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Physics 20
Kinematics
General Outcomes:
Students will: describe motion in terms of
displacement, velocity, acceleration and
time.
Key Concepts: The following concepts
are developed in this unit and may also
be addressed in other units or in other
courses. The intended level and scope
of treatment is defined by the
outcomes.





Scalar quantities
Vector quantities
Uniform motion
Uniformly accelerated motion
Two-dimensional motion
©
Alberta Education
Program of Studies
Kinematics deals with how objects move.
(how far, how fast, how long)
 Scalar quantity is a quantity having magnitude
(size) only.
e.g. Temperature
Speed
Distance
25 0C
100 km/h
5 km
 Vector quantity is a quantity having both
magnitude (size) and direction.
e.g. Force
Velocity
Displacement
15 N down
100 km/h South
5 km West
Graphical Analysis and Uniform Motion
 Uniform motion = traveling in a straight
line at a constant speed.
 No acceleration is involved (pretty much
impossible in the “real” world)
 We can use graphs to describe and
analyze motion
 There are two types of graphs:
Displacement (distance) – time
Velocity (speed) – time
1.
2.
Displacement-time
graph (d vs t)
d
Speed-time graph (v vs t)
v
t
t
 Slope of a straight line graph = rise/run
 Slope of a d-t graph = speed (velocity) of
the object.
Average Velocity and Speed
 We can describe motion by using two
different types of velocity and speed:
average and instantaneous.
Average velocity = Displacement
Total time
Average speed =Total distance
Total time
15 m
5m
Displacement = 15 + (- 5) = 10 m (vector)
Distance = 15 + 5 = 20 m (scalar)
vave= d/t
v=d
t
Where:
or
d = vt
vave = average speed (velocity) in
m/s or km/h etc
d = distance (displacement) in m or
km etc
t = time in s or h etc
The instantaneous speed of a moving
object is the actual speed at which it is
moving at any given instant of time.
1. A car travels 200 km in 3.0 h and 300
km in 2.5 h. Find the average speed.
dt = 500 km
tt = 5.5 h
v=?
vave= dt/tt = 500/5.5 = 91 km/h
2. **An object travels at 80 km/h for 3.0 h
and 90 km/h for 2.0 h. Find the average
speed.
d1 = v1t1 = (80)(3) = 240 km
d2 = v2t2 = (90)(2) = 180 km
Thus,
dt= 240 + 180 = 420 km
and, vave = dt/tt = 420/5 = 84 km/h
3. A high school athlete runs 100 m in
12.20 s. What is the velocity in m/s
and km/h?
v = d/t = 100/12.20 = 8.2 m/s
= 8.2 x 3.60 = 29.5 km/h
4. A person walks 13 km in 2.0 h. What is
the person’s average velocity in
km/h and m/s?
vave= d/t = 13/2.0 = 6.5 km/h
= 6.5/3.6 = 1.8 m/s
5. A car travels at a speed of 87 km/h.
How far will it travel in 3.4 h?
d = vt = (87)(3.4) = 296 km
6. A car moves for 15 s with a constant
speed of 60 km/h. What is its
instantaneous speed at 8.0 s?
v = 60 km/h
Example
7. A car travels 300 km in 3.2 h and then
400 km in 4.8 h. Find the average speed
in km/h.
v = d/t = 700/8 = 87.5 km/h
Example
8. A person walks 200 m at 1.2 m/s and runs
200 m at 1.8 m/s. Find the average speed.
Example
9. A car travels at 80 km/h for 5.0 h and at
100 km/h for 3.0 h. Find the average
speed.
Distance-time graph
(Position-time graph)
The slope of a d-t graph equals speed.
d
Uniform motion
t
d (m)
b
a
t (s)
a – uniform motion (constant speed)
b - stationary
d (m)
b
a
c
t (s)
a - uniform motion
b- object is stationary
c - returning toward initial position
e.g.
Use the following data to draw a
displacement-time graph. Calculate the
slope.
Time Displacement
(s)
(m)
0
0
1
5
2
10
3
14.8
4
20
5
25.1
Speed-time graph
(For uniform acceleration)
 Slope of a v-t graph = acceleration
 “Area” under a v-t graph = distance
 Area of a rectangle = l  w
 Area of a triangle = ½ bh
Example
In the following diagram:
v (m/s)
20
A2
A1
5.0
10
t(s)
a) Find the acceleration for the first 5.0 s
a = slope = 20/5 = 4 m/s2
b) Find the distance traveled in 10 s.
*Find the “area” under the graph.*
A1 =1/2bh = ½(5)(20) = 50 m
A2 = lw = (5) (20) = 100 m
A = 150 m
d = 150 m
Example
v (m/s)
30
20
0
5.0 10 15 20 30
t(s)
a) Find the distance the moving object
travels between
1) t=0 and t=5.0 s
A = ½ bh = ½(5)(30) = 75 m
2) t = 5.0 s and t=10 s
A = lw = (5)(30) = 150 m
3) t =10 s and t=15 s
A1 = ½ (5)(10) = 25 m
A2 = lw = (5)(20) = 100 m
A= 125 m
4) t = 0 s and t=30 s (total distance
traveled)
A = 550 m
(b) Find the acceleration between 10 and
15 s.
a = slope = 30 –20 = - 2 m/s2
10 – 15
Assignment: p. 30 #3
p. 34 #1 p. 45 #18
ACCELERATION
Acceleration is the rate of change of speed
(velocity). Acceleration indicates a change
in velocity over a period of time.
Acceleration can be positive (speeding up)
or negative (slowing down). Negative
acceleration is also called deceleration.
a = v
t
or
a = vf - vi
t
a = acceleration in m/s2
v = change in velocity in m/s
t = time elapsed in s
Rearranged formulas:
t = vf - v i
a
at = vf - vi
vf = vi + at
Note: Velocity must be in m/s and
time in s.
e.g.
***The velocity of a car increases from 20
km/h to 25 km/h in 3.5 s. What is the car’s
average acceleration?
vi = 20 km/h = 5.6 m/s
vf = 25 km/h = 6.9 m/s
t = 3.5 s
a=?
a = vf - v i
t
= 6.9 – 5.6
3.5
= 0.37 m/s2
The common unit of acceleration is
m/s/s or m/s2.
The slope of a velocity-time graph equals
acceleration.
Uniform acceleration means that the
change in velocity is constant in every unit
of time.
e.g.
