Final answer:
The force constant of the spring is 122.5 N/m.
Step-by-step explanation:
The force constant of a spring can be calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from equilibrium position. Hooke's Law can be expressed as F = -kx, where F is the force, k is the force constant, and x is the displacement. To find the force constant of the spring in this problem, we can use the equation F = kx and solve for k.
Given that the spring stretches 8.00 cm and the load is 10.0 kg, we can convert the displacement to meters (0.08 m) and use it in the equation F = kx, along with the known force (weight) of the load (F = mg, where g is the acceleration due to gravity and m is the mass).
So, the equation becomes mg = kx, where m = 10.0 kg, g = 9.8 m/s^2, x = 0.08 m. Plugging in these values, we can solve for k, which is the force constant of the spring.
k = mg/x = (10.0 kg)(9.8 m/s^2)/(0.08 m) = 122.5 N/m