Hooke's law states that the deformation of an elastic object is proportional to the applied force. For relatively small deformations, the amount of displacement is directly proportional to the deforming force. The concept is that the amount of force applied to a spring or elastic object is proportional to the amount of deformation. The greater the force applied, the more the object deforms through stretching or compression. Hooke's law is represented by the formula F=kx, where F is the applied force, k is the force constant, and x is the deformation.

1. HOOKE’S LAW

2. Hooke’s Law
• for relatively small
deformations of an
object, the
displacement or size of
the deformation is
directly proportional to
the deforming force or
load

3. Hooke’s Law
• Elastic behaviour of solids according to
Hooke’s law can be explained by the fact
that small displacements of their
constituent molecules, atoms, or ions from
normal positions is also proportional to the
force that causes the displacement

4. Hooke’s Law
• The concept of Hooke’s Law is that the
amount of force applied to a spring or
elastic object is proportional to the amount
of deformation (length of stretch or
compression). The greater the force
applied to an elastic object, the more
deformation (stretch or compression) there
is. With less force applied, there will be
less deformation in the spring.

5. the force on unit
areas within a
material that develops
as a result of the
externally applied
force.
Stress
Strain
the relative
deformation produced
by stress. For
relatively small

6. Formula
F=kx
Where:
F = force applied
k = force constant
x = amount of deformation
To determine whether there is a stretch or
compression in the spring, we use a method of signs. If
the spring is compressed, both “x” and “F” are
negative(-) and when stretched both are positive(+).

7. Sample problems
1. What is the force required
to stretch a spring whose
constant value is 100 N/m by
an amount of 0.50 m?

8. Solution:
1. What is the force required to stretch a spring whose
constant value is 100 N/m by an amount of 0.50 m?
Given:
F=?
k=100 N/m
x= 0.50 m
F=kx
F=(100N/m)(0.50m)
F=50 N

9. Sample problems
2. A force of 100 N is
stretching a spring by 0.2 m.
Calculate the force constant.

10. Solution:
2. A force of 100 N is stretching a spring by 0.2 m.
Calculate the force constant.
Given: Force F = 100 N,
Extension x = 0.2 m,
Force constant k = ?
k = F/x
k = (100N)/(0.2m)
k = 500 N/m.

11. END!