Final answer:
The force constant (k) of the spring is approximately (5.56 \times 10^5 \, \text{N/m}). This is found by calculating the acceleration from the change in velocity and using it to determine the force, then applying Hooke's Law ((F = kx)) to find (k) with the given displacement.
Step-by-step explanation:
To find the force constant k of the spring, we can use the formula F = kx, where F is the force, k is the force constant, and x is the displacement. In this case, the force is equal to the mass times the acceleration. The acceleration can be calculated using the equation vf^2 = vi^2 + 2ax, where vf is the final velocity, vi is the initial velocity, and a is the acceleration. Rearranging the equation, we have a = (vf^2 - vi^2) / (2x). Plugging in the given values, we have a = (0 - (0.500)^2) / (2 * 0.300) = -0.417 m/s^2. The negative sign indicates that the acceleration is in the opposite direction of the initial velocity. Next, we can calculate the force using F = ma. F = (4.00 × 10^5 kg) * (-0.417 m/s^2) = -1.67 × 10^5 N. Finally, we can find the force constant k by rearranging the formula F = kx to k = F / x. Using the given values, we have k = (-1.67 × 10^5 N) / (0.300 m) = -5.56 × 10^5 N/m. Since we're looking for a positive force constant, we can discard the negative sign and conclude that the force constant k of the spring is approximately 5.56 × 10^5 N/m.