form 4 physics chapter 2.5-2.9 - teacher's copy

7/30/2019 Form 4 Physics Chapter 2.5-2.9 - Teacher's Copy

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2.5 UNDERSTANDING THE EFFECTS OF A FORCE

A student is able to :

describe the effects of balanced forces acting on an object describe the effects of unbalanced forces acting on an object

determine the relationship between force, mass and acceleration i.e F = ma Solve problems using F = ma

1. Answer the following questions.

(a) What are the effects of a force when acting on an object?A force can.

(i) (iv)

(ii)

(iii)

(b) Force is a ( scalar / vector) quantity .

(c) What is the SI unit for force?

The S.I. unit for force is ____________ or ___________.

2. Fill in the blanks with appropriate answers for balanced and unbalanced forces.

(a) Forces acting on an object are said balanced when the net force is

___________.

(b) When the forces acting on an object are balanced, it is either at ___________

or moving with ___________ _____________.

(c) An object will ___________________ if the forces acting on it are not balanced.

3. (a) Draw the graph and state the relationship between acceleration, a, force, F andmass, m.

(i)

move a stationary object

stop a moving object

change the shape / size of an object

change the direction / speedof anobject

zero

rest

constant velocity

accelerate / decelerate

a

F

a

m

1

Newton, N kg m s-2


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(m is constant) (F is constant)

(ii) Relationship :

a is __________________ to F

Relationship :

a is __________________ to m

(iii) Combined the relationship:

F = k ; k is constant (k = 1)

F =

(b) The relationship between F, m and a is known as Newtons _____________ law

of ____________.

4. Solve the following.

(a) A force, F is required to move an object of mass 1000 kg with an acceleration of3 m s-2. Calculate F when(i) object is on a smooth surface(ii) object is on a surface where the frictional force is 200 N

Solution :

(i) F = ma = (1000) (3)

= 3000 N

(ii) F 200 = (1000) (3)F = 3000 + 200= 3200 N

FF

a =a = 3 m s-2

m =m = 1000 kgkg

F

Fm

1 a a

a

F m

Fm

m a

m a

second

motion

directly proportional inversely proportional

m a


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(b) A block of mass 20 kg is pulled along the ground by a force, F of 60 N. Thefrictional force is 10 N. Calculate the acceleration of the block.

Solution:F f = ma

60 10 = 20 a

a =20

50= 2.5 m s-2

(c) A car of mass 1200 kg which is travelling at 90 km h-1 comes to a stop in adistance of 50 m when the brakes are applied. What is the average braking forceof the car?m = 1200 kg , u = 90 km h-1, v = 0, s = 40 m, F = ?u = 90 km h-1 =25 m s-1

u=s

m

6060

1090 3

= 25 m s-1

v2= u2+ 2as0 = 252+ 2a(50)a = - 6.25 m s-2

F = ma= 1200 x 6.25= 7500.0 N

,


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2.6 ANALYSING IMPULSE AND IMPULSIVE FORCE


Explain what an impulsive force is Give examples of situations involving impulsive forces Define impulse as a change of momentum ie Ft=mv-mu Define impulsive force as the rate of change of momentum in a collision or

explosion i.e F= mv-mut

Explain the effect of increasing or decreasing time of impact on themagnitude of the impulsive force

Describe situations where an impulsive force needs to be reduced andsuggest ways to reduce it

Describe situations where an impulsive force is beneficial Solve problems involving impulsive forces

1. Define impulse and impulsive force by completing the following.

Situation Explanation

A ball of mass, m is kickedwith a force, F. The time ofcontact is t. The ballaccelerates from u to v.

From : F = ma

F = m

F =t

mv mu = change of

Impulsive force

Ft

1

F mumv

F =t

t

mumv=

Impulsive force is defined as the .........................

. in a collision orexplosion.

Impulse= force x time

Impulse = Ft == change of ..

SI unit for impulse is or ..

mv mu

t

mv mu

momentum

mv mu

rate of change of momentum

rate of change

of momentum

mv mumomentum

N s kg m s1


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2) (i) Fill in the blanks with appropriate answers.

Impulsive force is a force which acts over a very .. time interval

during .. and .

(ii) By using the figures given, determine whether the impulsive force acting is large orsmall.

