Under the action of a weight of 0.4 kg, the length of the dynamometer spring increased from L0 = 6 cm

Under the action of a weight of 0.4 kg, the length of the dynamometer spring increased from L0 = 6 cm to L1 = 10 cm. Determine the stiffness of the dynamometer spring

These tasks: m (mass of the load acting on the spring) = 0.4 kg; L0 (original length of dynamometer spring) = 6 cm = 0.06 m; L1 (length of the spring after the action of the load) = 10 cm = 0.1 m.

Constants: g (acceleration due to gravity) ≈ 10 m / s2.

The rigidity of the spring of the taken dynamometer is determined by the formula: F = m * g = -Fcont. = k * Δl = k * (L1 – L0), whence k = m * g / (L1 – L0).

Let’s perform the calculation: k = 0.4 * 10 / (0.1 – 0.06) = 100 N / m.

Answer: The spring rate of the taken dynamometer is 100 N / m.