what happens to the spring constant when the spring is cut in half

joej24 · Apr xxx, 2011

Homework Statement

A spring has a spring constant k. If the spring is cut into two equal parts, what is the spring abiding 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 + thou_new = k

k_new = 1000/2

Or,

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

the potential energy of the spring is cut past 4 because the length of the bound is cut by half.
0.five k ten^two = 2 * 0.five grand_new (x/2)²
k x² = 0.v* m_new 10^two
thousand = 0.5k_new
2k = m_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. And then, k_new + k_new = 1000

thou_new = 1000/2

Y'all should be able to find something more definite than a estimate that springs are analogous to circuits in this fashion.

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

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

the potential energy of the bound is cut by iv because the length of the jump is cutting by half.
0.5 1000 x² = two * 0.5 m_new (ten/2)²
k xⁱⁱ = 0.5* k_new 10²
k = 0.5k_new
2k = thousand_new

I don't know if I did this right.

Recall that 10 is how much the bound is compressed. It looks like yous are treating it as the length of the bound, in which instance it has a different value on each side of the equation.

kuruman

Suppose you hang a 100 N weight from the spring before you cutting it in one-half and the jump is in equilibrium. Say the bound stretches by 10 cm when yous hang the weight. This means that the top one-half of the spring stretches by five cm and so does the lesser one-half.

Question: How much force does the bottom one-half of the bound 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 = one thousand
The lesser one-half exerts kx force on the summit one-half. F = -kx = 1000*0.05 = l.

So, fifty Newtons.

Let's assume that y'all are correct and let'due south look at the bottom one-half of the jump. There is a down force of 100 N exerted on it past the weight. If, as y'all say, the bottom one-half exerts a force of fifty N down on the height half, then by Newton's 3rd law the meridian half will exert a force of l N upwards on the bottom one-half. Thus, the lesser half volition experience a internet strength of l N upward + 100 N downwardly = fifty N down. Practice you think that is correct? What must be true for the lesser half of the bound to be in equilibrium?

joej24

Thus, the bottom half will experience a internet forcefulness of 50 N up + 100 North downward = 50 N down. Practice you think that is right? What must be truthful for the bottom half of the spring to be in equilibrium?

No. That's non right..Sigh
Since it'southward in equilibrium, the internet force = 0. 100 North hangs down the lesser spring. The bottom half of the jump pulls up on the cake with 100 N because of the 3rd. But this doesn't bear witness us the force the bottom one-half of the spring exerts on the top half.

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

Perchance the bottom one-half of the bound pulls downward on the top one-half by 100 N?
I'thousand sort of dislocated on how to effigy this out.

joej24

I call up I got it. The weight pulls down the bottom spring past 100 N. Since all parts of the bound are in equilibrium, the bottom one-half must be in equilibrium equally well (Fnet = 0).

The meridian half of the spring pulls upwardly on the bottom half past 100 N because the weight pulls down on the bottom one-half past 100 N. By Newton'south 3rd, we find that the force the bottom one-half of the leap exert on the peak half is 100 North down.

Then, 100 N downwards.

kuruman

I think I got it. The weight pulls downwards the lesser spring past 100 N. Since all parts of the leap are in equilibrium, the lesser half must be in equilibrium as well (Fnet = 0).
The top half of the spring pulls upwardly on the bottom half by 100 N because the weight pulls downwardly on the bottom half past 100 N. By Newton's 3rd, we find that the force the lesser half of the spring exert on the meridian one-half is 100 Due north downward.

So, 100 North downward.

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

joej24

k has to exist twice as peachy as compared to the whole jump'due south spring constant since a half is stretched one-half every bit much as the entire spring and yet notwithstanding exerts the same amount of forcefulness.

kuruman

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

Correct. Congratulations, you have answered your original question. :approve:

what happens to the spring constant when the spring is cut in half

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

Homework Statement

Homework Equations

The Attempt at a Solution

Answers and Replies

The Attempt at a Solution

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