ppt for hookes law

1. Hooke’s Law and
Elasticity

2. Hooke’s Law
In the 1600s, a scientist called
Robert Hooke discovered a law
for elastic materials.
Hooke's achievements were
extraordinary - he made the
first powerful microscope and
wrote the first scientific best-
seller, Micrographia.
He even coined the word ‘Cell’.

3. Hooke's Law, elastic and plastic
behaviour
• If a material returns to its original size and shape when you remove
the forces stretching or deforming it (reversible deformation), we say
that the material is demonstrating elastic behaviour.
• A plastic (or inelastic) material is one that stays deformed after you
have taken the force away. If deformation remains (irreversible
deformation) after the forces are removed then it is a sign of plastic
behaviour.
• If you apply too big a force a material will lose its elasticity.
• Hooke discovered that the amount a spring stretches is proportional
to the amount of force applied to it. This means if you double the
force its extension will double, if you triple the force the extension will
triple and so on.

8. The elastic limit can be seen on the graph.
This is where it stops obeying Hookes law.

10. You can write Hooke's law as an
equation:
F = k ∆ x
Where:
• F is the applied force (in newtons, N),
• x is the extension (in metres, m) and
• k is the spring constant (in N/m).
• The extension ∆x (delta-x) is sometimes
written e or ∆l. You find the extension from:
• ∆x = stretched length – original length.

11. ___________ Law: The compression or elongation x of an
ideal spring from its ________________ position (x = 0)
is ____________________________to the applied force Fs.
Fs =
stretching or elongation: compression:
More F  more ____________ or __________________.
x = 0
x = 0
Fs =
Fs
x
Fs
x
directly proportional
kx
stretch compression
Hooke's
equilibrium

12. Hooke's Law is often written: Fs = -kx
This is because it also describes the force that the
_______________ exerts on an ___________ that is attached
to it. The negative sign indicates that the direction of
the spring force is always _____________ to the
displacement of the object
spring itself
stretched
spring:
compressed
spring:
object
opposite
undisturbed
spring
______________
position, Fs = __
equilibrium
-x
+x
Fs
Fs
x = 0
0
Fs ___ 0
Fs ___ 0
>
<

13. Ex. A weight of 8.7 N is attached to a spring that
has a spring constant of 190 N/m. How much
will the spring stretch?
Equation:
Given:
Unknown:
x
w/o weight
8.7
N
w/ weight
190 N/m
8.7 N
x = ?
k =
Fs =
Fs = kx
8.7 N = (190 N/m) x
x = 4.6 x 10-2
m

14. Key Definitions
• Hooke’s Law = The amount a
spring stretches is proportional
to the amount of force applied
to it.
• The spring constant measures
how stiff the spring is. The
larger the spring constant the
stiffer the spring.
• A Diagram to show Hooke’s
Law
F = k ∆ x

15. Fs = kx Fs
x
What quantity does the slope represent?
slope =
=
Compare to Fs = kx
Solve for Fs/x =
The slope represents _______________________________
direct
Ex: A force of 5.0 N
causes t0.015 m.
How far will it stretch
if the force is 10 N?
he spring to
stretch
2 (0.015 m)
= 0.030 m
y/x
Fs/x
the spring constant, k.
k
5
10
.015 ?

16. Fs
x
stiffer spring  _________ slope  _________ k
spring
A
spring
B
Ex. Comparing
two springs that
stretch different
amounts.
xB
Applying the same
force F to both springs
Which spring stretches more?
Which is stiffer?
xA
A
B
greater larger

17. ____________ PE - the energy stored in a spring when work
is done on it to stretch or compress it
PEs =
Ex. A spring with a spring constant of 370 N/m is
stretched a distance 6.4 x 10-2
m. How much elastic
PE will be stored in the spring?
PEs = (½)kx2
= (0.5)( 370 N/m)(6.4 x 10-2
m)2
= 0.76 (N/m)(m2
)
= 0.76 Nm
= 0.76 J
How much work was done to stretch the spring by
this amount?
W = PE = 0.76 J
(½)kx2
Elastic

18. • The spring constant measures how stiff the spring is.
• The larger the spring constant the stiffer the spring.
• You may be able to see this by looking at the graphs below:
k is measured in units of newtons per metre (Nm-1
).

19. Example
• A spring is 0.38m long. When it is pulled by a force of
2.0 N, it stretches to 0.42 m. What is the spring
constant? Assume the spring behaves elastically.
Extension, ∆x = Stretched length – Original length =
. 0.42m – 0.38m = 0.04 m
2.0N = k x 0.04m
So, k = 2.0 N
0.04 m
= 50 N m-1
F = k ∆ x

20. Elastic behaviour – Car Safety
• Elastic behaviour is very
important in car safety, as
car seatbelts are made from
elastic materials. However,
after a crash they must be
replaced as they will go
past their elastic limit.
• Why have seat belts that
are elastic?
• Why not just have very
rigid seatbelts that would
keep you firmly in place?
• The reason for this, is that it
would be very dangerous
and cause large injuries.
This is because it would
slow your body down too
quickly. The quicker a
collision, the bigger the
force that is produced.

23. Quiz no. 2
A. 0.10 B. 0.30 m C. 0.60 m D. 0.90 m
A. 0.09 J B. 0.30 J C. 0.60 J D. 0.90 J