Lesson plan

Topicmd2feddeefd6f8a41_1528449000663_0Topic

DynamicsdynamicsDynamics - summary

Levelmd2feddeefd6f8a41_1528449084556_0Level

Second

Core curriculummd2feddeefd6f8a41_1528449076687_0Core curriculum

Learning objectives - general requirements: 1) Using physical concepts and quantities to describe phenomena and to indicate their examples in the surrounding reality.

II. Motion and forces. The student:

13) describes the interaction of bodies using the third Newton’s law;

14) analyses the behaviour of bodies based on the first Newton’s law;

15) uses the concept of mass as a measure of inertia of bodies; analyses the behaviour of bodies based on the second Newton’s law and applies to the calculation the relation between force and mass and acceleration;

16) describes free fall as an example of uniformly accelerated motion;

17) uses the concept of gravity; uses the relation between force, mass and gravitational acceleration for calculations.

Timingmd2feddeefd6f8a41_1528449068082_0Timing

45 minutes

General learning objectivesmd2feddeefd6f8a41_1528449523725_0General learning objectives

Consolidation of knowledge and abilities concerning dynamicsdynamicsdynamics.

Key competencesmd2feddeefd6f8a41_1528449552113_0Key competences

1. Using Newton’s laws to describe the movement of bodies.

2. Calculating acceleration from the second Newton’s law.

Operational (detailed) goalsmd2feddeefd6f8a41_1528450430307_0Operational (detailed) goals

The student:

- describes the movement of bodies using Newton’s laws,

- calculates acceleration from the second Newton’s law.

Methodsmd2feddeefd6f8a41_1528449534267_0Methods

1. Student exercises.

Forms of workmd2feddeefd6f8a41_1528449514617_0Forms of work

1. Individual work.

2. Work with the whole class.

Lesson stages

Introductionmd2feddeefd6f8a41_1528450127855_0Introduction

Task 1

a) Provide the content of I, II and III Newton’s law.
b) Give what is the measure of the inertia of the bodies.
c) List the types of frictionfrictionfriction forces.
d) Specify what the friction force depends on.
e) Provide ways to change the friction force.

Answer:

First Newton's law:
If no force or forces act on the body, the body remains at rest or moves in a uniform linear motion relative to the inertial reference frame.

Second Newton's law:
If the body has a constant unbalanced forceunbalanced forceunbalanced force (or net force), the body moves in a uniformly accelerated motion with acceleration directly proportional to the applied force and inversely proportional to the mass of this body.

Third Newton's law:
When the body A acts on the body B with a certain force ${\vec{F}}_{A B}$ , the body B acts on the body A with the force ${\vec{F}}_{B A}$ with the same magnitude but opposite direction. These forces cannot be balanced because they are applied to two different bodies.

{\vec{F}}_{A B} = - {\vec{F}}_{B A}

The third Newton's law is often called action‑reaction law.

b) The measure of body's inertia is its mass.

c) We distinguish two basic types of frictionfrictionfriction: static friction and dynamic friction.

d) The friction force depends on how large is the force which presses the two surfaces together and on the type of the two surfaces.

e) The friction force can be changed by varying the force which presses the surfaces together or by changing the coefficient of friction by changing the roughness of the surfaces (e.g. by smoothing) or by using a lubricant to reduce friction, grease or oil.

Proceduremd2feddeefd6f8a41_1528446435040_0Procedure

Task 2

Check what the different letters in the formula mean:

[Interactive graphics]

Task 3

The Earth attracts a body with a mass of 2 kg with a force of 20 N, and a body with a mass of 0,5 kg with a force of 5 N. Calculate the acceleration that these bodies get.

Analysis:

Given:

m = 2 kg
F = 20 N
mIndeks dolny 1 = 0,5 kg
FIndeks dolny 1 = 5 N

Unknown:

aIndeks dolny 1 = ?
aIndeks dolny 2 = ?

Solution:

$a = \frac{F}{m}$

$a = \frac{20 N}{2 k g}$

$a = 10 \frac{N}{k g}$

$a_{1} = \frac{F_{1}}{m_{1}}$

$a_{1} = \frac{5 N}{0, 5 k g}$

$a_{1} = 10 \frac{k g \cdot \frac{m}{s^{2}}}{k g} = 10 \frac{m}{s^{2}}$

Answer:

Acceleration with which the body falls to Earth is $10 \frac{m}{s^{2}}$ , it is gravitational acceleration.

Task 4

The forces shown in the figure act on a block with a mass of 10 kg moving on a horizontal surface. Calculate the acceleration of this block.

[Illustration 1]

Analysis:

Given:

FIndeks dolny 1 = 10 N
FIndeks dolny 2 = 20 N
FIndeks dolny 3 = 30 N
m = 10 kg

Unknown:

a = ?

Solution:

$a = \frac{F_{w}}{m}$

$F_{w} = F_{2} + F_{3} - F_{1}$

$F_{w} = 20 N + 30 N - 10 N$

$F_{w} = 40 N$

$a = 4 \frac{k g \cdot \frac{m}{s^{2}}}{k g} = 4 \frac{m}{s^{2}}$

Answer:

The block acceleration is $4 \frac{m}{s^{2}}$ .

Task 5

The platform on which there is a 200 kg box moves with an acceleration of 2,5 $\frac{m}{s^{2}}$ . What is the frictionfrictionfriction force if the box does not move on the platform? What fraction of the normal forcenormal forcenormal force is the friction force in this case?

