Question

On the straight-line segment I in the figure we see that increasing the applied mass from 26 g to 44 g results in a reduction of the end-to-end distance from 21mm to 14 mm. What is the force constant in N/m on segment I?

Answer 1

Mass? Is that the right attachment? Because then I'm confused!

Answer 2

what?

Answer 3

In your post: "increasing the applied mass from 26 g to 44 g"
In the picture, I don't see a mass!

Answer 4

ummm

Answer 5

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Answer 6

Is there a multiple choice?

Answer 7

I think I understand it.

It's asking for a force constant in \(N/m\), meaning that you multiply it by a displacement (in meters) to find the force the nosestrip applies after it has been compressed to that displacement.

That is, how many newtons the strip applies per meter of compression.

Is that much understandable? Just that much?

Answer 8

Going on... You can ask about any part of this later,,,

That contraption, your last attachment, is something you must understand for this prob. This is how it works, ideally.
-The mass is forced down onto the nosestrip by gravity
-The mass is held still when the nosestrip force onto the mass is exactly opposite to the gravity on the mass (they cancel)
-So, the graph also shows how much force the nosestrip provides at a certain displacement.

Answer 9

Now, the graph's segment \(\text I\), given to you exactly in the question, looks straight, like a line segment.

Answer 10

Keep in mind that \(F_{gravity}\) is directly proportional to the mass, \(m\). Afterall, \(F_{gravity}=m\ g\)

Answer 11

Now, segment \(I\) is linear.

You gain this force value, then you MUST have decreased by this certain length. This slope is constant.

So, what is the slope? You have \(\Delta m\) and \(\Delta d\), you know \(\Delta F=\Delta m\ g\), what is \(\Large\frac{\Delta F}{\Delta d}\)? Look, it'd even have the right units.

By the way, the values for \(\Delta m\) and \(\Delta d\) come from the problem.

Answer 12

Post any questions!

Answer 13

The question looks like it is asking about hooke's law, which says that the force applied is proportional to the displacement.

The formula for hooke's law is F=-kx, where F is the force applied, k is the proportionality constant and x is the displacement. The minus sign means that the more force one applies, the displacement decreases.

I will check if this is indeed the case. In one of the graphs you provided, it shows that the values specified by the problem is within the range of this interval labelled as I. This segment is nearly straight, and it also says that if use a bigger weight, the x goes down, so far so good!

Now to solve it
\[F=-kx\]
Force F in this case is the weight of the object used to "crush" the springy bandage thingy.
\[F_{1}-F_{2} = -k(x_{1}-x_{2})\]

Letting F1= 26g and corresspondingly x1 = 21mm (take note of the order!)
I check what the question wants me to report the spring constant k is in, and it wants it in N/m, so I have to convert them. As an example of how I convert things so I don't get confused:

\[26g = 26g * 1 = 26g * \frac{1kg}{1000g} = \frac{26}{1000} kg\]

The 26g * 1 might seem trivial, but the next step that follows shows that it is not: I am indeed not changing it anything because I am multiplying it by "1" since 1000g = 1kg. And I multiply it in a way that the grams cancel out, leaving the kg.

Back to the question. Since weight = F = mg, \[(0.026kg - 0.044kg)*9.8/(0.021m-0.014m=-k)\]

Solving, k will be equal to 25.2N/m.
If you took proportionality as k and not -k, you will get a negative answer, but that's not how hooke's constant is defined.

The conversion of units might not be necessary here, since 1000g = 1kg and 1000mm = 1m. So the effect cancels out.