Hooke's law

Name’s of Groups :

1. Arventa2. Atika3. Aulia4. Imam5. M. Ababil6. Sebma

Basic Competition :

1. Analizing the connection of Hooke’s law with elasticity.

Indicator :1. Determining connection of elasticity

material’s character with Hooke’s law.

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 :

LA

LFE

If length increment of the spring is expressed by ∆L = ∆x, then the equation above can be expressed as follows :

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 ∆xWhere :k : : spring force constant (N/m)

xL

AEF

.

L

AE.

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.”

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.

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?

Solution :

1. Known : k = 100 N/mx = 0,5 mQuestion : Force (F)Answer : F = kx

= 100 x 0,5= 50 N

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.....

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?