A 1.00 kg mass is connected to a spring with a spring constant of 9.00 N/m.If the initial velocity is 4.00 cm/s and the initial displacement is 0.00 cm,then what is the maximum spring potential energy of the simple harmonic motion? A) 9.10 $\times$ 10^-4 J B) 8.00 $\times$ 10^-4 J C) 7.50 $\times$ 10^-4 J D) 6.70 $\times$ 10^-4 J E) 5.00 $\times$ 10^-4 J

Question

Quizplus · Accepted Answer

The maximum spring potential energy is equal to the initial kinetic energy, which is (1/2)mv^2 = (1/2)(1.00 kg)(0.04 m/s)^2 = 8.00 * 10^-4 J.

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