Question

Q20... help please I need it badly

Answer 1

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=(F)/(x)}`

`\Large\textbf{Unit:}`

The units of K are given as,

`\rightarrow{(N)/(m)}\\\\\rightarrow(D)/(cm) \\\\\rightarrow(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 =
`(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'

Answer 2

stress/strain

F/A / E/L

Step-by-step explanation:

2.5/A ×4/12

4A=30

A=30/4

A=7.5