Q20... help please I need it badly

stress/strain

F/A / E/L

Explanation:

2.5/A ×4/12

4A=30

A=30/4

A=7.5

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

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:

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.

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

The units of K are given as,

The dimension of K is,

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 =

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

F = kx

F = (0.625) x (12)

we need to know the time interval during which the force acts. impulse of force = momentum change

reserve is the answer