Two 4.2 kg masses are connected to each other by a spring with a force constant of 28 N/m and a rest length of 1.0 m. If the spring has been compressed to 0.80 m in length and the masses are traveling toward each other at 0.42 m/s (each), what is the total energy in the system (in J)?

The total energy of the system is 1.301 J

Explanation:

The total energy in the system is due to Kinetic of two 4.2 kg masses traveling at 0.42 m/s (each)  and Elastic potential energy in the spring.

Kinetic energy = ¹/₂mv²

Kinetic energy for the two masses = 2 ( ¹/₂mv²)

                                                         = 2( 0.5 * 4.2 *0.42²)

                                                         = 0.741 J

Elastic potential energy, P.E = ¹/₂kx²

where;

k is the force constant = 28 N/m

x is the extension in the spring = 1.0 - 0.8 = 0.2 m

Elastic potential energy, P.E = ¹/₂kx² = ¹/₂* 28 *(0.2)² = 0.56 J

Total energy of the system = P.E + K.E

                                           = 0.56 J + 0.741 J

                                           = 1.301 J

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