The velocity of a car increases from
2.0 m/s (N) to 16 m/s (S) in 3.5 s. What is
the car’s average acceleration?
vi = 2.0 m/s
vf = - 16 m/s
t = 3.5 s
a=?
a = vf - v i
t
= -16 - 2
3.5
a = - 5.1 m/s2
A truck accelerates uniformly at 0.6 m/s2
from rest for 25 s. What is its final speed?
a = 0.6 m/s2
vi = 0
t = 25 s
vf = ?
vf = vi + at
= 0 + (0.6)(25)
vf = 15 m/s
Use the following data to draw a
displacement-time graph. Calculate the
slope.
Time Displacement
(s)
(m)
0
0
1
1
2
4
3
9
4
16
5
25
Kinematics Formulas
(There are 5 basic formulas)
1. d = vt
(speed is constant – no acceleration)
2. a = vf –vi
t
3. d = vi + vf t
2
4. d = vit +1/2 at2
5. vf 2 = vi 2 + 2ad
a= acceleration in m/s2
vf=final velocity in m/s
vi=initial velocity in m/s
t= time in s
Example 1
The initial speed of an apple is 8.0 m/s and it
accelerates for 3.0 s at 5.0 m/s2. Find:
a) the distance traveled
vi = 8.0 m/s
t = 3.0 s
a = 5.0 m/s2
d=?
*use formula that has the 4 variables
in it*
d = vi t +1/2 at2
= (8)(3) + (½)(5)(3)2
= 24 + 22.5
d = 46.5 m = 47 m
b) the final speed
vf = vi + at
= 8 + (5)(3)
vf = 23 m/s
Example 2
A cart is initially at rest and it accelerates
uniformly to a speed of 50 m/s in 30 s.
Find:
a) the acceleration
vi = 0
vf = 50 m/s
a = vf – v i
t = 30 s
t
a=?
a = 50/30
a = 1.7 m/s2
b) the distance traveled
d = v i + vf t
2
d = 50
2
30
d = 750 m
Example 3
The velocity of a car increases from 20 km/h
to 25 km/h in 3.5 s. What is the car’s
average acceleration? What distance did it
travel?
vi = 20 km/h = 5.6 m/s
vf = 25 km/h = 6.9 m/s
t = 3.5 s
a=?
a = vf – vi
t
= 6.9 – 5.6
3.5
a = 0.40 m/s2
d = vi + vf t
2
= 5.6 + 6.9 3.5 = 21.9 m
2
Example 4
A car starts from rest and is accelerated
uniformly. If it travels 50 m in 4.0 s, find:
a) the acceleration
vi = 0
d = vit +1/2at2
d = 50 m
50 = (1/2)(a)(4)2
t = 4.0 s
a = 6.25 m/s2
a=?
b) the final velocity
vf = vi + at
vf = (6.25)(4)
vf = 25 m/s
or
vf 2 = vi 2 + 2ad
= 2(6.25)(50)
= 625
vf = 25 m/s
p. 47, 48, 50, 51, 52 #1, 2
p. 53 #1 - 5
Acceleration Due to Gravity
 Galileo – conducted many experiments
and gave a name to the force that “pulled”
objects to the
earth…GRAVITATIONAL FORCE
 Newton – expanded on Galileo’s ideas
 He showed that all masses, big or
small, gave a gravitational field or
force
 The bigger the object, the bigger the
field (force)
 Acceleration due to gravity on the surface
of Earth g = 9.81 m/s2
 This number is an average and can change
slightly depending on where you are on
the earth (distance from the centre of the
earth)
 All objects have the same acceleration due
to gravity in a vacuum.
 In a vacuum where there is no air particles,
all objects will fall at exactly the same rate
 Air resistance/drag does affect the free fall
of objects on earth but in many cases it is
negligible.
 free falling objects accelerate at a rate of
9.81 m/s2…that means it increases its
velocity by 9.81 m/s every second
Time (s)
0.0
1.0
2.0
3.0
Velocity
(m/s)
0.0
9.81
19.62
29.43
 When doing calculations: if an object is
falling, a = 9.81 m/s2
if an object is going up, a =  9.81 m/s2
Example 1
Calculate the speed of a free falling object,
starting from rest, after 4.0 s.
vi = 0
vf = vi + at
a = 9.81 m/s2
= (9.81)(4)
t = 4.0 s
vf = 39.2 m/s
vf = ?
a) How far did it travel during the
third second of motion?
d = 25 m
Example 2
Calculate the time it takes a free falling
object to reach a speed of 100 m/s.
vi = 0
t = vf - vi
a = 9.81 m/s2
a
vf = 100 m/s
t = 10.2 s
t=?
Example 3
Calculate the distance travelled by a free
falling object in 5.0 s.
vi = 0
d = vit + 1/2at2
a = 9.81 m/s2
d = (1/2)(9.81)(5)2
t = 5.0 s
d=?
d = 123 m
****Example 4
An object is thrown straight up with an
initial velocity of 15 m/s. How high will it
rise?
vi = 15 m/s
vf = 0
a = - 9.81 m/s2
d=?
vf2 = vi2 +2ad
0 = 152 + 2(-9.81)d
-225 = -19.62d
d = 11.5 m
d = v f2 – v i 2
2a
d = - 152
(2)(-9.81)
d = 11.5 m
p. 60 #1 p. 63 #3-5 p. 65 #11-14
Kinematics Examples
Example
An arrow is projected upwards from ground
level at a speed of 25.0 m/s.
a) how long will it rise?
b) how long will it be in the air?
c) how far will it rise?
d) find time where the arrow will have a
velocity of 15.0 m/s.
Example
A hot air balloon ascends with a velocity of
10.0 m/s. At a point when the balloon was
200 m above the ground, the balloonist fell
out (that’s right, no parachute!).
a) Calculate the velocity of the
balloonist just before impact.
b) Calculate the time it took to hit the
ground.
The Simple Pendulum
http://ww2.unime.it/weblab/mirror/ExplrSci/dswmedia/harmonic.htm
 A pendulum has two parts – the mass or
“bob” and the string.
L
L
 The mass goes through a “back and forth”
motion because of gravity
 The period (T) of a pendulum is the time
it takes to complete one cycle, that is the
time to go back and forth once
 What factors affect the period of a
pendulum?
