A \( 0.20-\mathrm{kg} \) object is oscillating on a spring with a spring constant of \( k=15 \mathrm{~N} / \mathrm{m} \). What is the potertial energy of the system when the object displacement is 0.040 m , exactly half the maximum amplitude? a. zero b. 0.0060 J a. 0.012 J d. 2.5 J

To find the potential energy stored in the spring, we use the formula for potential energy in a spring, which is given by \[ PE = \frac{1}{2} k x^2 \] where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position. Given \( k = 15 \, \mathrm{N/m} \) and \( x = 0.040 \, \mathrm{m} \): \[ PE = \frac{1}{2} \times 15 \, \mathrm{N/m} \times (0.040 \, \mathrm{m})^2 = \frac{1}{2} \times 15 \times 0.0016 = 0.012 \, \mathrm{J} \] Therefore, the potential energy of the system when the object is displaced by \( 0.040 \, \mathrm{m} \) is **0.012 J**, which corresponds to option c. When it comes to oscillations, it's fascinating to consider how energy transforms between kinetic and potential forms. At maximum amplitude, all the energy is potential, while at the equilibrium position, all the energy is kinetic. This beautiful dance of energy continues endlessly as the spring oscillates!