It takes 1200 J of work to stretch a spring from its natural length of 1 m to a length of 6 m . Find the force constant of the spring. The spring's force constant is \( \square \mathrm{N} / \mathrm{m} \). (Type an integer or a simplified fraction.)

1. The work done to stretch a spring is given by \[ W = \frac{1}{2} k (\Delta x)^2, \] where \( W \) is the work, \( k \) is the force constant, and \( \Delta x \) is the extension of the spring. 2. Here, the spring is stretched from 1 m to 6 m, so the extension is \[ \Delta x = 6 - 1 = 5 \, \text{m}. \] 3. Substitute the given work \( W = 1200 \, \text{J} \) and \( \Delta x = 5 \, \text{m} \) into the formula: \[ 1200 = \frac{1}{2} k (5)^2. \] 4. Simplify the equation: \[ 1200 = \frac{1}{2} k \times 25. \] \[ 1200 = \frac{25}{2} k. \] 5. Solve for \( k \) by multiplying both sides by \( \frac{2}{25} \): \[ k = 1200 \times \frac{2}{25}. \] \[ k = \frac{2400}{25}. \] 6. Simplify the fraction: \[ k = 96 \, \text{N/m}. \] The spring's force constant is \( 96 \, \text{N/m} \).