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2

2

2

2

v f = vo + 2 gH
Vo = v f − 2 gH
= (16 ft s ) − 2 (- 32 ft s 2 )(30 ft )
2

= 2176
vo = 46
...
6 ft s )
(2)(- 32)
= 33
...
9

maximum height reached by the ball

v f = vo + gt
gt = v f − vo
t=

v f − vo

g
0 − 16
=
- 32
= 0
...
5s

3
...
3

time to reach the highest point from a height of
30 ft above the ground

Motion on a Plane: Projectile Motion

Projectile motion is the motion of an object that is given an initial velocity from which it
travels freely under the action of gravity
...

Two coordinates must be used to describe the projectile’s motion since it moves horizontally
as well as vertically
...
3
...
1 Projectile Fired Horizontally ( = 0)
Let vo be the initial horizontal
velocity as it leaves its initial position,
gravity is the only force that acts on it
...
That is
...
3
...
2 Projectile Fired at an Angle
Let

2H
g

0

vx = horizontal component of vo
vy = vertical component of vo
H = maximum height
R = range

31

y

vx

A

H

vy
vx

B
x

A
R

At point A

x = (vo cos

y = (vo sin

)t
) t − 12 gt 2

At point B

y=0
0 = (vo sin

) t − 12 gt 2

2 vo sin
g
v sin
t1 = o
2
g
t=

total time of flight
half time

For the range R,

R = vxt = (vo cos
= (vo cos

)

)t

2 vo sin
g

2

=

2 vo sin cos
g
2

R=

vo sin 2
g

32

For the maximum height H,
H = (v o sin
= (v o sin

) t − 12 gt 2
)

v o sin
g

− 12 g

v o sin
g

2

v 2 sin 2
= o
2g

Sample Problems with Solutions:
1
...

s
a)
b)
c)

How many seconds will it fall to the ground?
How far from the base of the building will it hit the ground?
What is the velocity at the point in its path when it is midway from the top of the
building?

Solution:

2
...
08s
9
...
8 m s 2 )(10m ) = 14 m s

2H
= (8 m s )(4
...
64m
g

A bomb is dropped from an airplane traveling horizontally with a speed of 300

mi

h


...

Solution:
R = vot

vo = (300 mi h )

vo

5280 ft
1mi

1h
3600s

H

= 440 ft s

t=

2H
=
g

(2)(10,000 ft ) = 25s

R = (440 ft s )(25s )
= 11,000 ft

32 ft s 2

R

33

3
...


Find the range and the time of flight
...
1 ft

t=
=

2vo sin
g

(2)(150 ft s )(sin 450 )
32 ft s 2

= 6
...
8 ft
4
...

a) What was the initial velocity of the ball?
b) How high did it rise?
c) How long was it in the air?

34

Solution:

y

H

vy

300

vx
x

0

R = vxt = (vo cos

a)

(

)

R = 120 m

)t

120 = vo cos 30 t
t=

0

120
vo cos 300

y = (vo sin

) t − 12 gt 2
2
0 = (vo sin 300 ) t − ( 12 )(9
...
9)
0
vo cos 30
vo cos 300

2

(4
...
9)(120)2

= 1357
...
85 m s → initial velocity of the ball

2

b)

H =

vo sin 2
2g

=

H = 17
...
76s

(36
...
8

= 17
...
85)(sin 300 ) = 3
...
8

→

total time of flight of the ball

35

Lesson 3: Velocity and Acceleration
Assessment
Instruction; Show your complete and neat solution
...
See answer sheet format
...
An automobile accelerates from rest at 1 for 30 s, continues at constant speed for
2 min
...
What distance did it cover?

2
...
What
is his average speed for the entire trip?

3
...
It starts from rest and accelerates at 1
...


The bus continues at this speed until it

decelerates at 2 until it comes to a halt
...


4
...

(a) How long will it take the ball to reach the ground? (b) What will be its speed
when it strikes the ground?

