a weight of 12 N causes a spring to stretch 3.0 cm. what is the spring comstant k of the spring?

spring constant (k)

= (force)/(deflection)
In this case, the force is the weight, 12 Newtons.

If you want the answer in Newtons per meter, change the 3.0 cm to 0.030 m before dividing.

what is the deflection 3.0

F = k x like he said

Yes, deflection x = 3.0 cm
so
k = 12/3 Newtons / centimeter
which is of course 4 N/cm

Most of us would do this in meters rather than cm but I see in your other question posted N/cm were used so I am leaving it that way.

To find the spring constant (k) of a spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.

Hooke's Law: F = -kx

Where:
F = Force exerted by the spring (in Newtons)
k = Spring constant (in N/m)
x = Displacement from the equilibrium position (in meters)

In this case, the weight of 12 N is causing the spring to stretch by 3.0 cm (or 0.03 m). By substituting these values into Hooke's Law, we can solve for the spring constant (k).

12 N = -k * 0.03 m

To find k, we need to rearrange the equation:

k = -12 N / 0.03 m

k = -400 N/m

Therefore, the spring constant (k) of the spring is -400 N/m.

Note: The negative sign indicates that the force exerted by the spring is opposite to the displacement.