What is the force of the spring constant?

[tgould43]

If a spring is originally compressed by 6cm, what is the force of the spring constant?
A block of inertia m=0.2kg is launched from a spring onto frictionless surface. The velocity is 2.5m/s upon leaving the spring.
I used Hooke's law and got
k=mg/x
0.2kg x 9.8m/s / 0.06m = 32.7 N/m
The next question is if this spring was used to suspend a 3kg block from ceiling what would the displacement of the spring be?
I need some help with the ratio of what is needed to solve this problem

 

[tgould43]

would the equation change by the new equation of
3.0kg x 9.8 ms / .06 = 490 N/m

 

[tgould43]

My mind keeps thinking that there is some relationship between the displacement of the 2.5 m/s velocity of the first block after it left the spring. Is this something that is valid?

 

[Doc Al]

I moved your posts to a separate thread.

If a spring is originally compressed by 6cm, what is the force of the spring constant?
A block of inertia m=0.2kg is launched from a spring onto frictionless surface. The velocity is 2.5m/s upon leaving the spring.

You'll need to describe the problem more carefully. Is the mass launched horizontally?

I used Hooke's law and got
k=mg/x
0.2kg x 9.8m/s / 0.06m = 32.7 N/m

That would be true if the mass were placed on a vertical spring and allowed to come to equilibrium. Is that what's going on here?

 

[Nickie]

The force of the spring constant is a measure of the stiffness of the spring, and is represented by the symbol k. It is a constant value that relates the amount of force applied to a spring to the resulting displacement or stretch of the spring. In other words, it is the force required to stretch or compress a spring by a certain distance.

In the given scenario, the force of the spring constant (k) can be calculated using Hooke's law, which states that the force exerted by a spring is directly proportional to the displacement or stretch of the spring.

Using the formula k=mg/x, where m is the mass of the block (0.2kg), g is the acceleration due to gravity (9.8m/s^2), and x is the displacement of the spring (0.06m), we can calculate the force of the spring constant to be 32.7 N/m.

To determine the displacement of the spring when used to suspend a 3kg block from the ceiling, we can use the same formula and rearrange it to solve for x. Therefore, the displacement of the spring would be x=mg/k, where m is now 3kg and g is still 9.8m/s^2.

Plugging in the values, we get x=3kg x 9.8m/s^2 / 32.7 N/m = 0.914m. This means that the spring will be stretched by 0.914m when a 3kg block is suspended from it.