 # Understand Forces and Elasticity

In this worksheet, students will learn Hooke's law and apply both the equation and the graph. This worksheet will not cover the practical, but there is a related worksheet on the practical called 'Investigating Forces' Key stage:  KS 4

GCSE Subjects:   Physics: Single Subject, Physics: Combined Science

GCSE Boards:   OCR 21st Century, Pearson Edexcel

Curriculum topic:   Matter: Models and Explanations, Forces and Matter

Curriculum subtopic:   How Does the Particle Model Relate to Material Under Stress?, Forces and Matter

Difficulty level:   ### QUESTION 1 of 10

There is a separate activity on Hooke’s law experiment – we will not be covering the experiment in this activity. If you are looking for Hooke’s law experiment, then please search the required practical’s for forces experiment.

A bungee cord is basically a big rubber band, and it might help some of this to sink in if you have a rubber band to play with while you read this. If you apply just one force to a rubber band that it won't stretch, it will just move in the direction the force is applied. You need a minimum of 2 forces that will act in different directions to stretch anything. One of these forces could be stationary, you finger, and one could be moving, like the hand pulling the band backward. This will apply a force to the band and make it stretch. If you release the forces, then the band will return to its original shape. All good so far?

Okay, so Hooke’s law is a law that defines stretching things. It says that ‘The force applied to an object is proportional to the length of the extension’. This basically means that if you were to plot force and extension on a graph, it would make a straight line – just like the one shown below.

GRAPH IMAGE HERE

Hooke realized that he could make an equation out of these graphs that could tell him in any given circumstance what the extension of an object would be.

INSERT EQUATION HERE

F = force (Newtons (N))
k = spring constant – a number that tells you how hard it is to stretch the object (Newtons per meter (N/m))
e = Extension – how much the spring has stretched (meters (m))

So, the person putting the numbers into the computer at the start of the activity was just working out how much you would stretch the bungee using this equation.

Let’s have a go at this ourselves.

Q – A person stretches the bungee 15 m when the bungee cord has a spring constant of 40 N/m. Calculate the force applied by that person.

Step 1 – Highlight the number son the question:
A person stretches the bungee 15 m when the bungee cord has a spring constant of 40 N/m. Calculate the force applied by that person.

Step 2 – Write out the numbers:
F = ?
k = 40 N/m
e = 15 m

Step 3 – Put the numbers into the equation:

F = 40 x 15

F = 600 N

In order to squish a lump of plasticine, what is the minimum number of forces required?

1

2

3

4

Write down the equation that links Force, Spring Constant, and Extention.

Match the unit to the value

## Column B

Froce (F)
Newton per meter (N/m)
Spring Constant (k)
Meter (m)
Extention (e or x)
Newton (N)

Brooke stretches a rubber band, when she lets go it returns to its original shape. Paul does the same with a lump of plasticine, but it does not return to its original shape. Using the grind below, describe which was an elastic and which was an inelastic deformation.

 Rubber band plasticine Elastic deformation Inelastic deformation.

A spring has a spring constant of 50 N/m and is stretched 0.75 m. Calculate the force on the spring.

An elastic band has a spring constant of 1000 N/m and an original length of 0.25 m. When it is loaded with mass, the length increases to 0.45 m. Calculate the force added to the spring to cause this extension.

The data below shows extension and force data collected by a student. Using the data, describe if the object they where suing followed Hookes Law.

 Force (N) Extention (m) 100 0.2 200 0.4 300 0.6 400 0.8 500 1 600 1.2 700 1.3

In the previous question, the values started to change at the end. Suggest what is happening to the spring at this point.

 Force (N) Extention (m) 100 0.2 200 0.4 300 0.6 400 0.8 500 1 600 1.2 700 1.3

Tom has a crossbow that is able to fire bolts. He wants to work out the force with which he can fire the bolts. On the box, it says that the bow at the front of the crossbow has a spring constant of 3 kN/m. He measures the draw distance of the crossbow as 0.5 m. Calculate the force on the bolt.

A sping in a car suspension is compressed when it goes over a bump in the ground. The spring has an original length of 0.75 m but when it hits a bump, it compresses to a length of 0.65 m. The car is quoted as having a spring constant value of 5000 N/m. Calculate the force of the car going over the bump.

• Question 1

In order to squish a lump of plasticine, what is the minimum number of forces required?

2
EDDIE SAYS
You need to have AT LEAST 2 forces acting on an object in order to make it change its shape in any way - either squishing or stretching. If you want to squish it, you need a force pushing it one way and another force pushing it in the opposite direction. If you want to stretch an object, then you need 2 forces pulling in opposite direction. You need at least 2 forces is what we're saying here. 2 forces.
• Question 2

Write down the equation that links Force, Spring Constant, and Extention.

