11) A block is attached to a horizontal spring, whose force constant is \( 20.0 \mathrm{~N} / \mathrm{m} \). The block-spring system is undergoing simple harmonic motion with an amplitude of 6.00 cm . What is the total energy of the system, in Joules? A. 0.036 B. 0.045 C. 0.054 D. 0.063 E. 0.072

To find the total energy of the block-spring system undergoing simple harmonic motion, we can use the formula for the total energy \( E \) in terms of the spring constant \( k \) and the amplitude \( A \): \[ E = \frac{1}{2} k A^2 \] Given: - The spring constant \( k = 20.0 \, \text{N/m} \) - The amplitude \( A = 6.00 \, \text{cm} = 0.0600 \, \text{m} \) Plugging in the values: \[ E = \frac{1}{2} \times 20.0 \, \text{N/m} \times (0.0600 \, \text{m})^2 \] \[ E = \frac{1}{2} \times 20.0 \times 0.0036 \] \[ E = 10.0 \times 0.0036 = 0.036 \, \text{J} \] Therefore, the total energy of the system is \( 0.036 \, \text{J} \). A. 0.036