Hooke’s Law

Hooke’s Law In the 1600s, a scientist called Robert Hooke discovered a law for elastic materials. Hooke's achievements were extraordinary - he made the first powerful microscope and wrote the first scientific best-seller, Micrographia. He even coined the word ‘Cell’.

Hooke's Law, elastic and plastic behaviour 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. A plastic (or inelastic) material is one that stays deformed after you have taken the force away. If deformation remains (irreversible deformation) after the forces are removed then it is a sign of plastic 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.

If you measure how a spring stretches (extends its length) as you apply increasing force and plot extension (x) against force (F); the graph will be a straight line.

The elastic limit can be seen on the graph. This is where it stops obeying Hookes law. Elastic limit can be seen on the graph. Anything before the limit and the spring will behave elastically. This is where the graph stops being a straight line. If you stretch the spring beyond this point it will not return to its original size or shape.

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). The extension ∆x (delta-x) is sometimes written e or ∆l. You find the extension from: ∆x = stretched length – original length.

x = 0 x = 0 ___________ Law: The compression or elongation x of an ideal spring from its ________________ position (x = 0) is ____________________________to the applied force Fs. Hooke's equilibrium directly proportional Fs = Fs = kx compression: stretching or elongation: x = 0 x = 0 x x Fs Fs stretch compression More F  more ____________ or __________________.

Hooke's Law is often written: Fs = -kx This is because it also describes the force that the _______________ exerts on an ___________ that is attached to it. The negative sign indicates that the direction of the spring force is always _____________ to the displacement of the object spring itself object opposite -x compressed spring: Fs ___ 0 > Fs x = 0 ______________ position, Fs = __ equilibrium undisturbed spring +x stretched spring: Fs ___ 0 < Fs

8.7 N 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/o weight w/ weight Given: 8.7 N Fs = 190 N/m x k = Unknown: 8.7 N x = ? Equation: Fs = kx 8.7 N = (190 N/m) x x = 4.6 x 10-2 m

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 Compare to Fs = kx Solve for Fs/x = k Fs/x the spring constant, k. The slope represents _______________________________

stiffer spring  _________ slope  _________ k Ex. Comparing two springs that stretch different amounts. Fs spring B 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

____________ PE - the energy stored in a spring when work is done on it to stretch or compress it Elastic 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

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: The spring constant k is measured in Nm-1 because it is the force per unit extension. The value of k does not change unless you change the shape of the spring or the material that the spring is made of. k is measured in units of newtons per metre (Nm -1).

Example F = k ∆ x 2.0N = k x 0.04m = 50 N m-1 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

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.

This can be seen very plainly by comparing the effect of kicking a football, which squashes as you kick it giving a big collision time

Andy O Brian with both the Ball and his nose showing elastic behaviour

followed by kicking a brick followed by kicking a brick. The brick doesn't squash, giving a very quick collision time and a very painful foot. This is why airbags and crumple zones can reduce injuries (these are both parts of a car designed to squash rather than be rigid). So to reduce injuries in a collision, always slow down in as long a time as possible. This is why you bend your legs when landing after a jump and why parachutists roll when they hit the ground.

Key Definitions 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 F = k ∆ x