3. Two eggs of the same mass are released from the same height, as shown inthe figure below. EggA falls on a concrete floor while egg B falls on a thickfolded towel.

large short

collision explosion

large large

largelargesmall

small


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Using Figure A and Figure B, compare the conditions of the eggs after the falland explain therelationship between the material where the eggs landed and the forceacting on the eggs upon landing.

4. A tennis ball of mass 45 g travels at a velocity of 70 m s1 immediately after it isstruck by a racket. The time of contact between the racket and thetennis ball is 0.5 ms. Calculate (a) impulse and (b) impulsive force exerted on theball.

Solution :

(a) impulse = mv mu

= (0. 045)(70) (0. 045)(0)= 3.15 N s

(b) impulsive force, F = 3105.0

15.3

= 6.3 x 103 N

Figure A Figure B

Egg A cracked while egg B did not. Egg A fell on hard surface but egg B fell on

soft surface .The time of impact for hard surface is shorter and produce

a larger impulsive force.


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2.7 BEING AWARE OF THE NEED FOR SAFETY FEATURES IN VEHICLES


Describe the importance of safety features in vehicles

Safety features Importance

Padded dashboard To cushion an impact and increases the ______ __________ ofcollision so the __________ _______ produced is thereby reduced

Head rest To prevent injury to the necks when the car is hit.

Shatter-proof windscreen To prevent the windscreen from shattering.

Automatic air bagActs as a cushion to lessen the chance of head and body injuries inan accident.

________________________(ABS)

Shatter proof___________

Side impact bars

Automatic air______

Head______

Reinforced passenger compartment

Crumple zones

Safety seat________

Padded___________dashboard

bag

rest

belt

Anti lock braking

system

windscreen

time interval

impulsive force


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Safety seat belt To prevent the passenger from being thrown out of the car.

Side impact bar To minimize the force acting from a side- on collision.

Anti-lock braking system To prevent wheel lock and skidding, thus contributing to saferbraking

Crumple zoneTo increase the time interval of impact so that the resultantimpulsive force is reduced.

2.8 UNDERSTANDING GRAVITY


Explain acceleration due to gravity

State what a gravitational field is Define gravitational field strength Determine the value of acceleration due to gravity Define weight (W) as the product of mass (m) and acceleration due to gravity

(g) ie W=mg

Solve problems involving acceleration due to gravity

1. Complete the following by fill in the appropriate answer.

a. An object will .. to the surface of the earth because it is pulled down by

..

b. The pull or force of gravity also known as c. When an object falls without encountering any resistance and under the force

of . only, the object is said to be .

d. All objects freely with the same

acceleration regardless of their and

e. The constant .. of objects falling freely dueto the force of .. is knownas

, g.

f. Practically, a free . can only take place in

. where there is no air resistance.

fall

the force of gravity

Earths gravitational force

gravity

acceleration

gravity gravitational

fall

vacuum

vacuum

stone

featherfall constant

mass shape

free falling

acceleration


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f. The region around the earth is the . fieldof the earth.

g. The gravitational field strength is defined asthe .

acting on a 1 kg mass.

h. When an object is falling freely, acceleration due to gravity, g =

An object at the surface of the earth experiences gravitational force, g =

2. State the differences between weight and mass in the table below.

Weight Mass

1. the gravitational force acting on the object.

2. depends on the acceleration due to gravity

3. vector quantity

4. derived quantity, weight, W = mg

5. SI unit : Newton (N)

1. the quantity of matter in an object

2. is a constant quantity (everywhere)

3. scalar quantity

4. base quantity, mass = m

5. SI unit : kilogram, kg

Solve the problems below 3. A coconut falls from rest and hit the ground after 1.2 s.(a) What is the velocity of the coconut just before hits theground?(b) Calculate the height of the coconut before it fall.

