Question

question 16
1 pts
a spring is hung from the edge of a students desk. the student attaches a small box of mass 2 kilograms to the bottom of the spring causing it to stretch a distance of 0.4 meters downwards. assuming the box is at rest once the spring has stretched, calculate the spring constant \k\ of the spring.
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Explanation:

Step1: Analyze the forces

When the box is at rest, the force exerted by the spring \(F_s = kx\) (Hooke's Law) is equal to the gravitational force \(F_g=mg\). So, \(kx = mg\).

Step2: Solve for \(k\)

We know \(m = 2\space kg\), \(x=0.4\space m\), and \(g = 9.8\space m/s^{2}\). Rearranging \(kx = mg\) for \(k\) gives \(k=\frac{mg}{x}\). Substitute the values: \(k=\frac{2\times9.8}{0.4}\).

$$k=\frac{19.6}{0.4}=49\space N/m$$

Answer:

\(49\space N/m\)