Question

An ideal spring of negligible mass is 11.00 cm long when nothing is attached to it. When you hang a 3.10-kg weight from it, you measure its length to be 12.40 cm . Part A If you wanted to store 10.0 J of potential energy in this spring, what would be its total length

Answer 1

Answer: Total length is 15.7 centimeters.

Explanation:

The potential energy is modelled by applying Principle of Energy Conservation and Work-Energy Theorem:

$U_{2} - U_{1} = \Delta W\\\frac{1}{2} \cdot k \cdot (x_{2}^2-x_{1}^2) = \Delta W$

The spring constant can be found by using Statics:

$F_{spring} - W = 0\\k \cdot \Delta s - m \cdot g = 0$

Where $g = 9.81 \frac{m}{s^{2}}$.

$k =\frac{m \cdot g}{\Delta s}$

$k = \frac{(3.1 kg)\cdot (9.81 \frac{m}{s^{2}} )}{0.014 m}$

$k = 2172.21 \frac{N}{m}$

The total length is isolated from energy expression:

$x_{2}=\sqrt{x_{1}^{2}+ \frac{2 \cdot \Delta W}{k}}$

$x_{2}=\sqrt{(0,124 m)^{2}+\frac{2\cdot10J}{2172.21\frac{N}{m} } } \\x_{2}=0.157 m (15.7 cm)$