1. The document discusses Hooke's law, which states that the extension of a spring is proportional to the load applied, and defines it mathematically. 2. It provides examples of calculating spring force, potential energy, work done, and using Hooke's law to solve problems involving force, spring constants, and length changes. 3. Five sample problems are given applying Hooke's law to calculate unknown values like force required to stretch a spring a given amount or spring length under a specific force.

1. Name’s of Groups
:
1. Arventa
2. Atika
3. Aulia
4. Imam
5. M. Ababil
6. Sebma

2. Basic Competition :
1. Analizing the connection of Hooke’s
law with elasticity.

3. Indicator :
1. Determining connection of elasticity
material’s character with Hooke’s law.

4. Hooke’s Law
Based on the figure above, we will prove the
relationship between force and length increment.
For that purpose, recall the following equation :
F ⋅L
E=
A ⋅ ∆L

5. If length increment of the spring is expressed by
∆L = ∆x, then the equation above can be expressed as
follows :
 E. A 
F = ∆x
 L 
Where :
E : modulus of elasticity (N/m2)
A : cross-sectional area of the spring (m2)
L : initial length of the spring (m)
Because E, A, and L are constant in values, then
the equation above can be written as follows :
F = k ∆x
Where :
k :  E. A  : spring force constant (N/m)
 
 L 

6. According to the equation, we obtain the
relationship that length increment of a
spring is directly proportional to the force
acting upon it.
This equation is the mathematical
representation of Robert Hooke’s
statement, then known as Hooke’s law.
Robert Hooke was an English scientist who
stated “If the pulling doesn’t exceed the
spring elasticity limit, then the length
increment of the spring is directly
proportional of its pulling force.”

7. The force stored energy named the potential
energy.
Ep = 1/2 kx2
Definition :
Ep = potential energy (Joule)
k = spring contant (N/m)
x = stretch (m)
The difference of initial potential energy with
final potential energy known as work (w) to do by this
spring.
Ep2 – Ep1 = 1/2 kx2
Definition :
Ep1 = initial potential energy (Joule)
Ep2 = final potential energy (Joule)
w = work (J)
If initial potential energy value is 0, the effort will
have value as big as the final of potential energy.

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

9. Solution :
1. Known :
k = 100 N/m
x = 0,5 m
Question : Force (F)
Answer : F = kx
= 100 x 0,5
= 50 N

10. Exercise :
1. A spring will increase by 10 cm in length if
given a force of 10 N. What is the length
increment of the spring if given a force of 7
N?
2. The spring in the following figure is pulled
from 10 cm to 22 cm with a force of 4 N. If
the spring complies with the Hooke’s law, its
total length when a force of 6 N is applied on
it is.....

11. 3. A metal has the Young’s modulus of 4 x
106 N/m2, its cross-sectional area is 20
cm2 and its length 5 m. What is the
force costant of the metal?
4. What is the spring constant, if it is
given force of 400 N, and increases of
4 cm in length?
5. A spring of 15 cm in length is hung
vertically. Then it is pulled with a force
of 0,5 N so that its length becomes 27
cm. What is the length of the spring if
it is pulled with a force of 0,6 N?