Question

If a 25 kg load is applied on Spring A, it elongates 0.25 m, but it elongates 0.30 m after being attached to another Spring B in series. Determine the spring constant of both springs?

Answer 1

Let k₁ be the spring constant of spring A, k₂ be the spring constant of spring B.

From the Hook's Law

$$k_1\cdot0.25m=25kg\cdot9.81m/s^2$$

$$k_1=\frac{25kg\cdot9.81m/s^2}{0.25m}=981N/m$$

Since weights of springs are negligible, when spring A is attached to spring B, the force applied to both strings is the same. This force elongates spring A 0.25m, therefor it elongates spring B

0.30m - 0.25m = 0.05m,

$$k_2=\frac{25kg\cdot9.81m/s^2}{0.05m}=4905N/m$$

Answer: k₁ = 981N/m, k₂ = 4905N/m.