150 N compresses a spring by 0.2 m
- Force F = 150 N, displacement Δx = 0.2 m.
- k = F ÷ Δx = 150 ÷ 0.2 = 750 N/m.
- Interpretation: this is a moderately stiff spring suitable for heavier loads or suspensions.
science calculator
Determine spring stiffness (k) from applied force and displacement (Hooke’s law).
For a linear spring, Hooke’s law states F = kΔx, where F is force, k is spring constant, and Δx is displacement from the rest position.
We rearrange this relationship to solve for stiffness: k = F ÷ Δx.
By entering force in newtons and displacement in meters, we compute k in newtons per meter (N/m), which is the standard SI unit for spring stiffness.
k = F / Δx
Calculate spring constant k from force and displacement using Hooke’s law for quick lab, engineering, or DIY checks.
Enter force in newtons and displacement in meters to see stiffness in N/m and compare springs easily across projects.
Ideal for physics labs, suspension design, vibration isolation, and any situation where you need a fast stiffness estimate.
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Valid only within elastic range. Real springs may deviate when overloaded.