Question

what force is required to stretch a spring by 5 cm if its spring constant is 25 n/m?
200 n
125 n
1.25 n
not enough information to calculate.

Explanation:

Step1: Convert units

Hooke's law is \(F = kx\), where \(F\) is the force, \(k\) is the spring constant, and \(x\) is the extension. Given \(x = 5\ cm=0.05\ m\) and \(k = 25\ N/m\).

Step2: Calculate force

Substitute into the formula \(F=kx\). So \(F = 25\times0.05 = 1.25\ N\).

Answer:

1.25 N