Hooke's Law gives the relationship between applied forces m·g, 2·m·g and extensions of an elastic material x, 2·x based on its elasticity.
Reason:
Given parameter;
Spring constant = 30 N/m
Value of the extension, x = 0.5 m
Extension of the spring by mass, m = x
Extension of the spring by mass, 2·m = 2·x
Required:
To find the value of mass m.
Solution:
The measures of weight and extension from the diagram are;
[tex]\begin{array}{|l|cl|}\mathbf{Extension}&&\mathbf{Weight \ (Force), \, F}\\0&&0\\x&&m \cdot g\\2 \cdot x&&2 \cdot m \cdot g\end{array}\right][/tex]
The rate of change of the extension with the applied force are;
[tex]Between \ second \ and \ frirst\ row, \ \dfrac{\Delta F}{\Delta x} =\dfrac{m \cdot g}{x}[/tex]
[tex]Between \ third \ row \ and \ second \ row, \ \dfrac{\Delta F}{\Delta x} = \dfrac{2 \cdot m \cdot g - m \cdot g}{2 \cdot x - x} = \dfrac{m \cdot g}{x}[/tex]
Therefore;
The rate of change of the extension with the applied force, [tex]\dfrac{\Delta F}{\Delta x}[/tex], is a
constant equal to m·g, and the spring obeys Hooke's law.
Where;
F = The spring force
Therefore;
∴ m·g = -(-k·x) = 30 N/m × 0.5 m
Where;
g = Acceleration due to gravity = 9.81 m/s²
The mass, m ≈ 1.53 kg
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