Lesson video

Hello.

This lesson is an analysis of the stretching a spring experiment.

It's part of the physics unit energy of moving objects and my name is Mr. Fairhurst.

By the end of this lesson you should be able to describe and calculate the properties of the spring using hooke's law.

These are the key words that you're going to come across during the lesson.

If at any point you want to come back to this slide to check the definitions, then just pause the video and do just that.

In the lesson, we're gonna start by looking at what we mean by elastic and inelastic defamation.

We're then going to look at how we can calculate the supreme constant from a set of results and then we're going to finish by doing some calculations to find out various properties of different springs.

Okay, so let's make a start with the first part.

What I've got here is a metre ruler that's being clamped to a desk and one end that's sticking over.

The desk is free to be pulled up and down and if we bend that ruler it will spring back to its same place as it is now.

So we can add a weight, take the weight off and it will spring back to its same position as it was before.

We say that the ruler has undergone elastic defamation.

It's like an elastic band will spring back to the same shape it was before when you take away the force because it's elastic.

So that type of defamation is elastic defamation.

If the force is too big, however, then the ruler may undergo what we call inelastic defamation.

If you remember the two letters in the perceiver word, the prefix word, they mean the opposite.

So invisible is the opposite of visible.

You can't see it.

Inelastic is the opposite of elastic, so the ruler has stopped being elastic and if we take the weight off now, then the ruler will not return to its original shape.

It will remain a little bit bent.

Have a look at this question and see what you think.

A tennis ball undergoes elastic defamation as it bounces.

Is that true or false? Pause the video whilst you choose your answer and also choose part A or B to justify your answer.

Okay, so what did you think? It does undergo elastic defamation and the correct reason is because it shaped changes but only for a moment.

Okay, so well done if you got that right.

We've now got a second ruler and we can use that to measure how much this metre ruler bends when different forces are applied.

We've held that 30 centimetre of vertical in a clamp and stand.

Now you can get it exactly vertical by using a plumb line which is a heavy weight on the thread which will hang exactly vertical when you hold it at one end and we can gauge whether the 30 centimetre rule is exactly vertical by comparing it to the thread.

We've got zero at the top, which is exactly in line with the ruler when there's no forces added so that any reading we take from the ruler will measure the defamation of the ruler, but it will measure how far that ruler has bent.

In this case, that's 3.

7 centimetres and we're measuring from the top edge of the ruler.

This is a graph that shows all of the results for the ruler up to a force of 10 newtons applied and for each of these measurements, when we took the weights off, the ruler went back to its original shape.

It was not permanently deformed, it was elastic defamation.

And as you can see, all of those points are exactly in line in a straight line.

So we can say that the amount of bend is directly proportional to the force with applied.

So for example, if we choose a force of five newtons and look at the bend of the ruler, we find that was about 4.

5 centimetres.

Because it's directed proportion, if we use double the force 10 newtons, we would expect that to have twice the bend of the ruler and in this case that's nine Newtons, which which is exactly double 4.

5.

So the defamation is directly proportional to the force applied.

Okay, have a look at this question.

What happens to the amount a ruler bends elastically if the force bending it is made three times bigger? Pause the video why you choose your answer and then start again once you're ready.

Okay, so how did you get on? The correct answer is it bends exactly three times as far, you've got three times the force and the force is directly proportional to the amount it bends.

So if the force increases three times, so does the amount that it bends.

Well done if you got that right.

Now in this case we've applied even more force to the ruler and when we've removed the force this time the ruler has been permanently deformed, it's undergone inelastic defamation.

If we start with the same 10 points that we started with before, which we know are elastic defamation, if we extend that line further, that's what it would look like if the bend of the other for those extra points was directly proportional to the force, but clearly they are bending away from that line.

So we've now got a defamation that is not proportional to the force that's supplied and that happens when we've got inelastic defamation.

Have a look at this question, these three graphs here, which of these show a ruler that has been permanently changed? Pause the video while you choose your answer and start it again once you are ready.

Okay, how did you get on? If the shape of the ruler has been permanently changed, then the bend will not be directly proportional to the force.

So the graph will not be a perfect straight line through zero and the only graph that does that is graph C.

So well done if you chose that one.

What I'd like you to do now is to have a go at this task.

We've got three graphs that show how three different rulers bend when the force is applied, all of the graphs are drawn to the same scale.

There's two questions for you to answer.

Pause the video whilst you have a go at the task and start it again once you're ready with your answers.

Right then, so how did you get on? The first question was to state which ruler is the easiest to bend and explain how its graph shows that it's easiest and the answer is graph B and that's because for each small increase in force, each equal increase in force, the ruler bends by a greater amount, so decline goes up more steeply.

So for the same size force the bend is bigger.

Question two, ask you to state which rule is bent outta shape and explain again how its graph shows that it's been bent outta shape.

And the answer there was graph C.

The graph line curves up at the end which shows that the ruler is undergoing inelastic defamation and that its shape will be permanently changed when you take away the force.

It will keep some bend in the ruler when the force is removed.

So well done if you've got both of those correct.

We're now going to move on to the second part of the lesson and have a look at the spring constant.

