# Lesson Notes By Weeks and Term - Senior Secondary 1

Young Modulus

Term: 3rd Term

Week: 1

Class: Senior Secondary School 1

Age: 15 years

Duration: 40 minutes of 5 periods each

Date:

Subject:      Physics

Topic:-       Young Modulus

SPECIFIC OBJECTIVES: At the end of the lesson, pupils should be able to

1. Define stress and strain
2. Calculate the energy stored in an elastic string

INSTRUCTIONAL TECHNIQUES: Identification, explanation, questions and answers, demonstration, videos from source

INSTRUCTIONAL MATERIALS: Videos, loud speaker, textbook, pictures

INSTRUCTIONAL PROCEDURES

PERIOD 1-2

 PRESENTATION TEACHER’S ACTIVITY STUDENT’S ACTIVITY STEP 1 INTRODUCTION The teacher reviews the previous lesson on the elasticity in solids Students pay attention STEP 2 EXPLANATION He defines stress and strain and explains how the young modulus is calculated Students pay attention and participates STEP 3 DEMONSTRATION He solves exercises on how to find the energy stored in an elastic string Students pay attention and participate STEP 4 NOTE TAKING The teacher writes a summarized note on the board The students copy the note in their books

NOTE

YOUNG MODULUS

This is the ratio of tensile stress to tensile strain of an elastic material

Young Modulus =      Tensile stress

Tensile strain

STRESS: It is defined as the ratio of force to area

Stress =                     πΉππππ

πΆπππ π  ππππ‘πππππ ππππ

Its S.I. Unit is in NM-2

STRAIN: This is the ratio of extension to its natural length

Strain =             πΈπ₯π‘πππ πππ

πππ‘π’πππ πΏπππβπ‘

It does not have S.I. Unit

ENERGY STORED IN AN ELASTIC STRING

Energy stored = Work done during deformation of spring Energy stored (E) = Average force x extension

=       ½ F x e

Energy stored = ½ F e = ½ ke2

Example: A metal wire of length 2.5m and diameter 2.0mm is stretched by a force of 400N. If the force constant of the wire is 5000N/m and π = 227 calculate;

1. The extension of the wire
2. The tensile stress and strain on the wire
3. The young modulus
4. The work done by the wire

Solution:

L = 2.5m

d = 2.5mm = 0.0025m

F = 400N

K = 5000N/m

r = d        = 0.0025 = 0.00125m

2              2

(a)  F = ke

400 = 5000e

e = 0.08m

(b) Stress =      πΉππππ

πΆπππ π  ππππ‘πππππ ππππ

Stress     =       πΉ

ππ2

=       400

227 π₯ 0.001252

= 1.273 x 108N/m

(c) Strain  =     πΈπ₯π‘πππ πππ

πππ‘π’πππ πΏπππβπ‘

Strain =      0.08

2.5

= 0.032

(d) Young Modulus     = ππππ πππ ππ‘πππ π

ππππ πππ ππ‘ππππ

Young Modulus     = 1.273 π₯ 108

0.032

= 3.978 x 109Nm-2

(e) Work done = ½ Fe

= ½ x 400 x 0.08

= 16J

EVALUATION:   1. Define stress and strain

1. State the young modulus
2. What is the formular for finding stress and strain?
3. How do we calculate the energy stored?
4. A wire 2 m long and 2 mm in diameter, when stretched by weight of 8 kg has its length increased by 0.24 mm. Find the stress, strain and Young’s modulus of the material of the wire. g = 9.8 m/s²
5. A wire of length 2 m and cross-sectional area 10-4m² is stretched by a load 102 kg. The wire is stretched by 0.1cm. Calculate longitudinal stress, longitudinal strain and Young’s modulus of the material of wire.
6. A mild steel wire of radius 0.5 mm and length 3 m is stretched by a force of 49 N. Calculate

a) longitudinal stress

b) longitudinal strain

c) elongation produced in the body if Y for steel is 2.1 × 1011 N/m².

CLASSWORK: As in evaluation

CONCLUSION: The teacher commends the students positively