physics as unit2 05 hookeslaw01

8/10/2019 Physics as Unit2 05 HookesLaw01

1/12

1. To understand what happens when anincreasing force is applied to a wire or spring

2. To understand how springs in series and

parallel behave

3. To understand how to calculate the energy

stored in a stretched spring

4. Use this as a vehicle to check graph plotting

Book Reference : Pages 164-166


2/12

http://www.rdg.ac.uk/acadepts/sp/picetl/publish/ISEs/Forces2.htm

What happens when you apply increasing

tension to a spring or wire?

Hookes Law states

that...the change in

length produced by a

force on a wire orspring is directly

proportional to the

force applied.
http://www.rdg.ac.uk/acadepts/sp/picetl/publish/ISEs/Forces2.htmhttp://www.rdg.ac.uk/acadepts/sp/picetl/publish/ISEs/Forces2.htm


3/12

Extension Force Applied

L F

Well see

latter that

Hookes law

only applieswithin limits

L (m)

F

(N

)


4/12

To turn a proportionality into an equation

we need to introduce a constant ofproportionality...

L F

F = kL

We call k the spring constant and it defines

how stiff the spring is

How can we find K experimentally?

What should we do to minimise errors?

Take care to avoid confusion

between overall length &extension


5/12

Take the

gradient ofthe graph :

F2F1L2L1

F

L

L (m)

1. Make the gradient triangle as large as possible

2. Avoid outliers, choose data points which are on

the line

F

(N)


6/12

L (m)

When we undertake an experiment

we should only change one variable at

a time to make it a fair test. We call

this the independent variable

Quantities we measure, (and subsequently calculate) are

called dependent variables. All other variables which

are kept the same are called the control variables

Often graphs have the independent variable along

the bottom and the dependent up the side

Hookes law is a notable exception


7/12

C stops obeying Hookes law... After the limit of

proportionality the material behaves in a ductile

fashion. The material stretches more with a

small extra force.

Spring constant :

material A isstiffer than B & C

L (m)

F

(N)


8/12

Springs in parallel share the load &

have the same extension acting like a

single spring with a combined spring

constant

The force needed to stretch

springs p & q respectively is:-

Fp= kpL & Fq= kqL

Weight supported by both

springs W = Fp + Fq

p q

L


9/12

Cancel out L

k = kp+ kq

p q

L

So by substituting & expanding

W = kpL + kqL which can be

considered equal to kL where

k is the effective spring

constant

kL = kpL + kqL


10/12

Springs in series share the same

tension which is equal to W

The extensions in the two

springs is given by:-

Lp= W/kp& Lq= W/kq

The total extension is Lp+ Lq

W/k = W/kp+ W/kq = 1/kp+ 1/kq = 1/k

where k is the effective spring constant

p

qL


11/12

The stretched spring

has elastic potentialenergy. Work has been

done because the

force moves through adistance.

The distance moved by the force is L, the

force involved ranges from 0 to F and so the

average is F/2

Ep= FL

Only valid for

where Hookes

law is obeyed


12/12

1. Weve seen Hookes law and how we can useit to establish the spring constant

2. Weve discussed variables and how to

accurately establish the spring constantexperimentally

3. Weve seen how combinations of springs in

parallel and series can act as a single spring

4. We have related the elastic potential energy

in a stretched spring to the work done

stretching the spring

physics as unit2 05 hookeslaw01

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