Question

current attempt in progress
a 1.1 - kg object is suspended from a vertical spring whose spring constant is 160 n/m. (a) find the amount by which the spring is stretched from its unstrained length. (b) the object is then pulled straight down by an additional distance of 0.27 m and released from rest. find the speed with which the object passes through its original position on the way up.
(a) number
(b) number

Explanation:

Step1: Find the stretch in part (a)

When the object is in equilibrium, the force due to gravity \(F = mg\) is balanced by the spring force \(F = kx\). So, \(mg=kx\).
We know \(m = 1.1\space kg\), \(g = 9.8\space m/s^{2}\), \(k=160\space N/m\).

$$x=\frac{mg}{k}$$

$$x=\frac{1.1\times9.8}{160}$$

$$x=\frac{10.78}{160}=0.0674\space m$$

Step2: Use conservation of mechanical energy for part (b)

The initial mechanical energy \(E_{1}\) when the object is pulled down by \(A = 0.27\space m\) (relative to the equilibrium position) is the elastic potential energy \(E_{1}=\frac{1}{2}kA^{2}\) (since \(v = 0\) initially).
When the object passes through the original (equilibrium) position, the mechanical energy \(E_{2}=\frac{1}{2}mv^{2}\) (elastic potential energy at equilibrium is zero relative to the pulled - down position).
By conservation of energy \(E_{1}=E_{2}\)

$$\frac{1}{2}kA^{2}=\frac{1}{2}mv^{2}$$

$$v = A\sqrt{\frac{k}{m}}$$

Substitute \(A = 0.27\space m\), \(k = 160\space N/m\), \(m=1.1\space kg\)

$$v=0.27\sqrt{\frac{160}{1.1}}$$

$$v = 0.27\sqrt{145.45}$$

$$v=0.27\times12.06=3.26\space m/s$$

Answer:

(a) \(0.0674\space m\)
(b) \(3.26\space m/s\)