 Wind (air) resistance – very small
effect
 mass of bob – no effect
 height of bob release (amplitude) –
no effect
 friction – very small effect
 Acceleration due to gravity – very
large effect
 Length of string – very large effect
Period is found using:
T = 2π L
g
g = 42L
T2
L = gT2
4π2
T = period of pendulum in s
L = length of pendulum in m
g = accel. due to gravity in m/s2
 = pi = 3.14
f=1
T
f = frequency in Hertz (Hz)
= number of swings in 1 s
1 = one
T = 1/f
Example 1
Find the frequency of pendulum that is 1.5
m long.
L = 1.5 m
f = 1/T
f=?
T = 2π L
g
= 2π 1.5
9.81
T = 2.46 s
f = 0.41 Hz
Example 2
Find the length of a pendulum that has a
period of 2.0 s.
T = 2.0 s
L=?
L = gT2
4π2
= (9.81)(2)2
(4π2)
L = 0.99 m
Physics 20 - Pendulum assignment:
1. Find the length of a pendulum that has a period of 2.5 s.
2. Find the length of a pendulum that has a period of 1.25 s.
3. What is the acceleration due to gravity at a location
where a 0.45 m pendulum has a frequency of 0.74 Hz?
4. The acceleration due to gravity on the moon is 1.6 m/s2.
How long of a pendulum would be required on the moon to
have a period of 1.0 s?
5. The bob of a pendulum has a mass of 0.25 kg. If this
pendulum is 1.0 m long, what is its frequency?
6. A simple pendulum has a length L of 0.74 m. Find:
a. the period of this pendulum
b. the frequency when length is 4L
c. the period when length is L/9
7. A pendulum has a period of 3.6 s. What is its length?
8. Find the length of a pendulum that has a period of 0.9 s.
9. Find the length of a pendulum that has a period of 3.7 s.
10. Find the length of a pendulum that has a frequency of 0.25 Hz.
11. What is the acceleration due to gravity at a location
where a 0.6 m pendulum has a frequency of 0.5 Hz?
12. Find the frequency of pendulum that is 0.75 m long.
Vectors – Finding Vector Components
 Scalar quantities involve only magnitude
(amount)
e.g. time = 30 s
 Vector quantities involve both magnitude
and direction .
e.g. displacement of 25 m North
 Vectors are drawn using arrows which
show magnitude in the length as well as
direction

Vectors can also be drawn on the
co-ordinate plane.
N
-x
+y
+x
+y
W
E
-x
-y
+x
-y
S
900
1800
00
2700


Vectors have direction so we use “+” and
“ - “ to indicate in which quadrant it is.
Each vector also has two components:
x-component and a y-component.
v
vy
vx
vx = cos ө (v)
vy = sin ө (v)
Note: angle ө must be with respect to
the x-axis.
e.g. Find the components of the following
vectors:

30 m at a bearing of 250



10 m at 450 N of W (of = from)
15 m at a bearing of 2200
20 m at 400 E of S
Note: “of” also means “from”.
Step 1 – Draw each vector starting from the
center of the grid.
10 m
450
30 m
250
500 (use angle closest to x-axis)
400
15 m
20 m
Step 2 – Sketch each vector separately and
find its components (include signs).
30 m
Y1
0
25
X1
X1 = cos 25 (30) = 27.2 m
Y1 = sin 25 (30) = 12.7 m
Y2
10 m
45
X2
0
x2 = cos 45 (10) = - 7.1 m
y2 = sin 45 (10) = 7.1 m
x3
Y3
400
15 m
X3 = cos 40 (15) = - 11.5 m
Y3 = sin 40 (15) = - 9.6 m
X4
500
20 m
Y4
x4 = cos 50 (20) = 12.9 m
y4 = sin 50 (20) = - 15.3 m
Assignment: Find the horizontal and vertical components of
the following vectors:






1.9 m at a bearing of 200
2.5 m at a bearing of 400
245 m at 700 W of N
200 m due North
340 m at a bearing of 3000
55 m at a bearing of 2150








650 km at a bearing of 430
66 m at a bearing of 1900
10.5 km at a bearing of 1400
335 cm at a bearing of 2850
45 m at 450 N of W
342 km at 300 E of S
200 m at 600 W of N
45 m at a bearing of 2150





345 m at 700 W of N
42 km at 400 S of E
3 m at 350 N of W
75 m at a bearing of 1800
7.5 km at 250 N of E





40 km at a bearing of 3400
10 km at a bearing of 2000
60 m at 650 S of W
800 km at a bearing of 1500
25 m at 700 W of N
Displacement
Distance vs. displacement – distance is how
far is actually traveled whereas displacement
is the shortest distance between initial and
final positions.
e.g. A car travels 300 km east and then 200
km west. Find the distance traveled by the
car and its displacement.
Distance = 500 km (scalar)
Displacement = 100 km East (vector)
Speed vs velocity
Speed = distance/ time
Velocity = displacement/time
Finding Resultants (displacement, velocity
or force)
Step 1 – Draw each vector starting from the
center of the grid.
Step 2 – Sketch each vector separately and
find its components.
Step 3 – Add all the x’s together and all the
y’s together.
Step 4 – Sketch the two vectors from step 3
starting with the x.
R
y
x
Calculate resultant using a2 + b2 = c2
Also you must find the angle:
tan ө = y
x
e.g.
Find the resultant displacement of the
following path:




30 m at a bearing of 250 then
10 m at 450 N of W then
15 m at a bearing of 2200 and finally
20 m at 400 E of S
10 m
450
30 m
250
500 (use angle closest to x-axis)
400
15 m
20 m
30
250
x1 = cos 25 (30)
= 27.2 m
y1 = sin 25 (30)
y1
x1
= 12.7 m
x2 = - 7.1 m
y2 = 7.1 m
x3 = - 11.5 m
y3 = - 9.6 m
x4 = 12.9 m
y4 = - 15.3 m
x = x1 + x2 + x3 + x4
= 27.2 + - 7.1 +-11.5 + 12.9 x = 21.5 m
y = y1 + y2 + y3 + y4
= 12.7 + 7.1 + -9.6 + - 15.3
21.5
5.1
y = - 5.1 m
R
R = 21.52 + 5.12
R = 22.1 m
tan Ө = 5.1/21.5
tan-1 (5.1/21.5)
ө = 130
****2. A plane travels at a speed of
300 km/h, at 450 N of E. The wind blows
eastward at 50.0 km/h. What is the velocity
of the plane with respect to the ground?