Analysis:

Given:

m = 200 kg
a = 2,5 $\frac{m}{s^{2}}$
g = 10 $\frac{m}{s^{2}}$

Unknown:

FIndeks dolny T = ?
FIndeks dolny N = ?

Solution:

$a = \frac{F}{m}$

$F = a \cdot m$

$F_{T} = F = 2, 5 \frac{m}{s^{2}} \cdot 200 k g$

$F = 500 N$

$F_{g} = F_{N} = m \cdot g$

$F_{N} = 200 k g \cdot 10 \frac{m}{s^{2}}$

$F_{N} = 2000 N$

$\frac{F_{T}}{F_{N}} = \frac{500 N}{2000 N} = \frac{1}{4}$

$F_{T} = \frac{1}{4} F_{N}$

Answer:

The magnitude of the friction force is 500 N. The friction force is $\frac{1}{4}$ of the normal forcenormal forcenormal force.

Task 6

Passengers on a moving bus have shifted to the right, this means that:

a) the bus suddenly braked,
b) the bus suddenly moved forward,
c) the bus suddenly turned right,
d) the bus suddenly turned left.

Answer:

a) True. b) False. c) True. d) True.

Task 7

The body moves with a constant acceleration of 3 $\frac{m}{s^{2}}$ . This means that:

a) body speed increased by 1 $\frac{m}{s}$ in 3 seconds,
b) body speed increased by 6 $\frac{m}{s}$ in 2 seconds,
c) body speed increased by 3 $\frac{m}{s}$ in 3 seconds,
d) body speed increased by 3 $\frac{m}{s}$ in 1 second.

Answer:

a) False. b) True. c) False. d) True.

Task 8

If a body with a mass of 1 m acts on a body with a mass of 4 · m, with a certain force it means that a body with a weight of 4 · m acts on a body with a mass of 1 m with a force:

a) 4 times larger,
b) 4 times smaller,
c) the magnitude of the force is the same in both cases,
d) the magnitude of the force is the same, but opposite direction,
e) the magnitude of the force cannot be determined because there not enough information.

Answer:

a) False. b) False. c) True. d) True. c) True.

Task 9

Evaluate whether the sentences are true or false.
a) The direction of the friction force is always opposite to the direction of the velocity of the body.
b) If you want to stop the car, use force.
c) The friction that acts during movement is greater than the friction that must be overcome to set the body in motion.
d) When we stand on the Earth, we press on the Earth and the Earth does not press on us.
e) The car moves at a constant speed when the net force acting on it is constant and greater than zero.
f) The force of 10 N gave an acceleration of 2 $\frac{N}{k g}$ to a body with a mass of 0,5 kg.md2feddeefd6f8a41_1527752263647_0Evaluate whether the sentences are true or false.
a) The direction of the friction force is always opposite to the direction of the velocity of the body.
b) If you want to stop the car, use force.
c) The friction that acts during movement is greater than the friction that must be overcome to set the body in motion.
d) When we stand on the Earth, we press on the Earth and the Earth does not press on us.
e) The car moves at a constant speed when the net force acting on it is constant and greater than zero.
f) The force of 10 N gave an acceleration of 2 $\frac{N}{k g}$ to a body with a mass of 0,5 kg.

Answer:

a) True. b) True. c) False. d) False. e) False. f) False.

Task 10

Three blocks with a mass of m = 3 kg each, move without frictionfrictionfriction on a horizontal path. The acceleration magnitude in this movement is:

[Illustration 2]

a) $0, 25 \frac{m}{s^{2}},$
b) $4 \frac{m}{s^{2}},$
c) $6 \frac{m}{s^{2}},$
d) $12 \frac{m}{s^{2}},$
e) $24 \frac{m}{s^{2}} .$

Answer:

a) False. b) True. c) False. d) False. e) False

Lesson summarymd2feddeefd6f8a41_1528450119332_0Lesson summary

Dynamics describes the movement of a physical body under the influence of forces. The first Newton's law is that in the inertial reference frame a body can change its speed only an external force acts on it. The second Newton's law determines that the acceleration a which the body with the mass m under the influence of the unbalanced force F is experiencing is equal to:

a = \frac{F}{m}

The third Newton's law is that bodies always act on each other with equal forces but on opposite directions. Each action is accompanied by reaction equal in magnitude, but with opposite direction.
All bodies during their movement experience the forces of resistance to movement. Friction is the resistance to motion (force) associated with the interaction occurring on the contact surface of two bodies when a force acting on the body is trying to set the body in motion. To calculate the friction forces we can use the following formula:

FIndeks dolny T = f · FIndeks dolny N

where:
FIndeks dolny T [N] - friction force;
FIndeks dolny N [N] - normal force;
f - coefficient of friction.

Selected words and expressions used in the lesson plan

action forceaction forceaction force

balanced forcebalanced forcebalanced force

dynamicsdynamicsdynamics

frictionfrictionfriction

normal forcenormal forcenormal force

reaction forcereaction forcereaction force

resistance to motionresistance to motionresistance to motion

unbalanced forceunbalanced forceunbalanced force

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dynamics1

friction1

unbalanced force1

normal force1

action force1

balanced force1

reaction force1

resistance to motion1

Scenariusz

Topicmd2feddeefd6f8a41_1528449000663_0Topic

Lesson stages

Selected words and expressions used in the lesson plan