A 10-cm-long spring is attached to the ceiling. When a 2.0 kg mass is hung from it, the spring stretches to a length of 15 cm. (b) How long is the spring when a 3.0 kg mass is suspended from it?

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Hi, everyone in this particular practice problem, we're asked to actually determine the length of the spring. Then we'll have an ideal spring with an equilibrium length of centimeter. The spring is going to be suspended vertically from one of its ends. And the spring will extend to 18 centimeter. When a calibration weight of five g is attached to the spring's lower end. We're asked to find the length of the spring when an additional three g is actually added to the spring's lower end on top of the five g calibration weight. So I'm gonna start us off with actually drawing a system of our problem. So we will have here the ceiling and first, we have our original spring at equilibrium having this length right here. And I will have our new length, extended length when a calibration weight of five g is actually added. And I'm going to actually just model that as a particle here, which is exactly what we're going to do, which is assuming that our weight is a particle neglecting its dimension. So this is our y axis right here and we'll have uh our equilibrium position here originally. And we will also have our new extended final position. Sure, just like. So, so the difference between the length of the spring in the final position and in the equilibrium position is going to be delta Y which will equals to L F minus L E. From the problem statement, it is given that L F is 18 centimeters and L E is essentially 15 centimeters at equilibrium. So delta Y is thir uh three centimeters or um three times 10 to the power of -2 m just like. So, OK. So now we also know that this uh material or this particle right here will actually experience two different type of forces. First, it will experience the spring force F S. And at the same time, it will also have its own weight from the gravitational acceleration, which is M G or F G D F S is essentially just the spring constant multiplied by the um displacement or the expansion for the spring. From the problem statement, it is given that we are dealing with an ideal spring. So Hook's law is going to be applicable to our system. In order for us to solve this, we're going to use the combination of Hooks Law and also Newton's Law to the calibration weight to this particle right here in our system. In order for us to actually solve this problem, the calibration weight is subject to two forces which is the gravitational force or weight and the elastic force exerted by the spring F S. Uh the net force acting on the mass has to be zero. So Newton's uh according to Newton's law, this will then be equal to Sigma F equals to F G Plus FS Equals zero. And this will give us minus F G plus F S equals zero because the F G is pointing downward and the negative direction and F S is upwards in the positive direction. So this will then give us F G equals to F S which will be our first equation note that F S is pointing to the positive direction. And from Hooks Law Hooks Law F S is equals to minus K delta Y because we actually know in the way that we are defining our system that delta Y is going to equals to L F minus L E. And essentially, this will actually be a minus value due to how we define our Y axis. So we can determine that our Hooks Law itself is essentially going to be positive. So F S is just going to be K multiplied by delta Y. So because our um spring force is positive equals K multiplied by delta Y and the F G is just M multiplied by G. We can actually substitute those two into our equation. So this will be M G equals K delta Y. And this will give us our second equation substituted F S and substituted F G into the equation that we found from the new law. Now, we can actually rearrange this equation for us to determine the spring constant, the spring constant after rearrangement will equals to M G divided by delta Y. And we actually know all these values to find the spring constant from the first extension of the spring due to the five g of the calibration weight. So when the five g calibration weight is added, the spring is going to be extended or, or stretched with a delta Y equals to three centimeters. And from this, in uh information, we can find the spring constant. So the M is going to be the mass, which is uh the calibration mass, which is five g, which is equals to five times 10 to the power of minus three kg. And then the G is the gravitational acceleration 9.81 m per second squared. And the delta Y is the three centimeters or three times 10 to the power of minus two m. And that will give us a spring constant value of 1. newtons per meter. OK. Now that we find our spring constant, we can then calculate the total length of the spring by finding the total extension Cameed by the additional three g of the calibration weight added onto the spring's lower end. So because of this, the total mass will actually become three g plus five g, which will equals to eight g, which will equals to eight times 10 to the power of minus three kg. The total mass is now known, we can actually find the total length by finding the uh the length of the stretch by rearranging equation two. So rearranging equation two, we can find delta Y two equals to M G divided by the spring constant K. The M is going to be M taught our total mass G and the K is going to still be our spring constant found previously. And we can substitute our values into this formula right here. The total mass is eight g or eight times 10 to the power of minus three kg. The gravitational acceleration is still 9.81 m per second squared. And the K is the one that we just find which is 1. Newton for me there just like so and that will actually give us a delta Y value of 4. times 10 to the power of minus two m Which will give us a value of 4.8 cm just like. So OK. So Um this is the length caused by the additional or the elongation length caused by the additional three g. So the total length of the spring is going to be equals to the equilibrium length or L E plus the new delta Y. And that will give us a value of 5 15 centimeters plus the new delta Y which is 4.8 centimeter and that will give us a total length of 19.8 centimeters just like. So, so 19.8 centimeter is going to be the total length cost by The eight g total of mass added to the lower end of the spring. And that will correspond to option B on our answer choices. So option B is going to be the answer to our problem with the length of the spring when an additional three g is added to the spring's lower end being 19.8 centimeter. And they'll be all for this particular practice problem. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and they'll be all for this one. Thank you.

A 10-cm-long spring is attached to the ceiling. When a 2.0 kg mass is hung from it, the spring stretches to a length of 15 cm. (b) How long is the spring when a 3.0 kg mass is suspended from it?

Verified Solution

Key Concepts

Hooke's Law

Spring Constant

Equilibrium Position