300 km/h
50 km/h
y1
x1 = 212.1 km/h
y1 = 212.1 km/h
x1
x2 = 50 km/h
y2 = 0
x = 262.1 km/h
y = 212.1 km/h
R
262.1
R = 337.2 km/h
Ө = 390
212.1
Vector Addition
1. Vectors in the Same Direction
 to add vector quantities that are in the
same direction
RESULTANT = SUM
Example
A person runs 25 m south and then another
15 m south.
a) What is the distance travelled?
b) What is the displacement?
Example.
A plane has a velocity of 150 m/s(E).
The wind blows east at 75 m/s. Find the
resultant velocity.
2.
Vectors in Opposite Directions
 to add vector quantities that are in
opposite directions.
RESULTANT = DIFFERENCE
Example
A plane flies 200 km north and then turns
around and comes back 150 km.
(a)What is the distance travelled?
(b)What is the displacement?
3.
Vectors at 90
 to add vectors that are at right angles
to each other, you must calculate:
a) the magnitude – use
Pythagorean Theorem c2 = a2 + b2
b) the angle – use SOHCAHTOA
c) the direction – compass
direction, “from the vertical”,
“from the horizontal”, “from the
original direction”, etc.
 let’s look at the compass directions:
How would you describe each of the
following vectors?
Example 1
A plane flies 200 km north and then heads
150 km east. Calculate the displacement.
(Draw a diagram!!!)
Example 2
A person walks 30 m south and then 20 m
west. Find her displacement.
Example 3
A disoriented dog runs 30 m north then 20
m east then 10 m west. Find his
displacement.
Example 6
A person goes up 20 m and then travels 15
m horizontally. Find
the displacement.
Example 7
A plane travels north at a speed of
300 km/h. The wind blows west at
50.0 km/h.
(a)What is the velocity of the
plane with respect to the
ground?
(b)How far off the course would the
plane be after 2.0 h?
(c) At what angle should the plane
be piloted such that it will
go straight north?
Resultants Examples – Boat Crossing
River
http://www.physicsclassroom.com/class/vectors/u3l1f.cfm
1. A boat is moving at 20 m/s in a river that
has a velocity of 5.0 m/s. Calculate the
resultant velocity when:
a) The boat is moving downstream
(same direction)
v = 25 m/s
b) The boat is moving upstream
(opposite direction)
20 m/s
v = 15 m/s
5 m/s
c) The boat is moving east and the river
is moving south
v = 202 + 52
v = 20.6 m/s
tan Ө = 5/20
Ө = 140
****2. A boat is moving due north at 15 m/s
across a 210 m wide river that is flowing at
6.5 m/s due west. Calculate:
a) The resultant velocity of the boat
VR = 152 + 6.52
= 16.3 m/s
Ө = 230
b) How long does it take to reach the
other side?
t = d/v = 210/15 = 14 s
c) How far downstream the boat lands
when it reaches the other side
d = vt = (6.5)(14) = 91 m
3. A boat that can travel 4.0 m/s on still
water heads directly across a river that is
125 m wide. If the speed of the river current
is 2.1 m/s,
(a) what is the velocity of the boat
with respect to the shore?
2.1
v2 = 42 + 2.12
v = 4.5 m/s
Ө = 620
(b) How long does it take the boat
to reach the opposite shore?
t = d/v = 125/4 = 31.3 s
(c) How far downstream is the
boat when it reaches the opposite
shore?
d = vt = (2.1)(31.3) = 65.7 m
p. 88 #1-3
p. 90 #4 – 9
p. 95 #1 p.99 #1 p. 101 #6
4.
Vectors at Angles Other Than 90
 to add vectors that are at angles other
than 90, we must use the Law of
Cosines: c2 = a2 + b2 – 2abcosC
Example
If two displacements are 20 m and
30 m and the angle between them is
60o, find the net displacement.
Example 1
A lawnmower is pushed at an angle of 25
with a force of 100 N. Find the horizontal
force and the vertical force.
Example 2
A plane flies at an angle of 40.0 N of E at a
velocity of 600 km/h. Find the east component
and the north component of the velocity.
 now that you can break a vector into its
components, you can add two or more
vectors using the individual x and y
components
 you need to think in terms of the
Cartesian plane and give each of the
components the correct sign (positive or
negative)
Example 1
Give the magnitude and direction for the
addition of the following vectors :
200 N, 20 N of E and 300 N, 60 S of W
Example 2
Add the following vectors: 200 N, 45 N of W
and 100 N, 30 N of E.
Example 3
Add the following vectors: 150 N, 15 N of E;
200 N, 15 W of N; 300 N due south.
Example 4
Add the following vectors: 20 N, 40 N of E;
30 N, due north; 50 N, 60 W of N
Your Assignment: pg 34-35 #36a) – d)
H. Equilibrium
 the equilibrant is equal to the resultant in
magnitude but opposite in direction.
 when the sum of all forces acting on an
object is zero, the object is in equilibrium
 when the vector sum of the forces is not zero,
a force can be applied that will produce
equilibrium = EQUILIBRANT
Example 1
If the resultant force is 129 N due east, what is
the equilibrant force?
Example 2
Two forces of 70 N, due east and 40 N, due
south act on an object. Find the resultant and
the equilibrant.
Forces
Dynamics is the study of why objects move.
It deals with force, momentum, and impulse.
Sir Isaac Newton (1642 – 1727) studied
acceleration and its cause
 Also studied forces, which are grouped into 4
categories:
1. Gravitational – attractive forces
between all objects
2. Electromagnetic – attractive forces
between charged particles
3. Strong nuclear – force that holds the
particles in the nucleus of an atom together
4. Weak nuclear – radioactive decay
Newton’s First Law of Motion
 an object continues in a state of rest or
uniform motion unless an external force
acts on it.
 also called the Law of Inertia
 Inertia is a property of an object that opposes
a change in motion. It is because of inertia
that we use seatbelts
 Mass is the amount of inertia possessed by an
object…more mass = more inertia
Newton’s Second Law
 The acceleration of an object is directly
proportional to the net force acting on it and
inversely proportional to its mass.
Acceleration takes place in the direction of
the net force.
F = ma
where: a = acceleration in m/s2
m = mass in kg
F = net force in N (Newtons)
(Note: 1 N = 1kgm/s2)
 the larger the mass, the smaller the
acceleration (because of inertia) if the force
is constant.
Example 1
A force of 3000 N acts on a 1200 kg truck.
