The work done by the spring as it stretches from 14 cm to 18 cm beyond equilibrium is calculated using the formula for spring force work, resulting in 1.60 Joules.
To calculate the work W that the spring does on the object as it moves from 14 cm to 18 cm beyond its equilibrium position, we use the formula for the work done by a spring force, which is W = \(\frac{1}{2}kx_{B}^{2}\) - \(\frac{1}{2}kx_{A}^{2}\), where k is the spring constant and x_{A} and x_{B} are the initial and final displacements from equilibrium, respectively.
Given that k = 250 N/m, x_{A} = 0.14 m (14 cm), and x_{B} = 0.18 m (18 cm), we can find the work done by the spring:
W = \(\frac{1}{2} \times 250 N/m \times (0.18 m)^{2}\) - \(\frac{1}{2} \times 250 N/m \times (0.14 m)^{2}\)
W = \(\frac{1}{2} \times 250 \times 0.0324\) - \(\frac{1}{2} \times 250 \times 0.0196\)
W = 4.05 J - 2.45 J = 1.60 J
So the work done by the spring on the object when it is stretched from 14 cm to 18 cm beyond equilibrium is 1.60 Joules.