Download Since W = Fd, and v =d/t, we can also express power as

Vectors
Vectors are a useful concept in visually representing quantities of motion
that would be difficult to represent otherwise. Because of their concise
visual nature, we can use vectors to quickly represent and analyze a
motion situation.
Vectors are quantities consisting of magnitude and direction.
The
examples of vectors that we'll be using in this topic are:
Quantity
Example Magnitude
Example Direction
displacement
13 meters
30º West of North
velocity
8m/sec
45º South of East
force
155 kg
down
Vectors are represented by simply drawing an arrow:
The length represents the magnitude; therefore
it is necessary to choose a scale. We use
1 cm = __________ whatevers
Be careful to choose a scale that can "fit" the
values to be represented into the amount of
space available on your paper/page.
The direction of the vector is measured relative to north, south, east or
west.
Vector Addition: We add vectors to find the result of 2 or more vectors
of the same kind acting on an object. To do this, join he vectors head to
tail without changing their length or direction. The resultant (R) is drawn
so that its head touches a head. Some examples of vector addition are
shown on page 2:
1
Example 1:
+
=
R
Example 2:
R
+
+
=
Additional notes and examples:
2
Graphical Vector Problems
1. A plane flies east at 100km/hr. A wind blowing south at 30 km/hr
knocks the plane off course. Determine the resultant motion of the
plane.
2. A boat sails east across a river at 8 km/h. The current is moving north
(upstream) at 12 km/h and a breeze is blowing 25º N of E at 15 km/h.
Find the boat's resultant motion.
3. John drives 30 km in a direction 57º N of W and then 21 km in a
direction 20º E of N. What is his final displacement?
3
4. Three crazed shoppers are pulling a sweater (which is on sale) with the
following forces:
10 N at 30º E of S
14 N at 18º W of N
17 N at 62º W of S
5. A plane flies 45º N of W at a speed of 1500 km/h. A wind is gusting at
180 km/h in a direction 30º N of E. Find the resultant motion of the
plane.
6. A sailboat is headed downstream at 10m/s. A wind is blowing at
13m/s in a direction 56º E of S. Find the boat's resultant motion.
4
Vector Components: Once we have the resultant of a vector, we can
further analyze it by determining its components. The components of a
vector represent how much of the vector "acts" horizontally (vx) and how
much of it acts vertically (vy). Visually, the components are the legs of a
right triangle in which the resultant is the hypotenuse. The direction of
the components is simply the horizontal and vertical directions already
expressed in the resultant!!
Illustration of Vector components:
R
R
vy
vy
vx
vx
Problems: Determine the components of the resultants found in the
Graphical Vector Problems on Pages 3 and 4.
Trigonometry: When a vector problem involves only a resultant and its
components (which is often), the sketch of the problem will form a right
triangle. If this is the case, we can use trig to solve the problem. This is
convenient because rulers, protractors, and scales are NOT necessary!!
The trig functions will generate the correct answers!
The Trig functions we use are:
sin  = opposite/hypotenuse
cos  = adjacent/hypotenuse
tan  = opposite/adjacent
* Note opposite and adjacent are with respect to angle .
5
Trigonometry Problems
R

b

a
1. If  = 48º and a = 2, find b.
6. If  = 53º and R = 5, find a.
2. If  = 38º and b = 10, find R.
7. If a = 7 and R = 12, find .
3. If a = 5 and b = 6, find .
8. If b = 4 and R = 7, find .
4. If  = 37º and R = 6, find b.
9.If a = 9 and  = 17º, find R.
5. If a = 5 and R = 15, find .
10. If a = 3 and b = 7, find .
6
Physics/Trig Problems
1. A boat sails at 20 km/hr in a direction 58º S of E. Find its components.
2. A plane tries to fly east but a 50 km/hr southerly wind knocks the
plane off course by 17º. Find the plane's resultant speed and its
original speed.
3. Find the components of a push applied to a stalled car if the push is
125 N in a direction 30º N of E.
4. A boat crossing a river is thrown off course by 20º by a 10 km/hr
downstream current. Find the boat's original speed.
7
5. Suppose the displacement between your home and LHS is 4 km in a
direction 48º N of W. What are the components of your displacement
from home when you are at school?
6. A plane flying in a southwesterly direction has southern component of
speed of 200 km/hr and a resultant speed of 350 km/hr. Find the
plane's western component of speed and its exact direction.
7. A boat tries to cross a stream at 10 m/s. The current downstream is 3
m/s. Find the resultant speed and direction of the boat.
8. A plane tries to fly east at 137 km/hr but a north wind knocks the
plane off course by 32º. Find the speed of the wind.
8
PARALLELOGRAM METHOD: To solve using the parallelogram method
you can use the following equations - depending on what is given.
