if a force of 50 N stretches a spring 250 mm, what is the spring constant?

use the equation

F=-k(x)
F is the force
units:newtons (N)
K is the constant
units: newtons/meters (N/m)
x is the displacement
units: meters (m)

First you must convert the millimeters into meters
250mm=.25m
Since the spring is being pulled downward the displacement will also be negative making the ending product positive.

50=-k(-.25)
200 N/M=k

To find the spring constant, we can use Hooke's Law. Hooke's Law states that the force applied on a spring is directly proportional to the amount the spring is stretched or compressed. It can be defined as F = kx, where F is the force applied, k is the spring constant, and x is the displacement of the spring.

Given that a force of 50 N stretches the spring by 250 mm, we can substitute these values into Hooke's Law equation:

50 N = k * 250 mm

To solve for the spring constant, we need to convert the displacement from millimeters to meters (since the SI unit for force is Newtons and the SI unit for displacement is meters).

1 meter = 1000 millimeters

So, 250 mm = 250/1000 = 0.25 meters.

Now we can rewrite the equation:

50 N = k * 0.25 m

To solve for k, divide both sides of the equation by 0.25:

50 N / 0.25 m = k

k = 200 N/m

Therefore, the spring constant is 200 N/m.