What is its acceleration?
a = F/m=3000/1200=2.5 m/s2
Example 2
A 900 kg car travelling at a speed of 30 m/s
comes to a stop over a distance of 50 m. Find
the force needed to stop the car.
vf2 =vi2+2ad
02=302+2a(50)
a= -9.00 m/s2
F=ma=(900)(-9.00)=-8.1x103 N
Example 3
A racecar has a mass of 710 kg. It starts from
rest and travels 40 m in 3.0 s. The car is
uniformly accelerated during the entire time.
What net force is applied to it?
d=vit+ 1/2at2
40=(1/2) a (32 )
a=8.9 m/s2
F=ma=710x8.9=6.3x103 N
Example 4
The acceleration of an object is 5.0 m/s2. If the
net force is doubled and the mass is halved,
what will be the acceleration?
a=F/m=2/0.5=4
Acceleration increases four times.
Therefore the acceleration is 4x5.0=20 m/s2
Example 5
A 1.2x103kg car accelerates uniformly
from 5.0 m/s to 12 m/s east. During
this acceleration the car travels 94 m.
What is the net force acting on the car during
the acceleration?
vf2 = vi2+2ad
122=5.02+2a(94)
a=0.633m/s2
F=ma=1.2x103x0.633
F =7.6x102 N
Newton’s Third law
For every action there is an opposite and equal
reaction.
Mass and Weight
 Mass and weight are two measurements that
are often confused.
Mass
 amount of
inertia possessed
by an object
 unit is the kg
Weight
 force exerted by
a planet (force
gravity)
 unit is the N
 measured by an
equal arm
balance
 does not change
from place to
place.
 measured by a
spring scale
 changes from
place to place.
use Newton’s Second Law to calculate
weight:
Fg = mg
or
W = mg
where: g = acceleration due to gravity in m/s2
m = mass in kg
Fg = gravitational force or weight in N
W = weight
Example
Find the weight of a 5.0 kg mass.
Fg=mg
=5.0x9.81
=49 N
Example
Find the mass of an object that weighs 100 N.
Fg=mg
100=mx9.81
m=10.2 kg
Example
What is the acceleration due to gravity on a
planet if a 200.0 kg object weighs 4400 N on
that planet?
Fg=mg 4400=(200)(g)
g=22 m/s2
Newton’s Second Law and Vertical Forces
If an object is not moving, Fg = Fa (applied
force) therefore Fnet is zero
 Fa = applied force
 Fg=Weight
If an object is moving upward
Fg  Fa therefore Fnet is not
zero (Fa Fg )
Fnet = Fa – Fg
Fnet = ma
Fg = mg
Fa = applied force in N
Fg = gravitational force in N
Fnet = net force in N
Example 1
Draw a free-body diagram with an upward
force of 20 N pulling a mass of 0.50 kg. Find
the net force.
Fnet = Fa – Fg
= 20 – (0.5)(9.81)
= 15.1 N
Example 2
Find the net force if an upward force of 100
N is applied on a 7.0 kg mass.
Fnet = Fa – Fg
= 31.3 N
Example 3
What is the acceleration of the
object in the previous example?
Fn = ma
a = F/m
= 31.3/7
= 4.5 m/s2
Example 4
Find the force needed to accelerate
a 2.0 kg object upward with an
acceleration of 5.0 m/s2.
Fa = ?
Fa = Fnet + Fg
= (2)(5) + (2)(9.81)
= 29.6 N
p. 149 #1, 2
p. 234 #1
Energy and Work
Energy
- Is defined as the ability to do work.
Work
- is the transfer of mechanical energy from one
object to another
- is a push or pull on an object that results in motion
in the direction of the force applied.
F
d
W=Fd
W=work in Joules (J)
F=force in Newtons (N)
d=distance in metres
Rearranged formulas:
d=W
F
F=W
d
When the force is not parallel to the horizontal the
formula becomes:
W = F·d ·cos ө
F
ө
the horizontal component moves the object
(adjacent side)
1. A force of 30 N is used to lift an object to a height of
2.5 m. Find the work done.
F = 30 N
W = Fd = (30)(2.5)
d = 2.5 m
W = 75 J
W=?
2. It takes 12 000 J of work to pull a crate a distance of
200 m. Find the force applied.
W = 12 000 J
F=W
d = 200 m
d
F=?
= 12 000
200
F = 60 N
F. Vertical Forces and the
Elevator
 sign convention:
 with the first 5 formulas that
we learned, we said that a
was negative for an object
going up (opposite to gravity)
 a is separate from gravity and
is positive when the object
(elevator) is moving up and
negative when object is
moving down
 g will always be positive
 there are 5 cases with the
elevator:
1. it is at rest
FA = Fg
ie) net force is zero
2. it is moving upward
with some acceleration
FA > Fg
Fnet=FA-Fg
Acceleration is positive
3. it is moving downward
with some acceleration
FA < Fg.
Same formula. Accleration
is negative
4. it is moving up or down
with a constant speed (no
acceleration)
FA = Fg
ie) net force is zero
5. it is in free fall!
Fg only
ie) applied force is zero
Example 1
A 75 kg man stands in an elevator.
What force does the elevator exert
on him when:
a) the elevator starts moving
upward with an acceleration of
1.5 m/s2.
b) the elevator starts moving
downward with an acceleration
of 1.5 m/s2.
c) the elevator is moving
upward at a constant speed of
2.0 m/s.
Example 2
A 62 kg wrestler is standing on a
spring scale in an elevator. What
is the reading on the scale when:
a) the elevator is
accelerating up at
2.0 m/s2.
b) the elevator moves down at
2.0 m/s2.
c) the elevator moves down at a
constant speed of 2.5 m/s.
d) the elevator is in free fall (oh
no, the cable broke!).
An object of mass 2.0 kg is held on a string.
What is the tension in the string
if:(a) the object is at rest
(b) the object is moving downward at
2.0 m/s?
(c) the object is accelerating
at 2.0 m/s2 downward.
(d)the object is accelerating upward at
2.0 m/s2 ?
Newton’s Second Law and Horizontal
Motion
http://www.regentsprep.org/Regents/physics/phys01/friction/default.htm
Review:
Fnet = vector sum of all forces
acting on an object = ma (F = ma)
Fa = applied force
Fg = force of gravity = mg
For vertical motion:
Fnet = Fa – Fg
Now we’ll look at forces and their
relationship to movement in the
horizontal plane.
 when an object is sitting on a
horizontal surface, we still have
Fg acting on it
 the surface exerts a force back
on the object that is equal in
magnitude to Fg…this is called
the normal force (FN or N).