LAW of SINES
a
=
sin A
b
=
sin B
c
sin C
LAW of COSINES
a =  b2 + c2 - 2bc cos A
b = a2 + c2 - 2ac cos B
c = a2 + b2 - 2ab cos C
Example:
A sailboat is headed downstream (south) at 10 m/s. A wind is
blowing at 13 m/s in a direction 56 E of S. Find the boat's resultant
motion.
a
a = 10 b = 13 <C = 90 + (90-56)
= 124
C
c
c = a2 + b2 - 2ab cos C
=  102 + 132 - 2(10)(13)cos 124
b
= 20.3 m/s
Use the Law of Sines to find the angle and then the direction.
9
Linear Velocity
The most fundamental of all quantities in mechanics is velocity. This quantity is in
some way related to nearly every other quantity in mechanics. In this topic we will
show how velocity can be determined through formulas (easy) and to show how
information about displacement and velocity can be determined from graphs (a
preview to calculus).
First, a word of caution: Velocity and speed are NOT the same thing, although they
are related. Speed is simply how fast something is moving. Velocity is how fast
something is moving in a particular direction. This means that velocity is a vector,
while speed is not. For example, if one car moves north at 50 mph while another car
moves south at 50 mph, they have the same speed but different velocities. Get it?
Average Velocity Formula:
vave = d/t
(Note - you may recognize this formula as a form of d = rt, where we have replaced r,
the rate by vave)
Example 1 - The slowest animal on record is the Red Sea crab, which crawls at the
whopping rate of 5.7 km/year. How long would it take this crab to crawl the 2 meters
across the length of your table top?
Example 2 - Donovan Bailey can run 100 meters in 9.8 seconds. How many miles per
hour (mph) is this? (1 m/sec = 2.25 mph)
10
Average Velocity Problems
1. A woman can jog 3200 meters in 16 minutes. How far can she go in one hour at
this rate?
2. A car travels at 30 km/hr for 1 hour and then 40 km/hr for 2 hours. How far did
the car travel? What was its average speed for the entire 3 hours?
3. A car travels at 50 km/hr for 10 km, stops for 15 minutes, and then continues at
60 km/hr for 20 km. Find the car's average speed for the entire trip.
4. A cyclist rides 8 miles at 25 km/hr then goes 40 km/hr for 10 minutes, and then
for 12 km at 20 km/hr. What is her average speed overall?
5. A distance runner's workout consists of running 3 km at 11 km/hr, then running at
6 km/hr for 20 minutes, and then running 2 km for 15 minutes. What is his
average speed for the whole workout?
6. A swimmer swims 1000 meters in 15 minutes, then swims at 50m/min for 20
minutes and then swims at 40 m/min over a distance of 1500 meters. What is the
swimmer's overall average speed?
11
Average Velocity & displacement vs times Graphs:
Average velocity can also be found using displacement vs time graphs. The method
for doing this is very simple. Just draw a line connecting the initial and final points,
on the graph, of the time interval for which the average velocity can be found. The
slope of this line, as you should know by now, will yield the average velocity.
Example
Use the graph shown below to determine the average velocity for the interval t = 2
sec to t = 8 sec.
Solution
The line that connects the two points corresponding to the given time interval
contains the coordinates (2,4) and (8,10). Applying the slope formula yields:
vave = d/t = (10 - 4)/(8 - 2) = +1 m/sec
d (m)
(8,10)
10 __
8 __
d=6m
6 __
4 __
(2,4)
 t = 6 sec
2 __
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
11
t (sec)
Instantaneous Velocity
Instantaneous velocity is just that…..the velocity at a given instant. In other words,
the velocity when no time passes. So, we can't use the formula for average velocity
since t = 0 and we can't divide by 0. There is, however, a way to find vinst by using a
d vs t graph. Here's how:
vinst = the slope of the tangent to the d vs t graph at a given instant.
12
Why does this work? Because the tangent has the same slope as the graph has at the
instant. Since the slope = d/t, then it represents the velocity!! We use the tangent
because its slope can be measured a lot more easily than the slope of the teensy
weensy section of the graph that contains the instant in question. This happens to be
the foundation for something called the derivative in calculus.