Solution:

t = 1.2s , g = 10 m s-2, u = 0 m s-1 , v = ?

a) v = u + atv = 12 m s -1

b) s = ut + gt2= 0 + (10) (1.2)2

= 7.2 m

gravitational

gravitational force

10 m s-2

10 N kg-1


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4. A stone is thrown vertically upwards at a velocity of 20 m s-1. If g = 10 m s-2,calculate(a) the maximum height reached(b) the time taken for the stone to return to its original position.

u = 20 m s -1, v = 0 m s-1 , g = -10 m s -2

(a) v2= u2+ 2gs0 = (20)2+ 2 (-10) s

20s = 400s = 20 m

(b) g =t

uv

t =g

uv = s2

10

200=

time to go up and return = 4 s

2.9 IDEA OF EQUILIBRIUM FORCES


Describe situations where forces are in equilibrium State what a resultant force is Add two forces to determine the resultant force Resolve a force into the effective component forces Solve problems involving forces in equilibrium

1. Complete the following.

i) Forces that act on an object are said to be in .. when the object is. or is moving at .

ii) The net force that acts on an object when two or more forces act on it is knownas the .

iii) When . is reached, the resultant force acting on the objectis

iv) Newtons law of motion states that to every . there is an equalbut opposite ..

v) A force can be resolved into component which are to oneanother.

Horizontal component, Fx =

Vertical component , Fy =

F

Fx

Fy

equilibrium

stationary constant velocity

resultant force

equilibrium

zero

third action

reaction

perpendicular

F cos

F sin


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2. In the table below, mark and label all the forces that act on the object.

i) A box is at rest on a table.

Weight, W = Reaction R

ii) An object that is suspended by a rope.

Weight, W = Tension, T

iii) An airplane flying at a constant velocity.

Lift, R = Weight, WThrust F1 = Drag F2

iv) A trolley being pushed at a constant

velocity

Reaction, R = Weight, WPushing force F1 = friction f

v)

Solution :

(a) Fx = F cos = 60 cos 300

= 51.96 N

A box of mass 4 kg is being pulled by a forceof 60 N at an angle of 30 0 to the surface of afloor. Calculate(a) the component ot the force which causesthe box to move forward.(b) the acceleration of the box if the frictionalforce between the floor and the box is 5 N,

(b) Resultant force = 51.96 5= 46.96 N

46.96 = 4 aa = 11.74 m s -2

30 0

60 N

4 kg

W

R

W

T

F1F2

WW

R

f

F

R

Fy

Fx

f


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vi) A man pushes a box of mass 40 kg is up aninclined plane. The inclined plane makes an angleof 30 0 with the horizontal floor. The frictional forceacting between the inclined plane and the box is

120 N. If the man pushes the box with a force of400 N, calculate the acceleration of the box.

Solution : frictional force, f = 120 Nmg sin = 40 (10) sin 300= 200 N

Resultant force, F = 400 (200 + 120) F = ma= 80 N 80 = 40 a

a = 2 m s-2

vii)

Solution :

a) Resultant force, F = 60 15= 45 N

b) 45 = (6 + 4) aa = 45 / 10

= 4.5 m s -2

A 4 kg trolley is connected by a rope to a load of

mass 6 kg. The friction between the table and thetrolley is 15 N. The load is then released.

Assuming that the pulley is smooth and the rope isof negligible mass, find(a) the resultant force that act on the system(b) the acceleration of the system(c) the tension of the rope

c) 60 T = ma60 T = 6 (4.5)

T = 27 N

viii) Two loads of mass 3 kg and 5 kg are connected by arope which passes through a smooth pulley. If the

system is released from rest, calculate the accelerationof the 5 kg load.

Solution :

50 30 = (5+3) aa = 20 / 8

= 2.5 m s-2

6 kg

4 kg Smooth pulley

3 kg

5 kg

mg sin

W = mg

mg cos

400 N

W

T

T

f

30 N

T

T

f


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3 Forces In Equilibrium

i A block of mass 6 kg is suspended vertically by astring tied at O to the string AOB. If the system is inequilibrium, draw a vector diagram (triangle offorces) and hence determine the tension of stringOA and OB.

Solution :

sin 300=1

60

T

T1 =030sin

60

= 120 N

tan 300=2

60

T

T2= 030tan

60= 103.9 N

ii) A lamp of mass 1.5 kg is hung from a beamas shown in the diagram. Calculate the tensionin the rope.

Solution:

T = 2T sin 700

Therefore, mlampg = 2T sin 700

30 0A

OB

6 kg

50 N

T1

T2

W

T1

T2

W

T1 W


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T =0

lamp

2sin70

gm

= 02sin701.5(10) = 7.98 N

1.5 kg

form 4 physics chapter 2.5-2.9 - teacher's copy

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