Hooke’s Law & springs Objectives Week 12 Hooke’s Law & springs Objectives Describe how deformation is caused by a force. Describe the behaviour of springs and wires in terms of force, extension and elastic limit. Describe the behaviour of springs and wires in terms of Hooke’s law and the force constant. © Pearson Education Ltd 2008 This document may have been altered from the original

Week 12 Stretching a wire Tensile forces cause tension in an object, caused by equal and opposite forces on it at either end. Squeezing a spring Compressive forces cause an object to reduce in size, caused by two equal and opposite forces – spring doesn’t accelerate! © Pearson Education Ltd 2008 This document may have been altered from the original

Week 12 Experiment to stretch a wire – wire is stretched to less than the elastic limit. © Pearson Education Ltd 2008 This document may have been altered from the original

Week 12 Straight-line graph of tension against extension The area under this graph is equal to the work done to stretch the spring © Pearson Education Ltd 2008 This document may have been altered from the original

Hooke’s law F = -kx Applies to the straight line section of the curve. Week 12 Hooke’s law Applies to the straight line section of the curve. Force (F)  Extension (x) F = -kx F is the force causing the extension in the spring. –kx is negative as it indicates that it is a restoring force. © Pearson Education Ltd 2008 This document may have been altered from the original

Week 12 2 springs in parallel and series – what is the difference in extension? Experiment 8b will explain how to produce graphs for the extension of these! © Pearson Education Ltd 2008 This document may have been altered from the original

Week 12 If two samples are hung in series, they both have the same tension in them so the extension will be double. You should try to explain these 2 effects in terms of the force needed to extend the bonds. T L+L 2L+2L © Pearson Education Ltd 2008 This document may have been altered from the original

Week 12 If two samples are hung in parallel, they share the load. The result is half the extension with the same load. T T/2 L+L L+L/2 © Pearson Education Ltd 2008 This document may have been altered from the original

Week 12 Question. A sample which obeys Hooke’s law extends from 1.000m to 1.001m when a load of 600kg is applied. What will the extension be for a load of 700kg? Solution extension = 1.001 - 1.000 = 0.001m k = -F/x = 600x10/0.001 = 6x106Nm-1. x = -F/k = 700x10/6x106 = 1.17x10-3m © Pearson Education Ltd 2008 This document may have been altered from the original

The deformation of materials & the work done in changing the shape Week 12 The deformation of materials & the work done in changing the shape Objectives Define the terms elastic deformation and plastic deformation. Explain that the area under a force against extension/compression graph is equal to work done by the force. Use the equations for elastic potential energy. © Pearson Education Ltd 2008 This document may have been altered from the original

Week 12 Experiment to stretch a wire – in this case we stretch the wire until it breaks, having gone past it’s elastic limit © Pearson Education Ltd 2008 This document may have been altered from the original

Week 12 Graph plotted using data from stretch experiment – showing what happens beyond the elastic limit. © Pearson Education Ltd 2008 This document may have been altered from the original

Stretching a wire beyond its elastic limit Week 12 Stretching a wire beyond its elastic limit © Pearson Education Ltd 2008 This document may have been altered from the original

Week 12 Graphs of extension against tension for (i) copper, (ii) two sorts of steel, (iii) polyethylene and (iv) rubber. © Pearson Education Ltd 2008 This document may have been altered from the original

Possible extension (purple) and return (pink) curves for rubber Week 12 Possible extension (purple) and return (pink) curves for rubber © Pearson Education Ltd 2008 This document may have been altered from the original

1) Using a child’s spring toy Week 12 Compression of an open-wound spring – this can be demonstrated in two ways 1) Using a child’s spring toy 2) By firing a trolley along a ramp – converting elastic energy into kinetic rather than Ep © Pearson Education Ltd 2008 This document may have been altered from the original

Week 12 The work done in stretching the object is stored in it as elastic potential energy E. E = work done in acheiving the stretch E = average force x extension Average force = F/2 so E = (F/2)x E = (kx/2)x E = 1/2 kx2 E = 1/2 k x2 & E = 1/2Fx Total work done W = Fx , so where has the other half of the energy gone? © Pearson Education Ltd 2008 This document may have been altered from the original