What is the spring constant of both of the new springs?

[joej24]

Homework Statement

A spring has a spring constant k. If the spring is cut into two equal parts, what is the spring constant of both of the new springs?

Homework Equations

The Attempt at a Solution

I was wondering if this sort of problem relates to circuits. A spring cut into two is sort of like a parallel circuit. SO, k_new + k_new = k

k_new = k/2

Or,

By conservation of energy, the potential energy is conserved when the springs are in equilibrium.

the potential energy of the spring is cut by 4 because the length of the spring is cut by half.
0.5 k x² = 2 * 0.5 k_new (x/2)²
k x² = 0.5* k_new x²
k = 0.5k_new
2k = k_new

I don't know if I did this right.

 

[Fewmet]

The Attempt at a Solution

I was wondering if this sort of problem relates to circuits. A spring cut into two is sort of like a parallel circuit. SO, k_new + k_new = k

k_new = k/2

You should be able to find something more definite than a guess that springs are analogous to circuits in this way.

Were you given or did you derive an equation for the total k of two connected springs?

Or,

By conservation of energy, the potential energy is conserved when the springs are in equilibrium.

the potential energy of the spring is cut by 4 because the length of the spring is cut by half.
0.5 k x² = 2 * 0.5 k_new (x/2)²
k x² = 0.5* k_new x²
k = 0.5k_new
2k = k_new

I don't know if I did this right.

Recall that x is how much the spring is compressed. It looks like you are treating it as the length of the spring, in which case it has a different value on each side of the equation.

 

[kuruman]

Suppose you hang a 100 N weight from the spring before you cut it in half and the spring is in equilibrium. Say the spring stretches by 10 cm when you hang the weight. This means that the top half of the spring stretches by 5 cm and so does the bottom half.

Question: How much force does the bottom half of the spring exert on the top half?

 

[joej24]

Since F = -kx
100 = k(0.10)
1000 = k

The bottom half exerts kx force on the top half. F = -kx = 1000*0.05 = 50.

So, 50 Newtons.

 

[kuruman]

Since F = -kx
100 = k(0.10)
1000 = k

The bottom half exerts kx force on the top half. F = -kx = 1000*0.05 = 50.

So, 50 Newtons.

Let's assume that you are correct and let's look at the bottom half of the spring. There is a down force of 100 N exerted on it by the weight. If, as you say, the bottom half exerts a force of 50 N down on the top half, then by Newton's 3rd law the top half will exert a force of 50 N up on the bottom half. Thus, the bottom half will experience a net force of 50 N up + 100 N down = 50 N down. Do you think that is correct? What must be true for the bottom half of the spring to be in equilibrium?

 

[joej24]

Thus, the bottom half will experience a net force of 50 N up + 100 N down = 50 N down. Do you think that is correct? What must be true for the bottom half of the spring to be in equilibrium?

No. That's not right..Sigh
Since it's in equilibrium, the net force = 0. 100 N hangs down the bottom spring. The bottom half of the spring pulls up on the block with 100 N because of the 3rd. But this doesn't show us the force the bottom half of the spring exerts on the top half.

The forces that the bottom and top half of the spring exert on each other cancel each other out.

Perhaps the bottom half of the spring pulls down on the top half by 100 N?
I'm sort of confused on how to figure this out.

 

[joej24]

I think I got it. The weight pulls down the bottom spring by 100 N. Since all parts of the spring are in equilibrium, the bottom half must be in equilibrium as well (Fnet = 0).

The top half of the spring pulls up on the bottom half by 100 N because the weight pulls down on the bottom half by 100 N. By Newton's 3rd, we find that the force the bottom half of the spring exert on the top half is 100 N downward.

So, 100 N downward.

 

[kuruman]

I think I got it. The weight pulls down the bottom spring by 100 N. Since all parts of the spring are in equilibrium, the bottom half must be in equilibrium as well (Fnet = 0).

The top half of the spring pulls up on the bottom half by 100 N because the weight pulls down on the bottom half by 100 N. By Newton's 3rd, we find that the force the bottom half of the spring exert on the top half is 100 N downward.

So, 100 N downward.

Correct. So each half is stretched by 5 cm and yet exerts the same force as the whole spring that is stretched by 10 cm (twice that much). What does this imply about the spring constant of each half as compared with the spring constant of the whole spring?

 

[joej24]

k has to be twice as great as compared to the whole spring's spring constant since a half is stretched half as much as the entire spring and yet still exerts the same amount of force.

 

[kuruman]

k has to be twice as great as compared to the whole spring's spring constant since a half is stretched half as much as the entire spring and yet still exerts the same amount of force.

Correct. Congratulations, you have answered your original question.