5
...
Calculate (a) the total distance traveled and
m

s

6
...
The passenger train has an
acceleration of 80

cm 2
s


...


A ball rolls off the edge of a table top 1
...
Find (a) the time of flight, (b) the
initial velocity, and (c) the magnitude and direction of the velocity of the ball just
before it strikes the floor

8
...
22 m from the
ground
...
2 m high and 5
...
What should be the initiall velocity of the ball when it leaves the
boy’s hands so it will go into the basket?

36

LESSON 4: FORCE AND MOTION

Introduction
In the previous lesson, we described the motion of particles based on the definition of
displacement, velocity and acceleration
...

Learning Outcomes
After successful completion of this lesson, you should be able to:
To learn Newton’s laws of motion
To study the relationship between a force and the acceleration it causes
To solve problems related to the Newton’s second law of motion
To define frictional force and solve problems related to it

Discussion
4
...
If an object moves with
uniform motion (constant velocity), no force is required to maintain the motion
...

A force is an interaction that causes an acceleration of a body
...
A force
that accelerates that standard body by exactly 1 m s 2 is defined to have a magnitude of
one Newton (I N)
...
It is a
vector quantity
...
Some particular forces are described below:
1
...
The force is primarily due to an attraction, called gravitational
attraction, between the astronomical body and any object nearby
...


Normal Force — is the perpendicular force (perpendicular to the surface)
experienced by a body that is pressed against a surface, or pressed against
another body
...
2

3
...


4
...
A tension force pulls in the direction of the rope and is
exerted uniformly along its entire length
...
It is also defined as a measure of an object’s inertia
...

4
...
The three laws of motion are as follows:
1
...

In an equation form, if

F=0
then

a =0
2
...
The acceleration is directly proportional to the
vector sum of all the forces acting on the object and inversely proportional
to the object’s mass
...


Newton’s Third Law of Motion: The Law of Action and Reaction
Forces
Whenever one object exerts a force on another object, the second object
exerts a reaction force of equal magnitude but in opposite direction to the
first force
...

4
...

Kinematics is often used to relate an object’s acceleration to its changing velocity and
position
Problem-Solving Strategy
The following procedure is recommended when dealing with problems involving
the application of Newton’s Second Law:
1
...


2
...
Do
not include forces exerted by the object on its surrounding
...


3
...


4
...
You must have as many independent equations as
the number of unknowns
...


A 7
...
0 kg
...

a
...


What is the acceleration of the system?
What is the tension in the cord?

T

T

m2g

m1g

Given: m1 = 7
...
0 kg

Solution:
for m1:

for m2:
m1g – T = m1a

T – m2g = m2a

Therefore,
(m 1 g − T ) + (T − m 2 g ) = m 1 a + m 2 a

(m 1

− m 2 )g = (m 1 + m 2 )a
a =
=

(m 1

− m 2 )g

m1 + m 2

(7
...
0 kg )(9
...
6

2

7
...
0 kg
m

m1g −T = m1a

s

2

T = m1g − m1a

T = 57
...


An unbalanced force of 50 N acts on an object weighing 100 N
...


a=

F

F
,
m

F = 50 N,

W = mg

m=

Fg (50 N )(9
...
9 m s 2
w
100N

W
g

A constant horizontal force of 40 N acts on a body on a smooth horizontal surface
...

a
...
If the force ceases to act at the end of 5 s, how far will the body move in the
next 5 s?
S = 100 m
t=5s

F

m

Solution:
a)

F = ma

F =m a
S = v o t + 1 2 at 2 ,

m=

F
a

vo = 0

2 S 2 • 100m
=
= 8 m s2
2
t2
(5 s )
F 40N
m= =
= 5 kg
a 8 ms2
a=

b)

F=0

a=0
v = v o + at
= 0+8 m

s2

•5s

= 40 m s

4
...
What is

the tension in the supporting cable?

