Hooke's law

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Summary

Hooke's Law  

In a nutshell

Hooke's law is a proportionality principle in physics that states that up until the elastic limit of a material, the displacement of its extension or compression is directly proportional to the force applied on the material. 



Equations

DESCRIPTION

EQUATION

Hooke's Law
F=kxF=kx
Springs in series
1ktotal=1k1+1k2\dfrac {1}{k_ {total}} = \dfrac {1}{k_1} + \dfrac {1}{k_2}
Springs in parallel
ktotal=k1+k2k_ {total} = k_1 + k_2
Elastic potential energy
E=12FxE = \dfrac {1}{2} Fx

E=12kx2E = \dfrac {1}{2} kx^2​​



Variable definitions

​​QUANTITY NAME

SYMBOL

DERIVED 

UNIT

ALTERNATE 

UNIT

SI BASE

UNIT

forceforce
FF​​
NN​​
​ ​​
kg m s2kg \, m \, s^{-2}​​
displacementdisplacement​​
xx​​
mm​​

mm​​
elastic potential energyelastic\ potential\ energy​​
EE​​
JJ​​
WsWs​​
m2 kg s2m^{2} \, kg \, s^{-2}​​
force (spring) constantforce \space (spring) \space constant​​
kk​​
N m1N \, m^{-1}​​



Hooke's Law

Hooke's law is an important proportionality relation in physics that states that the displacement of a material's compression is directly proportional to the force applied to the material, until the material reaches the limit of proportionality. These two values are linked by the force constant (kk), which can also be called the spring constant when referring to springs.  


The force constant is used to evaluate how stiff a material is. The larger the value of the force constant, the more difficult it is to deform. 


From a point called the limit of proportionality, Hooke's Law will not apply. This is because the displacement of the material will no longer be proportional to the force exerted.


Hooke's law can be demonstrated through a force-extension graph.  


Physics; Materials; KS5 Year 12; Hooke's law
A.
force (N)force \ (N)​​
B.
extension (m)extension \ (m)​​
1.
Limit of proportionality


Example

A mass weighing 20 kg20\ kg​ is attached to one end of spring which has an original length of 1.0 m1.0\ m​. The new length of the spring is 1.2 m1.2\ m. What is the spring constant of the spring used?


Firstly, write down the known values: 


m=20 kgm=20\ kg​ 

x=1.2 m1 m=0.2 mx=1.2\ m - 1\ m = 0.2 \, m​ 

​​

Next, calculate FF​ by using F=mgF = mg:

​​

F=20×9.81F = 20 × 9.81

F=196.2 NF = 196.2 \ N​​​


Rearrange the equation for Hooke's law to make the spring constant the subject:


k=Fxk = \dfrac {F}{x}


Substitute in values:


k=196.20.2k = \dfrac {196.2}{0.2}​​

k=981 Nm1k = 981\ Nm^{-1}


The spring constant of the spring used is 980 Nm1\underline{980\ Nm^{-1}} (2.s.f).



Springs in series

Springs in series, means springs one after the other, just like components in a series circuit.   


Physics; Materials; KS5 Year 12; Hooke's law
When a weight is added onto springs in series, the total displacement of the whole system is:

xtotal=x1+x2+x3x_{total} = x_1+x_2+x_3
​​
Using Hooke's law, xx can be replaced with Fk\dfrac{F}{k}:

Ftotalktotal=F1k1+F2k2+F3k3\dfrac{F_{total}}{k_{total}} = \dfrac{F_1}{k_{1}}+ \dfrac{F_2}{k_{2}}+\dfrac{F_3}{k_{3}}​​

The force on each of the springs is the same so:

Ftotal=F1,F2,F3F_{total} = F_1, F_2, F_3

Therefore, the total spring constant of a system of springs in series is:

1ktotal=1k1+1k2\dfrac {1}{k_ {total}} = \dfrac {1}{k_1} + \dfrac {1}{k_2}​​


Example​​

Two different springs, one with a spring constant of 6.8 Nm16.8 \ Nm^{-1} and another with a spring constant of 2.6 Nm12.6 \ Nm^{-1} are arranged in a series placement. What is the total spring constant of this system? 

