A spring with a stiffness K of 45 lb/in is compressed by 21/64 inch. What is the resulting force in pounds?

A spring with a stiffness K of 45 lb/in is compressed by 21/64 inch. What is the resulting force in pounds?

• A. 25
• B. 14.765
• C. 60
• D. 45

Answer: (B)

Spring force follows Hooke’s law: F = kx. With a stiffness of 45 lb/in and a compression of 21/64 inch, the force is F = 45 × (21/64) = 945/64 ≈ 14.7656 pounds, about 14.77 pounds. This is the correct amount because the force scales linearly with both stiffness and displacement, so the given values yield this specific result. The other numbers would require a different product of k and x.

Spring force follows Hooke’s law: F = kx. With a stiffness of 45 lb/in and a compression of 21/64 inch, the force is F = 45 × (21/64) = 945/64 ≈ 14.7656 pounds, about 14.77 pounds. This is the correct amount because the force scales linearly with both stiffness and displacement, so the given values yield this specific result. The other numbers would require a different product of k and x.