Question

If the spring constant is 30 N/m and x

Answer 1

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.

• The value of the mass, m is approximately 1.53 kg.

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;

$\begin{array}{|l|cl|}\mathbf{Extension}&&\mathbf{Weight \ (Force), \, F}\\0&&0\\x&&m \cdot g\\2 \cdot x&&2 \cdot m \cdot g\end{array}\right]$

The rate of change of the extension with the applied force are;

$Between \ second \ and \ frirst\ row, \ \dfrac{\Delta F}{\Delta x} =\dfrac{m \cdot g}{x}$

$Between \ third \ row \ and \ second \ row, \ \dfrac{\Delta F}{\Delta x} = \dfrac{2 \cdot m \cdot g - m \cdot g}{2 \cdot x - x} = \dfrac{m \cdot g}{x}$

Therefore;

The rate of change of the extension with the applied force, $\dfrac{\Delta F}{\Delta x}$, 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

• $m = \dfrac{30 \, N/m \times 0.5 \, m }{9.81 \, m/s^2} \approx 1.53 \, kg$

The mass, m ≈ 1.53 kg

Learn more about Hooke's Law here:

brainly.com/question/4404276

brainly.com/question/2669908