D (m)
8 -6 -d
4 -t
2 -a closer look at t = 3 sec
|
|
|
|
|
|
1
2
3
4
5
6 t (sec)
d vs t graph
Finding vinst : use the slope formula and plug in the coordinates of the point on the
curve and the point where the tangent crosses an axis (either one!). In the example
shown below, the instantaneous velocity at time 3 sec, aka v3, is determined using the
method discussed above.
d (m)
7 -6 -5 -V3 = (4-0)/3-7) = -1m/sec
4 -3 -tangent line
2 -1 -|
|
|
|
|
1
2
3
4
5
13
|
6
|
|
|
7
8
9
t (sec)
Displacement vs Time Graph Problems
d (m)
12 __
10 __
8 __
6 __
4 __
2 __
|
1
|
2
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3
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4
|
5
|
6
|
7
|
8
|
9
|
10
|
11
t (sec)
1. Using the graph shown above, find:
a. the average speed between t = 2 sec and t = 5 sec
b. the average speed between t = 4 sec and t = 5 sec
c. the instantaneous speed at t = 2 sec, 4 sec, 6 sec and 8 sec
2. Use the graph above to explain when the motion is
a. Fastest
b. Stopped
c. turning around
d. constant
e. moving in the negative direction
14
d (m)
10 __
8 __
6 __
4 __
2 __
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
11
t (sec)
3. Use the graph shown above, find:
a. the average speed between t = 1 sec and t = 5 sec
b. the average speed between t = 3 sec and t = 9 sec
c. the instantaneous speed at t = 3 sec, 5 sec, 7 sec and 9 sec
4. Explain yours answers to the following questions concerning the graph above:
a. When is the motion constant?
b. When is the motion slowest?
c. Is the motion always in the same direction?
d. How many times does the motion change direction?
15
Finding Displacement from Velocity vs Time Graphs:
Since we've been able to find out information about velocity from examining
displacement vs time graphs, its probably not too surprising that the reverse is also
true…. We can use velocity vs time graphs to find out information about
displacement. Here's how:
displacement = the area "beneath" the v vs t graph and the time axis
Why does this work? Because when you find the area "beneath" the v vs t graph, you
are multiplying rate (v) times time, which is, of course, as you learned long ago, the
displacement!!
Example - Use the graph, shown below to find the displacement for the time
intervals:
a) t = 0 sec to t = 4 sec
b) t = 3 sec to t = 7 sec
c) t = 5 sec to t = 10 sec
v (m/s)
20 __
15 __
10 __
5 __
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
11
t (sec)
-10 __
16
Velocity vs Time Graph Problems
v (m/s)
20 __
15 __
10 __
5 __
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
11
t (sec)
-10 __
1. Using the v vs t graph above, find displacements for
a. t = 0 sec to t = 4 sec
b. t = 3 sec to t = 7 sec
2. Based on the graph above, explain the times at which the velocity is
a. Constant
b. Increasing
c. moving in the (-) direction
17
v (m/s)
20 __
10 __
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
t (sec)
-10 __
-20 __
3. Based on the v vs t graph shown above, find the displacement for
a. t = 1 sec to t = 5 sec
b. t = 2 sec to t = 7 sec
4. Refer to the v vs t graph above to give reasons for the following
a. Is the final displacement for the entire motion (+) or (-)?
b. Describe the motion between t = 2 sec and t = 4 sec
c. During which time intervals does the motion slow down?
18
Homework Problems
1. Given the d vs t graph below, find the following, and give reasons
when appropriate:
a. vinst at t = 3 sec and t = 6 sec
b. when the motion is constant
c. when the motion is negative
d. when the motion is fastest
e. when the motion turns around
f. when the motion stops
d (m)
10 __
8 __
6 __
4 __
2 __
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
11
t (sec)
19
2. Given the v vs t graph shown below, find the following and five reasons
when appropriate:
a. the displacement from t = 2 sec to t = 7 sec
b. when the motion is constant
c. when the motion is negative
d. when the motion is fastest
e. when the motion turns around
f. is the final displacement (+) or (-)
v (m/s)
20 __
10
|
1
|
2
|
3
|
4
-10 __
-20 __
20
|
5
|
6
|
7
|
8
|
|
9
10
t (sec)
|
Reference: Interpreting the Shapes of Motion Graphs
Description of the
v vs t graph
motion
d vs t graph
v
(+) and speeding up
d
1st Quadrant
Uphill, gets
Away from axis
steeper
v
(+) and constant
d
1st Quadrant
Uphill, stays
Parallel to axis
straight
v
(+) and slowing
down
1st
Quadrant
d
Uphill, gets
Towards the axis
flatter
v
(-) amd speeding up
4th
Quadrant
d
Downhill, gets
Away from axis
steeper
v
(-) and constant
4th
Quadrant
d
Downhill, stays
Parallel to axis
straight
v
d
(-) and slowing
4th Quadrant
Downhill, gets
down
Towards the axis
flatter
v
Turning around
Moving north (+) or
south (-)
Fast or slow
Constant or stopped
d
Peaks and
Crosses the axis
valleys
In 1st Quadrant or 4th Quadrant.
Highest point or point nearest the
time axis
Parallel to axis or on the axis
21
(+) slope or (-) slope
Steepest slope or flattest slope
Straight line or horizontal line
Linear Acceleration
You are probably already familiar with acceleration as meaning "to speed up".
