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Finding the stiffness of a spring

Lessons 25 · 1 lessons · P.Q. Habibullayev, A. Boydedayev, A.D. Bahromov, S.O. Burxonov. Physics Grade 7, revised and enlarged 4th edition. “O‘zbekiston milliy ensiklopediyasi” State Scientific Publishing House, Tashkent, 2017
25

Finding the stiffness of a spring (Lab 2)

Textbook: pp. 93–96
GoalLab 2: determine the stiffness of a spring experimentally using k = F/Δl.
New words
spring · prujinaload (weight) · yukelongation Δl · uzayish Δlaverage value · o‘rtacha qiymat
Explanation

The aim is to find the stiffness k of a spring. Hang the spring on a stand and measure its length l₀ without a load. Then hang loads of known mass m one by one, measure the length l each time and find the elongation Δl = l − l₀. When the load is in equilibrium the elastic force equals the weight force: F = mg (taking g = 10 m/s², a 100 g load gives F ≈ 1 N). Stiffness is calculated by k = F/Δl; use 3–4 loads and take the average. The elongation growing in proportion to the load confirms Hooke’s law. Work under the teacher’s supervision and put a soft pad under the stand in case a load falls.

Worked examples
Load 0.2 kg: F = 0.2 · 10 = 2 N, l₀ = 10 cm, l = 14 cm. Δl = 4 cm = 0.04 m, k = 2 : 0.04 = 50 N/m.
Another load of 0.4 kg: F = 4 N, Δl = 8 cm = 0.08 m, k = 4 : 0.08 = 50 N/m; the same result, so Hooke’s law holds.
Class activity

Fill the table: № | m (kg) | F = mg (N) | l₀ (m) | l (m) | Δl (m) | k (N/m). Compute k for three loads and average them. Safety: do not hold loads above your feet, use no load heavier than 0.5 kg, never overstretch the spring.

Practice
1
Which formula gives the stiffness in this lab?
2
A 0.3 kg load stretches a spring by 6 cm. Find F (N) and k (N/m) (g = 10 m/s²).
3
Three trials gave k = 48, 52, 50 N/m. Average k (N/m)?
4
Why does the elastic force equal mg when the load is in equilibrium?