If a spring constant doubles from 100 N/m to 200 N/m while stretched by 0.05 m, what is the new restoring force?

If a spring constant doubles from 100 N/m to 200 N/m while stretched by 0.05 m, what is the new restoring force?

• A. 5 N
• B. 10 N
• C. 15 N
• D. 20 N

Answer: (B)

Restoring force from a spring follows Hooke’s law: F = kx, where k is the spring constant and x is the displacement from equilibrium. If the displacement is 0.05 m and the spring constant is now 200 N/m, the force is F = (200 N/m)(0.05 m) = 10 N. This shows the force grows directly with the stiffness when the stretch is the same. For context, with the original stiffness the force would be 100 N/m × 0.05 m = 5 N; a stiffer spring like 300 N/m would give 15 N, and 400 N/m would give 20 N, illustrating how changing k changes the restoring force for the same stretch.

Restoring force from a spring follows Hooke’s law: F = kx, where k is the spring constant and x is the displacement from equilibrium. If the displacement is 0.05 m and the spring constant is now 200 N/m, the force is F = (200 N/m)(0.05 m) = 10 N. This shows the force grows directly with the stiffness when the stretch is the same. For context, with the original stiffness the force would be 100 N/m × 0.05 m = 5 N; a stiffer spring like 300 N/m would give 15 N, and 400 N/m would give 20 N, illustrating how changing k changes the restoring force for the same stretch.