41

Solution:

Fy = ma
T − mg = ma
T = m(g + a )
= (2000 kg ) (9
...


A 100 g mass lies on a frictionless table and a cord is attached to one end as
shown
...
Find
a)
b)

the acceleration, and
the tension in the cord
T
M
T

mg

Solution:
mg – T = ma
where T = Ma
mg – Ma = ma
mg = (m + M)a
mg
a=
m+M
T = Ma
10 g • 980 cm s 2
= 100 g • 89 cm s 2
=
10 g • 100 g
= 8900 dynes
= 49 cm s 2

42

6
...
The two bodies, 8 N and 10 N
are tied at the ends of a cord that passes over a massless, frictionless pulley
...
We find
that:

Mg sin 370 − mg sin 300 = (m + M )a
a=
=

(Mg sin 37

)

− mg sin 300 g
mg + Mg
0

(10 N )(sin 370 ) − (8 N )(sin 300 ) 9
...
1

2

8 N + 10 N
m

s2

43

Now,

T − mg sin 300 = ma
T = mgsin30 0 + ma

(

)

T = (8 N ) sin 300 +

8N
(1
...
8 m s 2

= 4
...


A traffic light, weighing 100 N, hangs from a cable tied to two other cables fastened
to a support as shown in the figure below
...

370

530
T2

T1

T3

Solution:
Free-body diagram:

F =0
T3

T3 − w = 0
=w
= 100 N

Vector Diagram:
w

44

Fy = 0
T1 sin 370 + T2 sin 530 − T3 = 0 → eqn (1)

Fx = 0
T2 cos 530 − T1 cos 370 = 0
T2 cos 530 = T1 cos 370
T2 =

cos 370
T1
cos 530

= 1
...
33T1 )sin 530 − 100 N = 0
T1 = 60 N
T2 = 1
...
8 N

8
...
If the block starts from rest at the
top and the length of the incline is 2 m, find
a)
b)

= 15°, as shown in

the acceleration of the block, and
its speed when it reaches the bottom of
the inclined plane

Solution:
S=2m

Vo = 0

a
W cos

Vf = ?
W
W sin

45

F = ma
ma = w sin
2

m

V f − Vo

2

2s
2

Vf − 0

= mg sin 150
= (9
...
185 m s

4
...
There are two common types of friction:
static friction and kinetic friction
...


Static Friction — static frictional
F
force exists when an object does not
f
slide along a surface on which it rests
even through a force is exerted to
make it slide
...
The
force of friction in this case is called static because the box remains
stationary
...
Roughness is
measured by the coefficient of static friction s
...

b)

2
...
The larger the normal force, the harder it is to make
the object move
...
The word kinetic signifies that the object is moving
...

After an object, initially at rest, that is pushed or pulled, starts to move, less
force is usually needed to keep the object sliding than the force required to
make the object move
...

46

The effect of friction on the motion of an object is accounted for by defining a
coefficient of kinetic friction, k, a number less than the coefficient of static
friction
...
If the component of F parallel to the surface increases,,
then fs also increases
...


That is, fs

where N is the magnitude of the normal force
...

sN

✓ Once a body begins to slide along a surface, the magnitude of the frictional
force rapidly decreases to a smaller constant value given by fk = kN, where
uk < us
...


A box weighing 100 N starts to move across a horizontal surface when a horizontal
force of 25 N is applied to it, but a force of only 20 N is needed to keep it moving
in uniform motion
...


Solution:
a)

fs

s

N,

F1 =

s

N

k

b)

k

F1
N
N=W
s

N–W=0
F1
s =
W
25 N
=
100 N
= 0
...
20
=

x – axis: Fcos300 – f = 0
Fcos300 = f = N
y-axis:

Fsin300 + N – W = 0
N = W – Fsin300
Fcos300 = N = (W – Fsin300)

F cos 300
=
W − F sin 300

N

F
300

f

W
s

=

F2 cos 300
W − F1 sin 300

25 N cos 30
100 N − 25 N sin 300
= 0
...
19

48

2
...
Determine the magnitude of the applied force if the coefficient of kinetic
friction between the surfaces in contact is 0
...


