Hooke's Law gives the relationship between applied forces <em>m·g</em>, <em>2·m·g</em> and extensions of an elastic material <em>x</em>, <em>2·x</em> based on its elasticity.
- The value of the mass, <em>m</em><em> </em>is approximately <u>1.53 kg</u>.
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 <em>m</em>.
Solution:
The measures of weight and extension from the diagram are;
![\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]](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7B%7Cl%7Ccl%7C%7D%5Cmathbf%7BExtension%7D%26%26%5Cmathbf%7BWeight%20%5C%20%28Force%29%2C%20%5C%2C%20F%7D%5C%5C0%26%260%5C%5Cx%26%26m%20%5Ccdot%20g%5C%5C2%20%5Ccdot%20x%26%262%20%5Ccdot%20m%20%5Ccdot%20g%5Cend%7Barray%7D%5Cright%5D)
The rate of change of the extension with the applied force are;


Therefore;
The rate of change of the extension with the applied force,
, is a
constant equal to m·g, and the spring obeys Hooke's law.
- According to Hooke's law, force applied to the spring, F = -K·x
Where;
F = The spring force
Therefore;
- The force applied by the weight of the mass, m·g = -F
∴ m·g = -(-k·x) = 30 N/m × 0.5 m
Where;
g = Acceleration due to gravity = 9.81 m/s²
The mass, m ≈ <u>1.53 kg</u>
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