By how much does each spring stretch?

[minitorpedo]

Homework Statement

Recall that the spring constant is inversely proportional to the number of coils in the spring, or that shorter springs equate to stiffer springs. An object is attached to the lower end of a 14-coil spring that is hanging from the ceiling. The spring stretches by 0.160 m. The spring is then cut into two identical springs of 7 coils each. Each spring is attached between the ceiling and the object. By how much does each spring stretch?

Homework Equations

F = -kx

The Attempt at a Solution

I don't remember an equation that involves the number of coils related to the k value or the distance it stretches. I thought if you cut the spring in half that they'll each stretch to half of the distance. I don't know if that's true. Help is appreciated.

 

[Hootenanny]

Welcome to the Forums,

The big clue is here in the question;

Recall that the spring constant is inversely proportional to the number of coils in the spring

So, if a spring has 14 coils we can say that;

F\propto -\frac{1}{14}x

Therefore, if we have the number of coils in the spring;

F\propto - \frac{1}{{\color{red}?}}x

 

[NWScience]

If you produce a spring built from the same material, having the same radius etc. as the referenced spring, than the spring constant is inversely prop. to the number of coils and therefore also inversely prop. to the length of the spring.

If you half the numbers of coils, you get the doubled spring const. If you attach both new springs the force is also doubled (superposition). If you want to look at this new system of springs like being just one spring, the spring const. k is doubled twice.

F=-4k x(new)

Assuming having the same force x(new) is just a 1/4 of the original stretch (40cm)

 

[minitorpedo]

so that would be -1/7x, and then do you make them equal to each other because the force is the same? I did that and got an answer of 0.08m for each spring. Is that how you do it?

 

[minitorpedo]

yea that makes NW science. Thanks a lot.

 

[minitorpedo]

New Problem

I just have one more question. there's two parts to it and i got the first part already.

A 30.0 kg block is resting on a flat horizontal table. On top of this block is resting a 15.0 kg block, to which a horizontal spring is attached. The spring constant of the spring is 315 N/m. The coefficient of kinetic friction between the lower block and the table is 0.555, while the coefficient of static friction between the two blocks is 0.925. A horizontal force F is applied to the lower block as shown. This force is increasing in such a way as to keep the blocks moving at a constant speed.
At the point where the upper block begins to slip on the lower block determine the following. (a) the amount by which the spring is compressed. (b) the magnitude of the force F.

For part a) the magnitude by which the spring is compressed is 0.432m.
For part b) i think there are many factors: the friction of the big block with the table, the friction of the big block with the small block, the force of the spring and how it varies as the force pushes in more. I don't know how to combine all of those into an equation.

 

[Hootenanny]

so that would be -1/7x

Correct!

and then do you make them equal to each other because the force is the same? I did that and got an answer of 0.08m for each spring. Is that how you do it?

Not quite. So now you have,

F \propto -\frac{1}{x}

However, since the mass is now suspended between two springs each spring shares half the load, i.e. each spring only experiences half the force, therefore;

\frac{1}{2}F \propto - \frac{1}{7}x \Rightarrow x \propto \frac{7}{2}x

Do you follow?

Edit: It is generally not advisable to move onto the next question until you have answered the current one correctly.

 

[minitorpedo]

I don't really get that last part with the 7/2x, but i did get the question right using NWsciences way. I am not sure if that's the same thing that u were trying to explain or not

 

[Hootenanny]

I don't really get that last part with the 7/2x, but i did get the question right using NWsciences way. I am not sure if that's the same thing that u were trying to explain or not

Yes, NWscience has explained the same thing a different way (possibly more lucid). However, note that 0.08 is not the correct answer.

 

[minitorpedo]

yea i tried 0.08 the first time but got it wrong, then 0.04 and got it right. Thanks for all your help. I posted up another problem, the only one i have left out of 53 other ones. If you could help on that one too id appreciate it greatly

 

[Hootenanny]

For your second question you have calculated the compression in the spring at the point where the block begins to slip. Now, if both blocks are traveling at a constant speed, what can we say about the net forces?

 

[minitorpedo]

the net forces are zero because the blocks are not accelerating

 

[Hootenanny]

the net forces are zero because the blocks are not accelerating

Correc! Now, if we assume both of the blocks move together (as is implied by the question) we can say that the horizontal force applied to the bottom block is equal to the horizontal force applied by the bottom block on the top block, since the applied force is not greater than the maximum frictional force between the two blocks. Do you follow?

In other words, we can treat the two blocks as a single block. Now, can you write an expression for the forces acting on the block.

 

[minitorpedo]

there's the force from the spring which is F = -kx
there's the force pushing on the 45kg blocks which is just F.which can't be greater than mgu so the blocks don't slip

 

[Hootenanny]

there's the force from the spring which is F = -kx
there's the force pushing on the 45kg blocks which is just F.which can't be greater than mgu so the blocks don't slip

Yes, so mathematically;

F_{applied} - F_{spring} = 0

 

[minitorpedo]

so the Force_applied = Force_spring
and the Force of the spring is -kx
and k is 315, and is x the answer to a?

 

[Hootenanny]

so the Force_applied = Force_spring
and the Force of the spring is -kx
and k is 315, and is x the answer to a?

Yes, assuming of course that you answered part (a) correctly.

 

[minitorpedo]

i got .432 for the first part, and my k value is 315
so the force should be 136.08 as I am calculating it. i put that in but got it wrong

 

[minitorpedo]

and i got part a) correct

 

[Hootenanny]

Try rounding to 3sf.

 

[minitorpedo]

i can't it only goes to 2 decimal places

 

[Hootenanny]

3 Significant figures would be 136.

 

[minitorpedo]

oh oh i thought u meant decimal places. sorry

 

[minitorpedo]

wait if the two forces are equal, wouldn't that mean that the blocks won't move at all?

 

[Hootenanny]

wait if the two forces are equal, wouldn't that mean that the blocks won't move at all?

What is Newton's First Law of Motion?

 

[minitorpedo]

An object in motion stays in motion?

 

[Hootenanny]

An object in motion stays in motion?

Good enough; however, it is actually An object at rest will remain at rest unless acted upon by an external and unbalanced force. An object in motion will remain in motion unless acted upon by an external and unbalanced force. Since, the question states it is moving at a constant velocity, the sum of all external forces must be zero.

 

[minitorpedo]

so using this formula of Force_applied = Force_spring, then the mass of the objects has no effect?

 

[Hootenanny]

so using this formula of Force_applied = Force_spring, then the mass of the objects has no effect?

Yes, it does. Remember you used the mass of the top block to calculate the compression in the spring, which in turn was used to calculate the F_spring in the above equation.

 

[minitorpedo]

im still gettin it wrong using 136