Question

A 5 cm spring is suspended with a mass of 1.057 g attached to it which extends the spring by 1.932 cm. The same spring is placed on a frictionless flat surface and charged beads are attached to each end of the spring. With the charged beads attached to the spring, the spring's extension is 0.286 cm. What are the charges of the beads? Express your answer in microCoulombs.

Answer 1

0.02182 μC

Explanation:

Given:

Mass (m) = 1.057 g = 0.001057 kg

Length spring (l) = 5 cm = 0.05 m

spring stretch (x) = 1.932 cm = 0.01932 m

spring stretch + beads (x1) = 0.286 cm = 0.00286 m

The first thing to calculate is the value of the spring constant k, just like this:

`\begin{gathered} F=k_sx \\ \\ F=mg \\ \\ mg=k_sx \\ \\ k_s=(mg)/(x) \\ \\ \text{ We replacing} \\ \\ k_s=((0.001057)(9.8))/((0.01932)) \\ \\ k_s=0.5361\text{ N/m} \end{gathered}`

When the charged beads are attached in equilibrium, therefore:

`\begin{gathered} (k\cdot q^2)/((l+x_1)^2)=k_s\cdot x_1 \\ \\ q^2=((k_s\cdot x_1)(l+x_1)^2)/(k) \\ \\ q=\sqrt{((k_sx_1)(l+x_1)^2)/(k)} \\ \\ \text{ We replacing:} \\ \\ q=\sqrt{(\left(0.5361\cdot0.00286\:\right)\left(0.05+0.00286\:\right)^2)/(9\cdot10^9)} \\ \\ q=2.182\cdot10^(-8)C\cdot\frac{1\text{ }\mu C}{1\cdot10^(-6)\text{ }C}=0.02182\text{ }\mu C \end{gathered}`

The charge of the beads is 0.02182 μC