Elasticity Chapter of Physics Class XI Non Medical
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Understanding Understanding Elasticity Elasticity Physics Chapter 2 , Form 4
Elasticity is .. the property of an object that enables it to return its original shape and dimensions after an applied external force is removed.
Forces Forces between atoms he T he
property of elasticity is caused by the existence of forces of repulsion and attraction between molecules in the solid material.
Forces between atoms in equilibrium condition
Explanation :
he T he
atoms are separated by a distance called the equilibrium distance and vibrate at it position. Force Force of repulsion re pulsion = Force of attraction
Forces Forces between between atoms in compression
Explanation
Force of repulsion takes effect. When the compressive force is removed, force of repulsion between the atoms pushes
Forces Forces between atoms in tension
xplanation E xplanation Force of attraction takes effect. When the compressive force is removed, force of repulsion between the atoms pushes the atom back to their equilibrium positions.
To investigate investigate the
relationship between force force and extension of a spring
Hookes Law that the extension extension of a spring is directly proportional to the
applied force provided that the elastic limit is not exceeded. exceeded. Elastic limit of a spring is defined as the maximum force that can be applied to spring such that the spring will return to its original length when the force released
The spring is said
to have a permanent extension extension :
when the length of the spring longer than the original
length even though the force acts was released and the elastic limit is exceeded. exceeded. The
elastic limit is not exceeded
o bey Hookes Hookes Law L aw.. When the spring obey he mathematical expression for Hookes Hookes Law is : T he F w x F = kx,
k = Force constant of the spring
Force constant, k =
F
x with unit N m-1, N cm-1 or N mm-1
Graph F against x
Spring
Constant, k
Example: A spring has an original length of 15 cm. With a load of mass 200 g attached, the length of the spring is extend to 20 cm. a. Calculate C alculate the spring constant. b. What is the length of the spring when the load is in increased by by 150 g? [assume that g = 10 N kg-1]
Solution
Elastic
potential energy
Elastic potential energy is the energy stored in
a spring when it is extended or compressed
Relationship
Area
between work and elastic potential energy
under the graph
= work done = ½ Fx
So,
lastic E lastic
potential energy = ½ Fx
example x = 15 8 = 7 cm = 0.07 m Force act to the spring, F = 5 x 10 = 50 N E lastic lastic
potential potential energy = ½
Fx = ½ 50 (0.07)
= 1.75 J
Factors
that effect effect elasticity
Summarise
the four factors that affect elasticity
Fact ctor or
Cha hang ngee in factor
Effec ectt on el elaast stiicity
Shorter spring
Less elastic
Longer spring
More elastic
Diameter of
Smaller diameter
Less elastic
spring
Larger diameter
More elastic
Diameter of
Smaller diameter
Mor oree elast stiic
spring wire
Larger diameter
Less elastic
Length
Type of
the elasticity changes with the type of
material
materials
Exercise 1 A 6 N force on a spring produces an extension of 2 cm. What is the extension when the force is increased to 18 N? State any assumption you made in calculating your answer.
SOLUTION : T o
solve the problem, determine the spring constant constant to to use the formula F = k x F = 6 N , x = 2 cm F = kx 6 = k (2) k = 3 N cm-1