Exercises 121 E10. To stretch a spring a distance of 0.30 m from the equilib- rium position, 180 J of work is done. a. What is the increase in potential energy of the spring? b. What is the value of the spring constant \( k \) of the spring?

When you stretch a spring, the work done on it is converted into potential energy stored in the spring. In this case, the increase in potential energy is equal to the work done, so the potential energy increase is 180 J. To find the spring constant \( k \), we can use Hooke's Law, which states that the work done (W) when stretching a spring is given by the formula \( W = \frac{1}{2} k x^2 \), where \( x \) is the stretch distance. Rearranging this formula gives us the spring constant: \( k = \frac{2W}{x^2} \). By plugging in the values \( W = 180 \, J \) and \( x = 0.30 \, m \), we find \( k = \frac{2 \times 180}{(0.30)^2} = 4000 \, N/m \).