The weight of the object hung from the spring is 6.27 N.
Hooke's law;
- Hooke's law states that the load applied to an elastic material is directly proportional to the extension of the material provided elastic limit is not exceeded.
F = Kx
[tex]k = \frac{F}{x} \\\\k = F (\frac{1}{x} )\\\\k = \frac{1}{slope}[/tex]
From the graph, we can use the following points to determine the slope of the graph;
(2, 0.8) and (4, 1.5)
[tex]k = \frac{1.5 - 0.8}{4-2} \\\\k = 0.35 \\\\\frac{1}{k} = \frac{1}{0.35} = 2.85 \ N/cm[/tex]
The extension of the spring is calculated as follows;
x = 19.2 cm - 17.0 cm = 2.2 cm
The weight of the object hung from the spring is calculated as follows;
F = kx
F = 2.85 x 2.2
F = 6.27 N
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