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 elastic potential energy of the spring? A) 9.10 × 10^-4 J B) 8.00 × 10^-4 J C) 7.50 × 10^-4 J D) 6.70 × 10^-4 J E) 5.00 × 10^-4 J

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Quizplus · Accepted Answer

The maximum elastic potential energy in a spring is given by the formula $E_p = \frac{1}{2}kA^2$, where $E_p$ is the elastic potential energy, $k$ is the spring constant, and $A$ is the amplitude of oscillation. However, since the initial displacement is 0 and we are given the initial velocity, we should use the conservation of mechanical energy principle. The total mechanical energy in a spring-mass system is conserved and is the sum of kinetic energy ($E_k$) and potential energy ($E_p$). At the maximum displacement, all the kinetic energy will have been converted into elastic potential energy.The kinetic energy given by the initial velocity can be calculated using $E_k = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the velocity. Substituting the given values (and converting the velocity to meters per second), we get:$$E_k = \frac{1}{2} \times 1.00 \, \text{kg} \times (0.04 \, \text{m/s})^2 = 8.00 \times 10^{-4} \, \text{J}$$This kinetic energy at the beginning will be entirely converted into elastic potential energy at the point of maximum displacement, making the maximum elastic potential energy equal to $8.00 \times 10^{-4} \, \text{J}$.

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