With a load of 200N, the dynamometer spring lengthened by 0.5 cm. How long did the spring lengthen with a load

With a load of 200N, the dynamometer spring lengthened by 0.5 cm. How long did the spring lengthen with a load of 700N? can’t figure out how to determine? force control = K delta L.

Initial data: F1 (initial load of the dynamometer spring) = 200 N; Δl1 (initial spring elongation) = 0.5 cm; F2 (final spring load) = 700 N.

The extension of the dynamometer spring under a load of 700 N is determined from the equality: k (stiffness coefficient) = F1 / Δl1 = F2 / Δl2, whence Δl2 = F2 * Δl1 / F1.

Let’s make a calculation: Δl2 = 700 * 0.5 / 200 = 1.75 cm.

Answer: With a load of 700 N, the dynamometer spring will lengthen by 1.75 cm.