This graph shows the extension of the spring, which is directly proportional to the force and that shows us that the spring has not passed what we call its limit of proportionality.

It's undergoing elastic defamation and the extension is directly proportional to the force that we're applying and we've got a straight line going through the origin that shows that this is the case.

Now each spring has got a spring constant.

This is a particular value for each spring and the spring constant will describe how hard it is to stretch a spring during elastic defamation.

If we've got a bigger spring constant, we're going to need more force to stretch a spring by a certain amount.

The greater the spring constant, the more needed to either stretch or indeed to compress a spring.

So a lot of springs can be stretched, some springs can also be squashed and made shorter.

So we call that compressing.

Now in this example, spring B has got a greater spring constant as spring A is harder to stretch, but we can see that because we've added the same size force to both springs, they both started at the same length, but spring A has been extended further than spring B, it's easier to extend.

So it's spring constant is smaller and spring b's spring constant is bigger because it's harder to stretch.

This is the graph for spring A, and this is the graph for spring B.

For the same sized force we can see clearly that the extension of spring B on the graph is smaller than spring A because it's harder to stretch.

It's got a bigger spring constant.

Have a look at this question.

Which of the springs represented by these graphs has got the greatest, the biggest spring constant? Pause the video whilst you work out your answer and start it again once you ready.

Okay, so how did you get on the greatest spring constant is for the spring, that's the hardest to stretch.

It's going to be the one where for the same size force it's got the smallest extension because it's hardest to stretch and the correct answer is graph A.

So well done if you chose that one.

Have a look at this graph.

The spring constant has got the symbol k, lowercase letter k.

It's measured in Newtons per metre.

The number of Newtons needed to stretch a spring by one metre.

We can work it out from the graph at three newtons.

It's got an extension of north 0.

125 metres and we can use that to calculate the supreme constant.

It's three newtons divided by the amount that it's been extended, which gives us a spring constant of 24 newtons per metre.

The spring constant is the same for any pair of measurements for the same spring.

So if we take a different measurement say for the same spring of five Newtons, the spring constant here, if you look at the graph scale is not 0.

21 metres.

We do the sums again, five newtons divided by 0.

21 metres and we get an answer of k equals 23.

8 newtons per metre, which if we round up to two significant figures will give us again 24 newtons per metre.

So we've got the same screen content no matter which pair of values we use to calculate it.

Have a look at this question, work out your answer and pause the video while you do so and start it again once you're ready.

Okay, so how did you get on 14 Newtons is needed to stretch a spring by 0.

05 metres.

What's the spring constant of that spring? We've got a force of 40 newtons divided by an extension of nor point nor five metres and that gives us an answer of 800 Newtons per metre.

Well done if you've got the right answer there.

I'd now like you to have a go at task B.

I'd like you to calculate the spring content for this spring represented by the graph and give your answer in Newton's per metre.

Pause the video while you do that and start it again once you're ready.

Okay, so how did you get on? You asked to calculate the spring content represented by this graph so you can take any pair of values.

I've taken 10 newtons because it's easy to work with, and that the extension here is 8.

4 centimetres.

So you've got a force of 10 Newtons extending by 8.

4 centimetres.

But you were asked if you remember to calculate this in Newtons per metre.

So we need to change 8.

4 centimetres into metres, so that's north 0.

084 metres.

We've divided the the centimetres by 100.

We can then do the calculation k is equal to 10 newtons divided by North 0.

084 metres, which gives an answer of 119.

0476 nuisance per metre.

If we look at the question the extension of the spring was to two significant figures and the fourth was to three.

We've got to give our answer to the smallest number of significant figures of any of the values we've used.

So we'll give our answer to two significant figures.

So we've rounded that up to 120 newtons per metre.

So well done if you've got the right answer there.

The final part of the lesson is about stretching calculations.

We want to use those ideas that we've developed so far to do some calculations.

We can calculate the spring concept by dividing the fourth stretching of spring by the extension to spring.

The spring concept K equals the fourth f divided by the extension x and if we multiply both sides by the extension, we get spring constant times the extension equals force divided by the extension times by the extension and on the right hand side both of those extensions cancel out and then we can swap sides and we get the force is equal to the spring constant times the extension.

That's known as hooke's law.

Hooke's law says that the extension of the spring of any elastic object is directly proportional to the force applied but only if the limited proportionality is not being exceeded.

In other words, hooke's law works if we haven't stretched an object outta shape we can write it down as using the equation that we've just mentioned.

Force is spring constant times the extension where the force is measured in Newton's, the extension is measured in metres and the spring constant is measured in Newtons per metre.

Let's have a look at some calculations.

A force of 16.

0 Newtons compresses a spring of length 8.

0 centimetres to a shorter length of 6.

0 centimetres.

Calculate the spring constant k.

While the extension is going to be how much longer the spring is than its original length, which is two centimetres which we need to change to metres because the spring constant is measured in Newtons per metre.

So we've got the extension is no 0.

020 metres, I'm being careful here to write it down with the correct number of significant figures.

Force is spring constant times extension so we can put in the numbers that we don't know.

16 new terms equals a spring constant times 0.

020.

