A property of matter that enables an object to return to

its original size and shape when the force that was acting on it is removed.

Elasticity

What can you observe in this figure?

Hooke’s LawState that:

“The extension of a spring is directly proportional to the stretching force acting on it provided the elastic limit of the spring is not exceeded”.

Elastic limit:

The maximum stretching force which can be applied to the spring before it ceases to be elastic.

The mathematical expression for Hooke’s law :

x

F = mg

F = k x F = Force on the spring

x = extension

k = Force constant of the spring (Nm-1)

Force constant of a spring: the force that is required to produce one unit of extension of the spring.

Example 1

A spring has an original length of 20 cm. with a load of mass 300 g attached to it, the length of the spring is extended to 26 cm.

x

F = mg

20 cm

26 cm

1. Calculate the spring constant.

2. What is the length of the spring when the load is

increased by 200 g? (assume that g=10 Nkg-1).

Solution

lo = 20 cm

l1 = 26 cm

x = l1 – lo = 26 – 20 = 6 cm = 0.06 m

m = 300 g = 0.3 kg

g = 10 N kg-1

F = mg = 0.3 x 10 = 3 N

k = F/x = 3 / 0.06 = 50 Nm-1

1

m = 300 g + 200 g = 500 g = 0.5 kg

F = mg = 0.5 x 10 = 5 N

x = F / k = 5 / 50 = 0.1 m = 10 cm

l1 = lo + x = 20 cm + 10 cm = 30 cm

Length of the spring is being 30 cm

2

Elastic Potential Energy

The elastic Potential Energy is the energy stored in a spring when it is extended or compressed.

The result of the work done to extend or compress the spring.

KATAPULT

ARROW

2

2

1xkEP

Ep = Elastic Potential Energy (J)

k = Force Constant of the spring (N/m)

x = extension of the spring (m)

F

x

Mass of the ball (m) = 300 g

The spring constant (k) = 200 Nm-1.

Spring extension (x) = 5 cm

What is the maximum velocity of the ball when the stretching

force is released?

Example 2

½ m v2 = ½ k x2

Maximum kinetic energy is equal to elastic potential energy

m = 300 g = 0.3 kgx = 5 cm = 0.05 m

V2 = (k x2) / mV2 = (200 Nm-1 x (0.05 m)2 ) / 0.3 kgV2 = ( 200 x 0.0025) / 0.3V2 = (0.5) / 0.3 = 1.666V = 1.29 ms-1

Solution

Factor Change in factor Effect on Elasticity

Length

Shorter spring Less elastic

Longer spring More elastic

Diameter of spring

Smaller spring Less elastic

Larger diameter More elastic

Diameter of Spring Wire

Smaller diameter More elastic

Larger diameter Less elastic

Type of MaterialThe elasticity changes with the type of materials

Factors that effect the elasticity

Summary

In this lesson, we learnt that:

1.Forces can change the shape of an object.

2.Objects that return to their original shape when the forces acting on them are removed are elastic.

3.Hooke’s Law state that the force on a spring is directly proportional to its extension, that is:

F = k x

F= force (N)

K = spring constant (Nm-1)

x = spring extension (m)