F

300

3 ft

4 ft

Solution:

3
= 370
4
The forces acting on the block are as shown:
= arctan

Fx = 0

N

F

x-axis: Fcos300 – fWsin370 = 0
f k = kN
Fcos300 –

kN

300

y

– Wsin370 = 0
f

0

x

0

y-axis: Fsin30 + N – Wcos37 = 0
N = Wcos370 – Fsin300

F cos 300 −

W cos 370 − F sin 300 − W sin 370 = 0

k

F cos 300 +

k

F sin 300 = W sin 370 +
F=

(

W sin 370 +
cos 300 +

k

W cos 370

cos 370
0
k sin 30
k

(

)

50 Ns sin 370 + (0
...
25 sin 300
= 40
...


A 50 kg block rests on the floor
...
70 and 0
...

a)
b)

What is the minimum force needed to
move the block?
If the same force continues to push the
block after it starts sliding, what will be its
acceleration?

N
y
x

F
f

W

Solution:
a)

The forces acting are as shown in the preceding page
...
Hence, the minimum force needed to
move the block is F = sW = s mg = 0
...
8 m s 2 = 34
...
7 − 0
...
8 m
= 1
...


s2

or

s2
2
...
The coefficient of kinetic friction
between the block and the floor is 0
...
If a horizontal force of 180 N acts on the
block for 3 s, find the velocity of the block at the end of that time
...
25) 9
...
2 m
5
...
Find

b)

= 0 + (1
...
6 m s

2

Three blocks are connected
as shown
...
8) − 10(2)
= 78 N

10

W1

Fx = ma
m 2 a = T1 − T2 − F f
Fy = 0
N − W2 = 0

a

N
T1

N = W2

T2
Ff
W2

since

→ (1)

→ (2)

=

Ff
N

F f = N = W2

→ (3)

substituting (3) into (1),

51

(5kg )(9
...
8)cos 250 − (3)(9
...
425 = T2 − 26
...
425 = T2 − 26
...
575 = -75
...
655
solving for T2 by using equation (4)
- 68 = -T2 − 49(0
...
9 N
6
...
3
...
The masses are released
from rest
...


T1

2 kg

T2

3 kg

10 kg

52

Solution:

a

F = ma
m1a = T1 − W1

common
acceleration

N
2 kg

T1

→ (1)

Ff = W1

a
W1
3 kg

T1

T2

F = ma
m2 a = T2 − T1

→ (2)

frictionless

a

F = ma
m3a = W3 − T2

10

→ (3)

Adding equations (1), (2) and (3)

m1a + m2 a + m3a = T1 − W1 + T2 − T1 + W3 − T2
a (m1 + m2 + m3 ) = W3 − W1
a=
=

W3 − W1
m1 + m2 + m3

10kg (9
...
3(2kg )(9
...
14

2kg + 3kg + 10kg
m

s

Solving for T1 and T2:
T1 = 18
...
6 N

53

Lesson 4: Force and Motion
Assessment
Instruction; Show your complete and neat solution
...

See answer sheet format
...
A force of 20 N acts upon a body whose weight is 8 N
...
Two masses of 8 kg and 12 kg are tied to each end of a string passing over a massless
and frictionless pulley
...

3
...
What is the tension in the
supporting cable if the elevator is a) ascending and b) descending?
4
...
What is the resulting acceleration of the two blocks?
5
...
a) What is the minimum force that
will make the block move if the coefficient of static friction is 0
...
10?
6
...


7
...
Take the coefficient of
static friction between A and the table top
to be 0
...


b)

Block C is suddenly lifted off A
...
15
...
Starting from rest, the box achieves a velocity of 1
...
5 s
...


54



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