We can divide both sides by the 0.

020 to get the spring constant is equal to 16 divided by 0.

02, which is 800 newtons per metre.

Have a look at this example yourselves, pause the video once you have a go.

Don't forget to show you're working out and start the video again once you've got your answer.

So how did you get on? Hopefully you started by working out the extension, which is the increase in the length of the spring from its visual length, which is five centimetres or 0.

050 metres.

You'd then use the equation for hooke's law and substituting the numbers that you know in this case we need to divide both sides by the 0.

05 in order to find the spring constant and the correct answer is 1,200 newtons per metre.

So well done if you've got the right answer and especially well done if you showed each step of your working out.

Let's have a look at another example.

In this case we've got a suspension spring on a car.

Now a car usually has a spring in each corner above each wheel to even out the bumps on the road so you don't get jolted when you're driving along.

Each of those springs has got a spring constant of 300,000 Newtons per metre.

Calculate how much it will be compressed by a force of 4,000 Newtons.

Now that is a typical force that the car might experience when it's driving along on a road.

Give your answer in centimetres.

We'll start by stating hooke's law equation.

We'll substitute the that we've got and then we can rearrange that.

We can divide both sides by 300,000 Newtons per metre to get extension on its own, which is the only unknown.

We can do the maths and we find that the extension 0.

013333is metres, or if we change that to centimetres it is 1.

3 centimetres to two significant figures in this case.

Have a look at this example yourselves, please show you're working out.

Pause the video whilst you work out your answer and start it again once you're ready.

Okay, so how did you get on? In this case we've got a Newton metre stretching when an applied force of 4.

6 newtons is put on it, the spring constant is 75 newtons per metre.

Start with hooke's law, substitute the numbers in that you know, and then rearrange the equation by dividing both sides by 75 newtons per metre to get the unknown on its own on the right hand side.

If we do the maths we find the extension is equal to 0.

061333 metres which rounds to 6.

1 centimetres.

So well done if you've got that and showed all of you working out.

And now I like you to have a go at these calculations, answer all of the questions and show all of you working out.

Pause your video where you have a go at these and start it again once you're ready.

Okay, how did you get on with those questions? Question one, was to calculate the force needed to give a spring and extension of five centimetres if the spring constant of the spring was 30,000 newtons per metre.

The extension we need to change to metres first of all.

So that's 0.

050 metres we've divided five centimetres by a hundred.

Force is equal to the supreme constant times extension, which is hooke's law.

We can substitute the values in that we're told and we get an answer of 1500 Newtons, 1,500 newtons.

Question two was to calculate the force needed to stretch a spring from eight centimetres to 10.

5 centimetres.

The spring constant is 2000 Newtons per metre.

Now in this case we need to work out the extension.

The increase in length of the spring from its original length, which is 2.

5 centimetres converted to metres 0.

025 metres, write down hooke's law.

It's always important to write down the equation 'cause on the exams you typically get a mark for writing down the correct equation.

If it's not stated already in the question, substitute the numbers in and work it out, we get an answer 50 newtons.

In question three, calculate the extension of a spring that has a spring constant of 10,000 newtons per metre if it is stretched by a force of 80 newtons.

In this case we start by writing down the hooke's law equation force is spring constant times extension, substitute the values in, and divide both sides by 10,000 to get the extension on its own.

And if we do the maths we find the extension is 0.

008 metres, which is 0.

8 centimetres.

So well done if you've got that one.

Question four, a spring of think 32.

0 centimetres has a spring constant of 5,000 newtons per metre, calculate how long the spring will be if it is stretched by a force of 250 newtons.

You've got to be a little bit careful with the wording of this question.

Will start again by stating hooke's law force, its spring constant times extension, substitute the numbers in, and divide by 5,000 so the extension is on its own.

We then calculate the extension as been 0.

05 metres, which is five centimetres.

You were asked though not for the extension, but for how long the spring will be, it'll be five centimetres longer than it was at the start, so that's 32 plus five, which is 37.

0 centimetres.

You'll notice in the question we had the spring length 32.

0, which was three significant figures.

The spring content was 5,000 newtons per metre, which was four significant figures.

So we give our answer to the smaller number of significant figures of what we're told in the question, which is three.

So very well done if you've got those questions correct and well done if you've got most of them right.

So we've now reached the end of this and very well done if you've got this far, this is a short summary slide that describes what we've covered during the lesson.

We've talked about hooke's law, which states that the extension of the spring is directly proportional to the force applied, but only if the limited proportionality has not been exceeded.

And the type of defamation that we've got is elastic defamation.

The equation for hooke's law is that the force applied equals the spring constant times the extension.

The force is most in Newtons, the extension in metres and the spring constant in Newtons per metre.

We know that the greater the spring constant of the spring, the greater the force needed to stretch or compress the spring by the same amount.

Stretching causes elastic defamation if the spring or any elastic object will return to its visual shape when the stretching force removed.

And if a spring is overstretched, there might be what we call inelastic defamation and the spring will then not return to each individual shape because it's being stretched outta shape even when the force has been removed.

So again, well done for getting to the end of the lesson.

I do hope you join me next time.

Goodbye.