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IP2.11.5 Hooke’s Law © Oxford University Press 2011 Hooke’s Law.

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Presentation on theme: "IP2.11.5 Hooke’s Law © Oxford University Press 2011 Hooke’s Law."— Presentation transcript:

1 IP2.11.5 Hooke’s Law © Oxford University Press 2011 Hooke’s Law

2 IP2.11.5 Hooke’s Law © Oxford University Press 2011 In 1676 the British scientist Robert Hooke investigated the relation between the force applied to an elastic object and the change in its length (called the extension).

3 IP2.11.5 Hooke’s Law © Oxford University Press 2011  He found the force applied to an object was directly proportional to its extension, up to a certain point.  When he doubled the force on a spring he found the extension would also double. If applying 20 N caused a spring to extend 0.5 m then applying 40 N would cause it to extend 1.0 m.  This relationship continues up to the limit of proportionality. Above this point the spring starts to deform – it will no longer return to its original length.

4 IP2.11.5 Hooke’s Law © Oxford University Press 2011 Can you think of an application that uses this relationship? A newtonmeter or weighing scales use this relationship. The force of gravity caused by an object’s mass (its weight) can be measured by using the extension of a spring. The spring has to calibrated first for a known weight and then other weights can be measured. Before Hooke discovered this relationship weighing had to be done using a balance, with known weights used on one side to balance the weight of the object you were weighing.

5 IP2.11.5 Hooke’s Law © Oxford University Press 2011  How much a certain spring extends depends on the force applied and the spring constant.  This is a measure of the stiffness of the spring and it is measured in newtons per metre (N/m).  The higher the spring constant the stiffer the spring and the less it extends for a given force.

6 IP2.11.5 Hooke’s Law © Oxford University Press 2011 The relationship between force, spring constant and extension is represented in this equation. force = spring constant  extension (newtons, N) (newtons per metre, N/m) (metres, m) If the force is called F, the spring constant k and the extension e, then F = k  e


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