Fair • Fg Free body diagram
GRADE 11 SUBJECT Physical Sciences WEEK 4 TOPIC Newton second law-Acceleration: Time 60 mins Lesson
LESSON SUMMARY FOR: DATE STARTED: DATE COMPLETED:
LESSON OBJECTIVES
At the end of the lesson learners should be able to:
• Calculate the acceleration of a single object on which several forces act simultaneously • Calculate the acceleration of two objects that are joined together by a string
TEACHING and LEARNING ACTIVITIES 1. TEACHING METHOD/S USED IN THIS LESSON:
Induction method, Demonstration method, Question and Answer method
2. LESSON DEVELOPMENT 2.1 Introduction
a) PRE-KNOWLEDGE learners need understanding of the following:
(i) Newton’s laws of motion (ii) Free body diagram
b) BASELINE ASSESSMENT (educator to design a worksheet/ transparency or write questions on the board [preferably a worksheet to save time] to gauge the learners memory of their relevant prior knowledge)
QUESTIONS for the BASELINE ASSESSMENT [5 min]
i) State Newton’s second law in two different ways ii) What does constant resultant force do to an object
iii) What is the relationship between force and acceleration? Elaborate Answers
i) Newton's Second Law of Motion states that the rate of change in momentum of the body is directly proportional to the net force applied
Constant resultant force produces acceleration in the direction of force. Acceleration is directly proportional to the applied net force, Acceleration is inversely proportional to mass.
ii) Produces acceleration in the direction of force
2.2 Main Body (Lesson presentation) [30 min] From the equation
t
v
m
F
net∆
∆
=
, it can be noted thatF
net=ma
sincet
v
a
∆
∆
=
. This is only applicable when the mass is constant. When a resultant force is applied to a body it produces an acceleration that is directly proportional to the resultant force and inversely proportional to the mass of the body But the former equation based on the rate of change of momentum is applies to both. From F= ma, Newton second law can be stated .An unbalanced or resultant force produces an acceleration in the direction of force. This acceleration is directly proportional to the force that produced it, and acceleration is inversely proportional to the mass of an object. It defines the relationship of Force, Mass and Acceleration.
Force = mass x acceleration F = ma
A force of one newton will give a mass of one kilogram an acceleration of 1 m∙s-2. Examples
1. A trolley with a mass of 15 kg is accelerated at 3 m∙s -2. What force is needed to do this? solution F = ma
= 15 x 3 = 45 N
2. A force of 20N is used to accelerate a 250 g mass. What is its acceleration? F = ma 20 =
1000
250
x a a = 80 m∙s-22.2 Learners Activities [15 min]
2.2.1 A car of mass 800kg is pulling a trailer of mass 200kg along a straight road. The car is connected to the trailer by a tow bar which can be modeled by a light inextensible rod. The driving force of the car is 2.6kN and the total resistance to motion of the car and trailer is 600N.
a) Determine the acceleration of the system.
b) Given that the resistances of the car and trailer are proportional to their masses. Determine the tension in the tow bar
c) The driver sees an accident ahead and applies the brakes causing the car to decelerate at 2
3m⋅s
− . Determine the braking force applied and the force acting in the tow bar.2.2.2 The dung beetle below is accelerating the dung ball at 0.1 m∙s-2. What is the force the dung beetle would have to apply to accelerate this 0.001 kg dung ball at 0.1 m∙s-2?
.
www.wikipedia.org
2.2.5 Mass of weight 1.2kg and 1.0kg hang at the ends of a light rope passing over a light frictionless pulley. Find the acceleration of the mass and the tension in the rope. Corrections
2.2.1 The total mass is .
2.2.4 An inventive child wants to reach an apple in a tree without climbing the tree. Sitting in a chair connected by a rope that passes over a frictionless pulley, he pulls on the loose end of the rope with such a force that the spring scale reads 250. N. The child's true weight is 320. N, and the chair weighs 160. N
a) Show that the acceleration is upward and find its magnitude b) Find the force the child exerts on the chair.
Newton’s Second Law states that
Thus Acceleration
As the resistance to motion of the car and trailer are proportional to the masses, then The Resistance of the car is
The Resistance of the trailer is
Therefore the tension in the tow -bar is Instead of considering the Trailer we could have applied Newton's Second Law to the Car
2.2.2 F = ma = 0,001 x 0,1
= 1,00 x 10-4m∙s-2 to the direction of motion 2.2.3 F = ma
625 = 25a
a = 25 m s-2 forward
2.2.4 a) Since the total weight of the system is 480.N, the total mass of the system = 480. N / 9.81 m/s2 = 49.0 kg
Taking upward as positive, the acceleration of the system is found using the second law. ∑Fy = 2T - mg = may
Solving for ay,
ay = [(250. N + 250. N) - 480. N] / 49.0 kg ay = + 0.408 m/s2
(positive sign indicates upward)
b) The downward force the child exerts on the chair has the same magnitude as the upward normal force exerted on the child by the chair.
Applying the second law, ∑Fy = T + Fn - mchildg = mchilday Fn = mchilday + mchildg - T
Reflection/Notes:
Name of Teacher: HOD:
Sign: Sign: Date: Date: Fn = (320. N / 9.8 m∙s-2)(0.408 m∙s-2) + (320. N) - (250. N) Fn = 83.3 N 2..2.5 2.3 Conclusion
Activity to Re-enforce lesson (Educator may summarise the main aspects of the lesson) [5 min.]
HOMEWORK QUESTIONS/ ACTIVITY (educator must give learners a few questions to answer at home by either writing them on the chalkboard or giving an exercise from the prescribed textbook) [30 min]