Question

a 2 - n force is applied to a spring, and there is displacement of 0.4 m. how much would the spring be displaced if twice the force were applied? (1 point) 4 m 0.2 m 0.8 m 0.6 m

Explanation:

Step1: Find the spring constant

According to Hooke's law \(F = kx\), where \(F\) is the force, \(k\) is the spring constant, and \(x\) is the displacement. Given \(F_1=2N\) and \(x_1 = 0.4m\), we can find \(k\) as \(k=\frac{F_1}{x_1}\).

$$k=\frac{2}{0.4}=5N/m$$

Step2: Calculate the new displacement

If the force is doubled, \(F_2 = 2F_1=4N\). Using \(F = kx\) again, we solve for \(x_2\). Since \(k = 5N/m\) and \(F_2=4N\), then \(x_2=\frac{F_2}{k}\)

$$x_2=\frac{4}{5}=0.8m$$

Answer:

\(0.8m\) (the third option)