Final answer:
The spring constant is found using the potential energy formula for a spring. Given that the spring stores 42.0 J of potential energy at a displacement of 7.20 cm, the spring constant is calculated to be approximately 16195 N/m.
Step-by-step explanation:
To find the spring constant, we can use the formula for the potential energy stored in a compressed or stretched spring: \(U = \frac{1}{2}kx^2\), where \(U\) is the potential energy, \(k\) is the spring constant, and \(x\) is the displacement from the equilibrium position. In this case, the potential energy \(U\) is given as 42.0 J, and the displacement \(x\) is 7.20 cm, which we convert to meters (since the spring constant \(k\) will be in units of N/m) as 0.072 m. Rearranging the formula to solve for \(k\), we get \(k = \frac{2U}{x^2}\).
Now we can plug in the given values:
\(k = \frac{2 \times 42.0 J}{(0.072 m)^2} \approx \frac{84.0}{0.005184} N/m \approx 16194.57 N/m\).
Therefore the spring constant for the given spring is approximately 16195 N/m.