How Is Spring Stiffness Calculated in Physics Problems?

[ohheytai]

what is the spring stiffness?!??!

In outer space a rock of mass 4 kg is attached to a long spring and swung at constant speed in a circle of radius 6 m. The spring exerts a force of constant magnitude 720 N. What is the speed of the rock?

i found the speed it is 32.8 m/s

part 2:
The relaxed length of the spring is 5.7 m. What is the stiffness of this spring?

thats what i have no idea about could someone help lead me in the direction of solving it or show me how? please and thanks!

Homework Equations

i used F=mv^2/r to get speed

 

[pat666]

Hooke's law:
F=-Kx

x is the change in length of the spring due to a force F.
K is the the spring constant or "stiffness"

 

[ohheytai]

thanks!

 

[pat666]

what are you talking about it changes from 6m to 5.7m, there is no material in the world with no elasticity.

 

[rwisz]

and F=-kx to find stiffness

The spring stiffness, also known as the spring constant, is a measure of how much force is required to stretch or compress a spring by a certain distance. It is represented by the variable k and has units of Newtons per meter (N/m). It is a characteristic property of a spring and is dependent on the material and design of the spring. The higher the spring stiffness, the harder it is to stretch or compress the spring.

To solve for the stiffness in this scenario, we can use the equation F=-kx, where F is the force exerted by the spring, x is the displacement from the relaxed length, and k is the stiffness. In this case, we know that the force exerted by the spring is 720 N and the relaxed length of the spring is 5.7 m. So, we can plug these values into the equation and solve for k:

720 N = -k(6 m - 5.7 m)

720 N = -0.3k m

k = 720 N / 0.3 m

k = 2400 N/m

Therefore, the stiffness of this spring is 2400 N/m.