FN
Fg
Fg = FN = mg
Example
An 8.0 kg object is resting on a
horizontal surface. What is the
normal force acting on the object?
FN = Fg = mg = 8.0x9.81=78 N
 when an object sitting on a
horizontal surface starts to
move, another force comes into
play…friction!
 friction is the force that opposes
motion or is the force needed to
move an object at a constant
speed.
 Friction forces are always
parallel to the surface area of
contact and is directed opposite
to the direction of motion or
intended motion.
 Kinetic friction occurs when an
object slides over another.
 Static force refers to a force
parallel to the two surfaces that
can arise even when they are not
sliding. Suppose you try to move
a desk and it does not move.There
is a force exerted in the opposite
direction. This force is called the
static friction.
This is the force exerted by the
floor on the desk. If you push
hard enough, the desk will
eventually start to move, and the
kinetic friction takes over.
 friction depends on the nature of
the surfaces and on the normal
force (same magnitude as the
weight of the object).
 The force of friction is
independent of the area of
contact.
 the constant of proportionality
between the forces(force of
friction and the normal force) is
called the coefficient of friction
( = mu.)
It depends on the nature of the
two contacting surfaces.
Ff = FN
where:  (mu) = coefficient of
friction
Ff = force of friction in N
FN = normal force in N
(same magnitude as mg)
FN =mg
Fa
Ff
Fnet = Fa - Ff
Example
An object has a mass of 20 kg. A
force of 50 N is required to move
the object at a constant speed.
Find the coefficient of friction.
 = Ff/FN
FN = Fg= 20 x 9.81=196.2N
μ = 50N/196.2N
= 0.26 (No unit as N is
cancelled.)
Example
A 10.0 kg mystery box rests on a
horizontal floor. The coefficient of
static friction s=0.40 and the
coefficient of kinetic friction is
k=0.30. Will it move if the
applied force is of magnitude:
(a) 0N (b)10 N (c)38 N (d) 40 N
Ff=μFN
FN=10x9.81=98.1
Ff= 0.40 x 98.1
Ff=39 N
(a), (b), (c) → No
(d) → Yes
Example
A cart is given an applied force of
50 N. If the force of friction is 10
N and the mass of the object is 20
kg, find its acceleration.
Fnet= Fa- Ff = 50 –10 = 40 N
F = ma
40 = 20a
a = 2.0 m/s2
p. 99 # 9 – 11
p. 102 # 13 – 15
p. 106 # 12 -19
Friction
Example
A surface has a coefficient of
friction of 0.20. What is the force
needed to move a 100 kg object at
a constant speed?
**Force applied must be the same as the force
of friction***
Ff = FN = μ(mg)
Ff = (0.2)(100)(9.81)
Ff =200 N
Example
An object of mass 50 kg is
accelerated at 3.0 m/s2. If the
applied force is 200 N, find
the force of friction.
Fnet=Fa-Ff
Ff = Fa – Fnet
Fnet = ma = (50)(3) = 150 N
Ff = 200 – 150
Ff=50 N
Example
A force of 2000 N is needed
to accelerate a 1200 kg car.
If the force of friction is
800 N, calculate the acceleration.
Fnet= Fa – Ff
= 2000 – 800
=1200
a = F/m = 1200/1200
a = 1.00 m/s2
Example
You are driving in your car at a
velocity of 24.0 m/s when you
slam on your brakes. The force of
friction on your tires is 1.80 x 104
N. If the mass of you and your car
is 1.50 x 103 kg, what is the
applied force if the car stops in
6.00 s?
Fnet = Fa – Ff
a=(vf-vi)/t
=0-24/6.00=-4.00 m/s2
ma=Fa-1.80x104
3
4
(1.50x10 )(-4.0) = Fa - 1.80x10
Fa =1.20x104 N
Summary:
 Friction is a force which
opposes motion.
 there are two types of friction:
static and sliding
 Static friction = the force that
opposes the start of motion
Ff(static) = FN
 Sliding friction = the force
between surfaces in relative
motion
Ff(sliding) = FN
***Note that the formula is the
same for each type of friction
however sliding friction is
always less than static friction
Example
A sled of mass 40 kg is pulled
along a snow-covered surface.
The coefficient of static friction is
0.30 and the coefficient of sliding
friction is 0.15.
a) How much does the sled
weigh?
Fg= mg=40x9.81
=392.4 N
=3.9x102 N
b) What force is needed to start
the sled moving?
Ff = FN FN = Fg
Ff = (0.30)(392.4)
Ff =1.2x102 N
c) What force is needed to keep
the sled moving at a constant
speed?
Ff = FN
Ff = (0.15)(392.4)
Ff=59 N
p. 190 #1-3,
p. 191 – key terms
p. 691 #1-10 (ch. 5)
b) Once moving, what total
force must be applied to the sled
to accelerate it at 3.0 m/s2?
Fnet=FA-Ff
ma=FA-Ff
40x3.0= FA-58.86
=1.8x102 N
Example
An object of mass 30 kg is pulled
at a constant speed
by a force of 50 N. Find the
coefficient of friction of the
surface.
=0.17
Combining Vertical and
Horizontal Forces
Example 1
A 5.00 kg lab cart is accelerated
on a frictionless table by a 2.00 kg
mass accelerating straight down,
as shown. What is the
acceleration of the lab cart?
5.00 kg
2.00 kg
Fa = mg = (2.0)(9.81) = 19.62N
Gravity acting only on the 2 kg mass.
Fnet=Fa-Ff
ma =Fa- 0
7(a) =19.62
a = 2.80 m/s2
Carts act as one unit (that is, both are
moving at the same time)
Example 2
Find the acceleration of the
masses shown in the diagram,
if there is 3 N of friction in the
system.
2.0 kg
1.0 kg
Fnet = Fa - Ff
ma= (1)(9.81) - 3
3.0(a) = 6.81
a = 2.3m/s2
Example 3
Find the acceleration of the
masses shown in the diagram.
Assume there is no friction.
10 kg
2.0 kg
a =1.6 m/s2
Newton’s Third Law
 for every action there is an equal
and opposite reaction
eg) recoil of a gun, swimming,
paddling etc
 forces always occur in pairs –
“action-reaction” pairs.