EDDIE SAYS
These are common questions in exams - they want to know that you can remember the equations, so get your remembering hat on! If it comes to it in an exam and you can't remember, just write them down with a = and either a times or a divide in them! It'll be something like that!
• Question 3

Match the unit to the value

## Column B

Froce (F)
Newton (N)
Spring Constant (k)
Newton per meter (N/m)
Extention (e or x)
Meter (m)
EDDIE SAYS
So, hopefully, by now you are getting the hang of why we are doing these types of questions. It helps you to spot them when it comes to picking out the numbers in a maths question. In this case, I'm going to say remember the capital N for Newtons - it was his name after all!
• Question 4

Brooke stretches a rubber band, when she lets go it returns to its original shape. Paul does the same with a lump of plasticine, but it does not return to its original shape. Using the grind below, describe which was an elastic and which was an inelastic deformation.

 Rubber band plasticine Elastic deformation Inelastic deformation.
EDDIE SAYS
To remember this, all you need to do is think about another name for a rubber band - elastic band. If something is elastic, then it will return to its original shape, just like an elastic band. If something is not elastic, then it will not return to its original shape - it is inelastic. This is the difference between elastic and inelastic deformation! Boom!
• Question 5

A spring has a spring constant of 50 N/m and is stretched 0.75 m. Calculate the force on the spring.

37.5
EDDIE SAYS
This is a simple find the numbers and plug them into the equation job! Let's go through the steps of the question together. Step 1 - find the numbers and put them next to their symbols: F = ? k = 50 N/m e = 0.75 m Step 2 - put them into the equation: F = 50 x 0.75 Step 3 - do the calculation: F = 37.5 N DONE!
• Question 6

An elastic band has a spring constant of 1000 N/m and an original length of 0.25 m. When it is loaded with mass, the length increases to 0.45 m. Calculate the force added to the spring to cause this extension.

200 N
EDDIE SAYS
This one is a little more ticky then the others. For this one, we need to think about extention of the elastic band in a little more detail. A common mistake you might make is to calculate this with the length being 0.45 m, but it isn't. Let's take a look: Extention is how much the elastic band has moved when the mass is put on it. It is originally 0.25 m long, but when it is stretched it becomes 0.45 m long. How far has it moved? Well 0.45 - 0.25 = 0.2, so it has stretched 0.2 m. Let's go through the steps together: Step 1 - Find the numbers and write them down: F = ? k = 1000 N/m e = 0.45 - 0.25 = 0.2 m Step 2 - put the numbers into the equation: F = 1000 x 0.2 Step 3 - do the maths: F = 200 N DONE!
• Question 7

The data below shows extension and force data collected by a student. Using the data, describe if the object they where suing followed Hookes Law.

 Force (N) Extention (m) 100 0.2 200 0.4 300 0.6 400 0.8 500 1 600 1.2 700 1.3

EDDIE SAYS
For this question, you have a lot of information to take in and process - that is why it is a three mark question. Let's break it down into parts: 1 - It is asking you if you can remember what Hooke's Law says. It says that a graph of extension vs force will be a straight line. This is true in this case because the numbers on the left go up by 100 each time and the numbers on the right go up by 0.2 each time. 2 - It is asking you if you can use data. Can you spot the pattern in the data? Did you see that the numbers went up by 0.2 each time in the extension? This is what they are really asking you about here. It's just a case of breaking the question down - looking for what they really want you to answer.
• Question 8

In the previous question, the values started to change at the end. Suggest what is happening to the spring at this point.

 Force (N) Extention (m) 100 0.2 200 0.4 300 0.6 400 0.8 500 1 600 1.2 700 1.3

EDDIE SAYS
Suggest means you need to think of your own answer that you haven't been taught and it will come up in your exam at some point. In this question, you needed to think about what would happen to an elastic band if you put a whole load of force onto it. It should get to a point where it will not be able to stretch any more then it already has. This is why it is good to play with elastic bands - you will get to know where that point is. Play away - you have our permission!
• Question 9

Tom has a crossbow that is able to fire bolts. He wants to work out the force with which he can fire the bolts. On the box, it says that the bow at the front of the crossbow has a spring constant of 3 kN/m. He measures the draw distance of the crossbow as 0.5 m. Calculate the force on the bolt.

1500
1,500
EDDIE SAYS
In this question, there is one key thing that separates it from the others - kN/m. This means that you need to convert from kilo Newtons in Newtons. 3 kN/m = 3000 N/m. Apart from that the procedure is the same: F = ? k = 3000 N/m e = 0.5 F = 3000 x 0.5 F = 1500 N Boom, done!
• Question 10

A sping in a car suspension is compressed when it goes over a bump in the ground. The spring has an original length of 0.75 m but when it hits a bump, it compresses to a length of 0.65 m. The car is quoted as having a spring constant value of 5000 N/m. Calculate the force of the car going over the bump.

500
EDDIE SAYS
This is a fairly simple question, with one exception - it has given you a starting length and a finishing length instead of an extension. It is also a compression instead of an extension, but this makes no difference to our maths. Step 1 - calculate the extension: e = 0.65 - 0.75 = -0.1 Step 2 - write down the numbers: F = ? k = 5000 N/m e = 0.1 m Step 3 - Put them into the equation. F = 5000 x 0.1 F = 500 N 