By how much is the energy stored in a Hooke's law spring increased when its stretch is increased from 8.00 cm to 16.0 cm ? 1) \( 100 \% \) 2) \( 200 \% \) 3) \( 300 \% \) 4) \( 400 \% \) 5) The correct answer is not given.

To determine the increase in energy stored in a Hooke's law spring when its stretch changes, we can use the formula for elastic potential energy, \( U = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the stretch. First, calculate the energy at 8.00 cm (0.08 m): \[ U_1 = \frac{1}{2} k (0.08)^2 = 0.0032 k \] Now, calculate the energy at 16.0 cm (0.16 m): \[ U_2 = \frac{1}{2} k (0.16)^2 = 0.0128 k \] To find the increase in energy: \[ \Delta U = U_2 - U_1 = 0.0128 k - 0.0032 k = 0.0096 k \] To find the percentage increase, calculate: \[ \text{Percentage Increase} = \left( \frac{\Delta U}{U_1} \right) \times 100\% = \left( \frac{0.0096 k}{0.0032 k} \right) \times 100\% = 300\% \] The answer is: 3) \( 300 \% \)