Hey There!
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Answer:
[tex]\huge\boxed{\textbf{F = 7.5N}}[/tex]
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[tex]\Huge\textbf{Hooke's Law}[/tex]
[tex]\Large\textbf{Statement:}[/tex]
"The restoring force is directly proportional to the displacement from the equilibrium position under elastic limits"
[tex]\Large\textbf{Mathematical Expression}[/tex]
Consider a block on a horizontal, frictionless surface is connected to a spring. If the spring is either stretched or compressed a small distance x from its mean position, it exerts on a block a force:
[tex]\Rightarrow K \propto -\ x \\\Rightarrow K = -\ kx\\\Rightarrow\boxed{K = -kx}[/tex]
K is positive constant called the force\spring constant of the spring. Negative sign indicates that F and x always have opposite directions with reference x-axis as the direction of force is always towards mean position.
[tex]\Large\textbf{For K:}[/tex]
Using the magnitude of force, in hooke's law,
[tex]\rightarrow{F = Kx}\\\\\rightarrow\boxed{K=\frac{F}{x}}[/tex]
[tex]\Large\textbf{Unit:}[/tex]
The units of K are given as,
[tex]\rightarrow{\frac{N}{m}}\\\\\rightarrow\frac{D}{cm} \\\\\rightarrow\frac{lb}{ft}[/tex]
[tex]\Large\textbf{Dimension:}[/tex]
The dimension of K is,
[tex]\rightarrow\boxed{MT^{-2}}[/tex]
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[tex]\huge\textbf{Question:}[/tex]
DATA:
Force = F = 2.5N
Displacement = x = 4.0cm
Spring constant = k = ?
Final displacement = x = 12 cm
Load = F = ?
SOLUTION:
F = -kx
substitute the variable,
2.5 = k x 4
Rearrange the equation,
k = [tex]\frac{2.5}{4}[/tex]
[tex]\boxed{k = 0.625}[/tex]
Use the spring constant with the extension 12cm to find the load,
F = kx
F = (0.625) x (12)
[tex]\boxed{F = 7.5N}[/tex]
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Best Regards,
'Borz'
