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Hooke’s Law ?

Hooke’s Law ?. Hooke’s Law. In 1660, a scientist called Robert Hooke discovered a law for elastic materials. He was trying to improve the type of clocks that were used on ships (pendulum clocks) and theorized that a spring would work better on a pitching ship.

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Hooke’s Law ?

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  1. Hooke’s Law ?

  2. Hooke’s Law In 1660, a scientist called Robert Hooke discovered a law for elastic materials. He was trying to improve the type of clocks that were used on ships (pendulum clocks) and theorized that a spring would work better on a pitching ship.

  3. We are familiar with spring scales Not used in industry, b/c can’t be calibrated accurately and can become overstretched

  4. Hooke's Law, elasticity • If a material returns to its original size and shape when you remove the forces stretching or deforming it (reversible deformation), we say that the material is demonstrating elastic behaviour. • If you apply too big a force a material will lose its elasticity. • Hooke discovered that the amount a spring stretches is proportional to the amount of force applied to it. This means if you double the force its extension will double, if you triple the force the extension will triple and so on.

  5. The elastic limit can be seen on the graph. This is where it stops obeying Hookes law.

  6. You can write Hooke's law as an equation: F = k ∆ x Where: • F is the applied force (in newtons, N), • x is the extension (in metres, m) and • k is the spring constant (in N/m). • ∆x = stretched length – original length.

  7. H___________ Law: The compression or elongation x of an ideal spring from its e________________ position (x = 0) is d____________________________to the applied force Fs. Hooke's equilibrium directly proportional Fs = kx compression: stretching or elongation: x = 0 x = 0 x x Fs Fs More F  more s____________ or c__________________. stretch compression

  8. Hooke's Law is often written: Fs = - kx This is because it also describes the force that the s _____________ exerts on an o___________ that is attached to it. The negative sign indicates that the direction of the spring force is always o _____________ to the displacement of the object spring itself object opposite

  9. A weight of 8.7 N is attached to a spring that has a spring constant of 190 N/m. How much will the spring stretch?

  10. w/o weight x Ex. A weight of 8.7 N is attached to a spring that has a spring constant of 190 N/m. How much will the spring stretch? w/ weight Given: 8.7 N Fs = 190 N/m k = Unknown: 8.7 N x = ? Equation: Fs = kx 8.7 N = (190 N/m) x x = 4.6 x 10-2 m

  11. Fs = kx Fs direct Ex: A force of 5.0 N causes t0.015 m. How far will it stretch if the force is 10 N? he spring to stretch 10 5 2 (0.015 m) = 0.030 m .015 ? x What quantity does the slope represent? slope = = Dy/Dx k Fs/x the spring constant, k. The slope represents _______________________________

  12. Ex. Comparing two springs that stretch different amounts. spring B Fs spring A Applying the same force F to both springs x xB xA Which spring stretches more? Which is stiffer? A B greater stiffer spring  _________ slope  _________ k larger

  13. Elastic ____________ PE - the energy stored in a spring when work is done on it to stretch or compress it PEs = (½)kx2 Ex. A spring with a spring constant of 370 N/m is stretched a distance 6.4 x 10-2 m. How much elastic PE will be stored in the spring? PEs = (½)kx2 = (0.5)( 370 N/m)(6.4 x 10-2 m)2 = 0.76 (N/m)(m2) = 0.76 Nm = 0.76 J How much work was done to stretch the spring by this amount? W = DPE = 0.76 J

  14. Hold on a minute, K? Spring Constant?! • The spring constant measures how stiff the spring is. • The larger the spring constant the stiffer the spring. • You may be able to see this by looking at the graphs below: k is measured in units of newtons per metre (Nm -1).

  15. A spring is 0.38m long. When it is pulled by a force of 2.0 N, it stretches to 0.42 m. What is the spring constant? Assume the spring behaves elastically. Extension, ∆x = Stretched length – Original length = . 0.42m – 0.38m = 0.04 m F = k ∆ x 2.0N = k x 0.04m So, k = 2.0 N 0.04 m = 50 N m-1

  16. Elastic behaviour – Car Safety • Elastic behaviour is very important in car safety, as car seatbelts are made from elastic materials. However, after a crash they must be replaced as they will go past their elastic limit. • Why have seat belts that are elastic? • Why not just have very rigid seatbelts that would keep you firmly in place? • The reason for this, is that it would be very dangerous and cause large injuries. This is because it would slow your body down too quickly. The quicker a collision, the bigger the force that is produced.

  17. Key Definitions F = k ∆ x • Hooke’s Law = The amount a spring stretches is proportional to the amount of force applied to it. • The spring constant measures how stiff the spring is. The larger the spring constant the stiffer the spring. • A Diagram to show Hooke’s Law

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