1. 1.To understand what happens when an increasing 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. http://www.rdg.ac.uk/acadepts/sp/picetl/publish/ISEs/Forces2.htm What happens when you apply increasing tension to a spring or wire? Hooke’s Law states that...the change in length produced by a force on a wire or spring is directly proportional to the force applied.

3. Extension  Force Applied  L  F We’ll see latter that Hooke’s law only applies within limits  L (m) F (N)

4. To turn a proportionality into an equation we need to introduce a constant of proportionality...  L  F F = k  L 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. Take the gradient of the graph : F 2 – F 1 L 2 – L 1 FF L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.  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 Hooke’s law is a notable exception

7. C stops obeying Hooke’s 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 is stiffer than B & C  L (m) F (N)

8. 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:- F p = k p  L&F q = k q  L Weight supported by both springs W = F p + F q pq LL

9. Cancel out  L k = k p + k q pq LL So by substituting & expanding W = k p  L + k q  L which can be considered equal to k  L where k is the effective spring constant k  L = k p  L + k q  L

10. Springs in series share the same tension which is equal to W The extensions in the two springs is given by:-  L p = W/k p &  L q = W/k q The total extension is  L p +  L q W/k = W/k p + W/k q = 1/k p + 1/k q = 1/k where k is the effective spring constant p q LL

11. The stretched spring has elastic potential energy. Work has been done because the force moves through a distance. The distance moved by the force is  L, the force involved ranges from 0 to F and so the average is F/2 E p = ½F  L Only valid for where Hooke’s law is obeyed

12. 1.We’ve seen Hooke’s law and how we can use it to establish the spring constant 2.We’ve discussed variables and how to accurately establish the spring constant experimentally 3.We’ve 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