Transcript Elasticity - GEOCITIES.ws

Elasticity
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.
What can you observe in this figure?
Hooke’s Law
State 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 :
F=kx
F = Force on the spring
x = extension
k = Force constant of
the spring (Nm-1)
x
F = mg
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.
20 cm
26 cm
x
F = mg
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
1
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
m = 300 g + 200 g = 500 g = 0.5 kg
F = mg = 0.5 x 10 = 5 N
2
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
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
1 2
EP  k x
2
Ep = Elastic Potential Energy (J)
k = Force Constant of the spring (N/m)
x = extension of the spring (m)
Example 2
x
F
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?
Solution
Maximum kinetic energy is equal to elastic potential energy
½m
2
v
=½k
2
x
m = 300 g = 0.3 kg
x = 5 cm = 0.05 m
V2 = (k x2) / m
V2 = (200 Nm-1 x (0.05 m)2 ) / 0.3 kg
V2 = ( 200 x 0.0025) / 0.3
V2 = (0.5) / 0.3 = 1.666
V = 1.29 ms-1
Factors that effect the elasticity
Factor
Change in factor
Effect on Elasticity
Shorter spring
Less elastic
Longer spring
More elastic
Smaller spring
Less elastic
Larger diameter
More elastic
Smaller diameter
More elastic
Larger diameter
Less elastic
Length
Diameter of spring
Diameter of Spring
Wire
Type of Material
The elasticity changes with the type of
materials
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=kx
F= force (N)
K = spring constant (Nm-1)
x = spring extension (m)