Question

Q20... help please I need it badly

Answer 1

Answer:

stress/strain

F/A / E/L

Explanation:

2.5/A ×4/12

4A=30

A=30/4

A=7.5

Answer 2

Hey There!

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Answer:

$\huge\boxed{\textbf{F = 7.5N}}$

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$\Huge\textbf{Hooke's Law}$

$\Large\textbf{Statement:}$

"The restoring force is directly proportional to the displacement from the equilibrium position under elastic limits"

$\Large\textbf{Mathematical Expression}$

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:

$\Rightarrow K \propto -\ x \\\Rightarrow K = -\ kx\\\Rightarrow\boxed{K = -kx}$

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.

$\Large\textbf{For K:}$

Using the magnitude of force, in hooke's law,

$\rightarrow{F = Kx}\\\\\rightarrow\boxed{K=\frac{F}{x}}$

$\Large\textbf{Unit:}$

The units of K are given as,

$\rightarrow{\frac{N}{m}}\\\\\rightarrow\frac{D}{cm} \\\\\rightarrow\frac{lb}{ft}$

$\Large\textbf{Dimension:}$

The dimension of K is,

$\rightarrow\boxed{MT^{-2}}$

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$\huge\textbf{Question:}$

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 = $\frac{2.5}{4}$

$\boxed{k = 0.625}$

Use the spring constant with the extension 12cm to find the load,

F = kx

F = (0.625) x (12)

$\boxed{F = 7.5N}$

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Best Regards,

'Borz'