02 J = 20 mJ. The P.E when at a positio of 3.5cm is given by                                                          U = (1/2) k x2                                                               =0.5 * 40 * (3.5 * 10-2)2                                                               =0.03 J  = 30 mJ.                                                               =0.5 * 40 * (3.5 * 10-2)2                                                               =0.03 J  = 30 mJ.    Note: you can observe that at 3.5cmdistance, the sum of U and K is equal to E which is the energyconservation. all the values are rounded to 2 decimals. Hope this helps you. All the best.

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