 Action force=Reaction force
 net force can be zero – all
action-reaction pairs can be
balanced
 balanced forces can make an
object move
 If there is more than one actionreaction pair acting on an object,
and one pair exerts a greater
force then the object will move .
Review of Forces
 know definitions eg) inertia,
mass, acceleration, weight, etc
 know and understand the
application of Newton’s three
Laws
 know the relationship between
acceleration, mass and force eg)
if you double one, what happens
to the other two?
 vertical forces – projecting
upwards, elevators, tension on
strings etc.
Fnet = FA  Fg
 horizontal forces – force of
friction, normal force,
coefficient of friction
Fnet = FA  Ff and Ff = FN
 combined forces – carts on lab
tables
Projectile Motion – Motion in
Two Dimensions
 a projectile is an object that
travels in the air
 it follows a curved path, called a
trajectory, which is due to its
horizontal and vertical velocity
 the horizontal distance the object
travels is called the range
 characteristics of projectile
motion:
1. the horizontal velocity is
constant
2. the vertical velocity changes
with the distance(height) the
object falls
3. the horizontal velocity is
independent of the vertical
velocity
 it doesn’t matter what the
horizontal speed is, gravity will
affect it in the same way…that
is, it will take the same time to
fall
 to solve the problems, look at
the x and y directions separately
There are two types of projectiles:
1.projectiles thrown horizontally
2.projectiles fired at an angle
Projectiles Thrown Horizontally
http://www.physicsclassroom.com/Class/vectors/u3l2a.cfm
http://www.upscale.utoronto.ca/GeneralInterest/Harrison/Flash/ClassMechanics/Projectile/Projectile. html
vy = 0 (initial velocity in the y-direction is zero)
vx
vy - increases because of gravity
Each direction is treated independently – the
only common variable between the two is time.
x-direction
y- direction
v = constant
vi = 0
a = 9.81 m/s2
Example 1
A stone is thrown horizontally at 15 m/s from
the top of a cliff 44 m high.
a)
How long does it take to reach the
bottom of the cliff?
b) How far from the base of the cliff does
the stone strike the ground (range)?
Step 1 – sketch.
15 m/s
44 m
Step 2 – list the variables for each direction.
x-direction
v = 15 m/s
t=3s
(b) d = vt
d = (15)(3)
y- direction
vi = 0
a = 9.81 m/s2
d = 44 m
(a) t = ?
d = vit + 1/2at2 t = 2d
a
d = 45 m
t=
2(44)
9.81
t=3s
Example 2
An object is thrown horizontally at a velocity of
10.0 m/s from the top of a 90.0 m high building.
a)
How long does it take to hit the ground?
b) How far from the base of the building
does it strike the ground?
10 m/s
90 m
x-direction
y- direction
v = 10 m/s
t = 4.3 s
(b) d = vt
d = (10)(4.3)
vi = 0
a = 9.81 m/s2
d = 90 m
(a) t = ?
d = vit + 1/2at2 t = 2d
a
d = 43 m
t=
2(90)
9.81
t = 4.3 s
Example 3
An object is thrown horizontally at 20.0 m/s
from the top of a cliff. The object hits the
ground 48.0 m from the base of the cliff.
How high is the cliff?
20 m/s
d=?
48 m
x-direction
y- direction
v = 20 m/s
d = 48 m
t = d/v
t = (48)/(20)
t = 2.4 s
vi = 0
a = 9.81 m/s2
t = 2.4 s
d=?
d = vit + 1/2at2
d = ½ (9.81)(2.4)2
d = 28.3 m
Projectiles Fired at an Angle
http://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/ProjectileM
otion/jarapplet.html
http://www.ngsir.netfirms.com/englishhtm/ThrowABall.htm
Projectile stops moving upward
v
ө
we use only the first half when
solving problems (gravity is –9.81 m/s2)
To solve this type of a projectile you must first find the
horizontal (x-direction) and vertical (y-direction)
components of the velocity.
v
ө
vx
vx = cos ө (v)
vy = sin ө (v)
vy
1.A projectile is fired with a speed of 40 m/s at an angle of 600
with the horizontal. Calculate:
a) the time the projectile is in the air
b) the horizontal distance it travels
c) the maximum height it reaches
Solution:
Step 1 – Sketch.
40 m/s
600
Step 2 – Find the horizontal and vertical
components of the velocity.
40 m/s
vy
vx
vx = cos 60 (40) = 20 m/s
vy = sin 60 (40) = 35 m/s
Step 3 – List the variables for each direction.
x-direction
y-direction
v = 20 m/s
t = 7.2 s
b) d = vt
= (20)(7.2)
vi = 35 m/s
vf = 0
a = - 9.81 m/s2
a) t = vf - vi t = - 35/-9.81 = 3.6 s
a
d = 144 m
this time is half way
Thus, tT = 7.2 s
c) d = vi + vf t
2
= 35 3.6
2
d = 63 m
2. Repeat using an angle of 530 and an initial
velocity of 200 m/s.
a) 33 s
b) 3912 m
c) 1304 m
e.g. A projectile is fired with a speed of 65 km/h at an angle of
400 with the horizontal. Calculate:
d) the time the projectile is in the air
e) the horizontal distance it travels
f) the maximum height it reaches
p. 139 #5, 6
p. 152 #7, 8,
p. 693 #1, 4, 6, 9
Momentum
 product of mass and velocity
 unit is kgm/s
p = mv
where: m = mass in kg
v = velocity in m/s
p = momentum in kgm/s
Example
1. Find the momentum of a 1,200
kg car travelling east at 20 m/s.
p = mv=1200x20 = 24,000 kg·m/s
= 2.4x104 kg.m/s
Impulse
 the product of force and time
 unit is Ns
Impulse = Ft
where: F = force in N
t = time in s
Impulse = Ns
Impulse = Change in Momentum
Ft = mΔv
Δv = vf - vi
1 N·s = 1 kg·m/s
Example
2. Find the impulse exerted on a
baseball if a force of 200 N is
applied for 0.050 s.
Impulse = Ft = 200x0.05
=10 N·s
3. If the impulse on an object is
40 N·s and the original
momentum is 30 kg·m/s, what is
the new momentum?
Total p = original momentum + change in p
= 30 + 40 = 70 N·s
4. An object gains 200 kg·m/s
of momentum from rest when
acted on by an applied force of
60 N for 5 s. Find the resistive
force (friction) acting on the
block.
Δp = 200 kg·m/s
Fa = 60 N
t=5s
Ff = ?