 

Firstly, write down the known values: 


k1=6.8 Nm1k_1 = 6.8\ Nm^{-1}

k2=2.6 Nm1k_2 = 2.6\ Nm^{-1}​​​


Write down the equation for calculating spring constant in series:


1ktotal=1k1+1k2\dfrac {1}{k_ {total}} = \dfrac {1}{k_1} + \dfrac {1}{k_2}

​​

Substitute the known values into the equation:


1ktotal=16.8+12.6\dfrac {1}{k_ {total}} = \dfrac {1}{6.8} + \dfrac {1}{2.6}


1ktotal=0.5317...\dfrac {1}{k_ {total}} = 0.5317 ...


Find the reciprocal to get ktotalk_{total}:


ktotal=10.5317...=1.88 N m1k_{total} = \dfrac {1}{0.5317...} = 1.88 \, N \, m^{-1}​​

The total spring constant of the system is 1.9 Nm1\underline{1.9\ Nm^{-1}} (2.s.f).



Springs in parallel

For springs in parallel, the force is shared between the springs.


Physics; Materials; KS5 Year 12; Hooke's law

This means that:


Ftotal=F1+F2+F3F_{total} = F_1 +F_2+F_3​​


Substituting kxkx in for FF:


ktotalxtotal=k1x1+k2x2+k3x3k_{total}x_{total}= k_{1}x_{1}+k_{2}x_{2}+k_{3}x_{3}​​


The springs will all be extended the same amount so: 


xtotal=x1,x2,x3x_{total} = x_1, x_2, x_3


Therefore, the total spring constant of a system of springs in parallel is:


ktotal=k1+k2k_ {total} = k_1 + k_2


​​Example 

Two different springs, one with a spring constant of 7.9 Nm17.9 \ Nm^{-1} and another with a spring constant of 5.4 Nm15.4 \ Nm^{-1} are arranged in a parallel placement. What is the total spring constant of the system? 


Firstly, write down the known values: 

k1=7.9 Nm1k_1 = 7.9\ Nm^{-1}

k2=5.4 Nm1k_2 = 5.4\ Nm^{-1}​​

Substitute the known values into the equation:


ktotal=7.9+5.4k_{total} = 7.9 + 5.4

ktotal=13.3 Nm1k_{total} = 13.3\ Nm^{-1}


The total spring constant of the system is 13 Nm1\underline{13\ Nm^{-1}} (2.s.f). 

​​


Elastic potential energy

When a material is extended or compressed, work done on a material is transferred to elastic potential energy. This energy is represented as the area under the graph in a force-extension graph, and therefore can be written as follows:  


E=12FxE = \dfrac {1}{2} Fx


This equation can be written in another form as we know that F=kxF = kx


E=12Fx=12(kx)×xE = \dfrac {1}{2} Fx = \dfrac {1}{2} (kx) × x

E=12kx2E = \dfrac {1}{2} kx^2​​​​

Example

The elastic potential energy of a spring that is extended by 0.60 m0.60\ m​ is 48 J48\ J. Calculate the spring constant of the spring.


Firstly, write down the known values: 

x=0.6 mx = 0.6\ m

E=48 JE = 48\ J​​​

Next, write down the equation for elastic potential energy:


E=12kx2E = \dfrac {1}{2} kx^2​​


Rearrange the equation to make kk the subject:


k=2Ex2k = \dfrac {2E}{x^2}​​

​​​

 Substitute the known values into the rearranged equation:


k=2×480.62k = \dfrac {2 ×48}{0.6^2}


k=266.67 Nm1k = 266.67 \ Nm^{-1}


The spring constant is 270 Nm1\underline{270\ Nm^{-1}} (2.s.f). 


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Learn with Basics

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Unit 1

Forces and elasticity: Hooke's Law

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Unit 2

Elasticity and the spring constant

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Hooke's law

Unit 3

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