Question

1. (5pts) a mass, ( m = 0.002 mathrm{~kg} ), is attached to an initially unstretched, vertically oriented spring of spring constant ( k = 2 mathrm{~kg} / mathrm{s}^{2} ). if the mass is attached to the spring and allowed to come to rest, how far will it stretch? remember that the force due to gravity is ( m g ) and the magnitude of the spring force is ( k x ). use ( g = 10 mathrm{~m} / mathrm{s}^{2} ) for the magnitude of the gravitational acceleration.

Explanation:

Step1: Analyze the forces

When the mass is in equilibrium, the spring force \(F_s = kx\) is equal to the gravitational force \(F_g=mg\). So, \(kx = mg\).

Step2: Solve for \(x\)

We need to find \(x\), and from \(kx = mg\), we can express \(x=\frac{mg}{k}\).
Given \(m = 0.002\space kg\), \(g = 10\space m/s^{2}\), and \(k = 2\space kg/s^{2}\).
Substitute the values into the formula: \(x=\frac{0.002\times10}{2}\).
First, calculate \(0.002\times10=0.02\).
Then, \(\frac{0.02}{2}=0.01\space m\).

Answer:

The spring will stretch \(0.01\space m\).