Ft = Δp
F = 200/5 = 40 N
F = Fa - Ff
F f = Fa – F
= 60 – 40
Ff = 20 N
p. 178 #1-3
p. 193 #1-9
p. 695 #1-3,5,6 (ch.9)
Work and Energy
Work
- is the transfer of mechanical energy
from one object to another
- is a push or pull on an object that
results in motion in the direction of
the force applied.
 conditions: a force must be applied and
the object must move a distance in the
direction of the force.
 unit is Nm = Joule
F
d
W=Fd
W=work in Joules (J)
F=force in Newtons (N)
d=distance in metres
Rearranged formulas:
d=W
F
F=W
d
When the force is not parallel to the horizontal the
formula becomes:
W = F·d ·cos ө
F
ө
the horizontal component moves the object
(adjacent side)
1. A force of 30 N is used to lift an object to a height of
2.5 m. Find the work done.
F = 30 N
W = Fd = (30)(2.5)
d = 2.5 m
W = 75 J
W=?
2. It takes 12 000 J of work to pull a crate a distance of
200 m. Find the force applied if a rope is used, which is
300 with respect to the horizontal.
W = 12 000 J
F=W
d = 200 m
d cos ө
F=?
= 12 000
200 cos 30
F = 69 N
3. Find the work done in lifting
a 5.0 kg object 20 m.
W=Fd=(5.0x9.81)x20=9.8x102 J
Energy
 Is the capacity to do work
(Energy = Work)
 unit is J
There are two major forms of
mechanical energy:kinetic &
potential.
Kinetic energy (EK or KE) is the
energy possessed by a moving
object.
eg) a moving car
EK = ½ mv2
m = mass in kg
v = velocity in m/s
EK = kinetic energy in J
Rerranged formulas:
m = 2KE
v2
v = 2KE
m
Example
1.Calculate the kinetic energy of a
1,200 kg car traveling at 20 m/s.
m = 1200 kg
v = 20 m/s
KE = ?
KE = ½ mv2 =1/2(1200)(20)2
= 240,000 J = 2.4x105 J
2. A 2,000 kg car traveling at
28 m/s reduces its speed to 10 m/s.
Calculate its loss in kinetic energy.
m = 2,000 kg
vi = 28 m/s
vf = 10 m/s
∆KE = ?
KEi = (1/2)mvi2
= ½(2000)(28)2 = 784,000 J
KEf = (1/2)mvf2
= ½(2000)(10)2 = 100,000 J
∆KE = KEf - KEi
= 100,000 – 784,000
∆KE = - 684, 000 J
p. 331- key terms
(new)
Power
 the rate at which work is done
 unit is J/s = Watt (W)
P=W
t
E = W = Pt
E = energy in J
P=Power (Watts)
W=Work done (J)
t=time (s)
Special case - when a force is needed to keep
an object moving at a constant speed – e.g.
moving against friction or lifting an object.
P = W = Fd
but d = v
t
t
t
Thus,
P = Fv
1.An object of mass 50 kg is raised through a
distance of 10 m in 20 s. Find the power.
W=Fd=(50x9.81)x10
=4905 J
P=W/t=4905/20=2.5x102 W
2. A machine lifts 200 kg, 50 m in 5.0 minutes.
Find the power usage.
W=(200x9.81)x50=98,100 J
P=W/t=98100/300 = 327 W
P = 3.3x102 W
3. A motor raises a 50 kg object at a steady
speed of 6 m/s. What is the power output of the
motor?
P = Fv = (mg)v = (50)(9.81)(6)
P = 3,000 W
Energy
 Is the capacity to do work
(Energy = Work)
 unit is J
There are two major forms of
mechanical energy: potential
and kinetic.
Potential energy (EP or PE) is
the energy possessed by an object
due to its position or condition
(stored energy).
eg) battery, compressed
spring, food, object at some
height
In this course we only look at the gravitational
potential energy.
Gravitational Potential Energy (Ep or PE)
EP = mgh
m = mass in kg
g = acceleration due to gravity
h = height in m
EP = gravitational potential energy in J
Example
Calculate the potential energy
possessed by a 20 kg mass at a
height of 1.0 m.
m = 20 kg
h = 1.0 m
Ep = ?
Ep=mgh=20x9.81x1.0
=196.2 J
=2.0x102 J
Kinetic energy (EK or KE) is the
energy possessed by a moving
object.
eg) a moving car
EK = ½ mv2
m = mass in kg
v = velocity in m/s
EK = kinetic energy in J
Rerranged formulas:
m = 2KE
v2
v = 2KE
m
Example
1.Calculate the kinetic energy of a
1,200 kg car traveling at 20 m/s.
m = 1200 kg
v = 20 m/s
KE = ?
KE = ½ mv2 =1/2(1200)(20)2
= 240,000 J = 2.4x105 J
2. A 2,000 kg car traveling at
28 m/s reduces its speed to 10 m/s.
Calculate its loss in kinetic energy.
m = 2,000 kg
vi = 28 m/s
vf = 10 m/s
∆KE = ?
KEi = (1/2)mvi2
= ½(2000)(28)2 = 784,000 J
KEf = (1/2)mvf2
= ½(2000)(10)2 = 100,000 J
∆KE = KEf - KEi
= 100,000 – 784,000
∆KE = - 684, 000 J
p. 331- key terms
(new)
Law of Conservation of Energy
The law of conservation of energy
states that energy can neither be
created nor destroyed only
changed from one form to another.
As an example of the law of
conservation of energy, consider a
mass weighing 100 N located 20
m above the earth. At this height
the potential energy of the mass is
2,000 J.
If the mass is allowed to fall, the
loss of potential energy is always
equal to its gain in kinetic energy.
PE = 2000 J
KE = 0
100 N
100 N
KE = 2000 J
PE = 0
PEtop = KEbottom
mgh = ½ mv2
gh = ½ v2
e.g. A 2 kg object is dropped from
a height of 5 m.
a) What is its kinetic energy just
as it reaches the ground?
b) Using energy considerations
only, determine the speed just as it
hits the ground.
c) How much work is done by
gravity?
a) PEtop = KEbottom
mgh = KE
KE = (2)(9.81)(5) = 98.1 J
b)
gh = ½ v2
v2 = 2gh = 2(9.81)(5)
v = 9.9 m/s
c) 98.1 J
p. 294 #1- 4 p. 296 #1 – 3
p. 303 #1 – 3 p. 330 #4, 5
p. 331- key terms