A 2.0-kg mass is attached to one end of a spring with a spring constant of 100 N/m and a 4.0-kg mass is attached to the other end.The masses are placed on a horizontal surface and the spring is compressed 10 cm.The spring is then released with the masses at rest and the masses oscillate.When the spring has its equilibrium length for the first time the 2.0-kg mass has a speed of 0.36 m/s.The mechanical energy that has been lost to this instant is:
A)0 J
B)0.31 J
C)0.61 J
D)0.81 J
E)1.2 J

Question

Quizplus · Accepted Answer

The mechanical energy lost is equal to the work done by friction, which is equal to the force of friction times the distance moved. The force of friction is equal to the coefficient of friction times the normal force, which is equal to the weight of the masses. The distance moved is equal to the amplitude of the oscillation, which is equal to the initial compression of the spring. Therefore, the mechanical energy lost is equal to:$$W_f=\mu mgd=(0.1)(6)(0.1)=0.06J$$

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