# Chapter 7: Matter and Materials

Density
Density is defined as the mass per unit volume of a substance.
Ƿ =m/v

Pressure
Pressure is defines as the normal force per unit cross sectional area.
P = F/A
Pressure in a fluid
It is given by:
Weight of water = mass × g = ƿ × A × h × g
Pressure = force/area = ƿ×A×h×g/A
= ƿgh
Hence,
Pressure (p) = ƿgh

Hooke’s Law
A material follows Hooke’s law if the extension produced in it is proportional to the applied force.
If we take a metal wire or a spring and hang it from the ceiling it will have a neutral, upstretched length of l meters, if we attach masses to the bottom of the wire, it will begin to stretch. The amount of length it has increased by we will call it extension and represent it by e.
If the extension increases proportionally to the force applied then it follows Hooke’s law:
The force needed to stretch a spring is directly proportional to the extension of the spring from its natural length. So it takes twice as much force to extend a spring twice as far.
So, F α e
F = ke
Where k is constant.
The spring constant
If we rearrange Hooke’s law we get,
K= F/e
Spring constant is measured in Newton per meter, N/m.

Strain
Strain is the fractional increase in the original length of wire. It has no unit.
Strain =extention/(original length) = x/L

Stress
Stress is defined as the force applied per unit cross sectional area of wire. Its unit is Pa.
Stress= force/(cross sectional area) = F/A

Young Modulus
It is the ratio of stress to strain of a material
Young modulus =stress/strain
= σ/ϵ
= (F/A)/(e/L)
=FL/eA

Elastic potential energy Elastic potential energy is the energy stored in a stretched or compressed material. As long as the elastic limit mas not exceeded, the energy can be recovered.

Area= ½ base × height

So,

Elastic potential energy = ½ F x

E = ½ F x

We can also do it by using Hooke’s law,

According to Hooke’s law, applied force F and extension x are shown by

F= kx    where k is constant

Elastic potential energy =½ Fx = ½ × kx × x

Elastic potential energy = ½ kx2