Answer
3.1 N/cm
Work Step by Step
In Problem 13b, we found that $\Delta U_{s}$ = –0.98 J, and from the given information, we know that the spring was compressed $x_{i}$ = 8.0 cm = 0.08 m. We also know that:
$\Delta U_{s} = \frac{1}{2} k (x_{f}^{2} – x_{i}^{2})$
Since the spring is left at its relaxed length after the marble is released, $x_{f}$ = 0.
Solving for $k$ and then plugging in our values, we get:
$\Delta U_{s} = \frac{1}{2} k (x_{f}^{2} – x_{i}^{2})$
$2\Delta U_{s} = k (x_{f}^{2} – x_{i}^{2})$
$\frac{2\Delta U_{s}}{(x_{f}^{2} – x_{i}^{2})} = k$
$k = \frac{2(–0.98 J)}{(0^{2}–(0.08 m)^{2})}$ = 310 N/m or 3.1 N/cm