In
physics, acceleration means more than that. Acceleration is any change in velocity
over an interval of time. So, an object can be said to accelerate when it speeds up,
slows down or changes direction (as in circular motion, for example).
The formula which defines acceleration is as follows
a = v/t
(unit is m/sec2)
Formulas for a Uniform Linear Acceleration:
vf = vi + at
d = vit+ (1/2)at2
vf2 = vi2 + 2ad
vavg = d/t = (vf + vi)/2
Note - In using these formula, you must be given 3 values. Be on the lookout for
"hidden givens" based on the wording of the problem, if you are not given 3
values directly.
22
Uniform Linear Acceleration Problems
1. How far will an object fall in the first 4 seconds of free fall?
2. An arrow is shot standing up in the air with a speed of 20 m/sec.
a. How long is it in the air?
b. How high dies it rise?
3. A stone is thrown down from a bridge with a speed of 30 m/sec.
a. How far will it have fallen after 2 seconds?
b. How far is it moving after 4 seconds?
4. A penny is dropped from the top of a 400 meters high building. How fast is it
moving when it hits the ground?
5. How fast are Olympic platform divers moving when they hit the water?
23
6. What is the acceleration of a car that goes from 0 to 27 m/sec (i.e. 60 mph) in 6
seconds? Compare your answer to the acceleration of gravity.
7. A cyclist accelerates at 2 m/s2 for 10 seconds, starting with a speed of 5m/sec.
a. What speed does she reach at the end of the 10 seconds?
b. How far did she ride in this time?
8. A car traveling at 9m/s slams on its brakes, coming to a stop in 2 seconds. Find
how far the car traveled in coming to a halt.
9. Repeat the last problem for the same car (therefore, you can assume the same
acceleration) if it was
a. going half as fast (4.5 m/s)
b. going 3 times as fast (27 m/s)
24
10.
The best sprinters in the world reach a speed of 10 m/sec sprinting 10 meters,
starting from rest. Compare their acceleration to the acceleration of gravity.
11.
A ball rolling at 8 m/sec approaches an incline. If it takes the ball 3 seconds to
stop by rolling up the incline, determine the length of the incline.
12.
A ball, starting from rest, rolls down an incline that is 3 meters long. It requires
4 seconds to do this. (see illustration below)
a. Find the ball's speed at the bottom of the incline
b. Find how far the ball rolled during the final second
0 sec
1 sec
2 sec
3 sec
4 sec
25
Graphical Analysis of Uniform Acceleration:
You may have recognized that the expression for acceleration back on the first page
of this topic as being in the "form" of a slope. In this case, acceleration would ve the
slope of a v vs t graph. Uh oh!!! They're back….graphs!!! Before you panic, keep a
couple of things in mind:
1. Since we are only dealing with uniform (or constant) acceleration, then the
slopes on the v vs t graphs must be constant. In other words, the v vs t
graphs we'll be examining will consist only of straight line segments. You
may find the reference page of this topic to helpful in visually coordinating d
vs v graphs with the type of acceleration that they represent.
2. The sign (+ or -) of the acceleration has 2 meanings physically.
This is
related to the motion in the quadrant in which the velocity has a (+) or (-)
slope. We'll examine this more.
Graphical Analysis Problems
1. For the v vs t graph shown below,
a. Find the acceleration from t = 2 sec to t = 5 sec
b. Find the acceleration from t =7 sec to t = 9 sec
c. Describe the motion in terms of acceleration
v (m/s)
20 __
15 __
10 __
5 __
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
11
t (sec)
-5__
-10 __
-15 __
26
2. For the v vs t graph shown below,
a. Find the acceleration between t = 0 sec and t = 3 sec
b. Find the acceleration between t = 5 sec and t = 9 sec
c. Describe the motion in terms of acceleration
v (m/s)
40 __
30 __
20 __
10 __
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
11
t (sec)
-10__
-20 __
-30 __
3. For the v vs t graph shown below,
a. Find the acceleration from t = 1 sec to t = 4 sec
b. Find the acceleration from t = 5 to t = 7 sec
c. Describe the motion in terms of acceleration
v (m/s)
6 __
4 __
2 __
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
t (sec)
-2 __
-4 __
-6 __
27
Reference: Visually Identifying Acceleration on v vs t and d vs t Graphs
Types of Acceleration
v vs t graph
d vs t graph
v
d
NONE: The slope of v
t
vs t = 0 and/or there is
t
no change in the slope
of d vs t graph
v
d
v
d
v
d
v
d
v
d
POSITIVE: The slope of
v vs t is (+) and/or the
slope of d vs t is
getting more (+) (i.e.,
speeding up north) or
less (-) (i.e., slowing
down south)
NEGATIVE: The slope of
v vs t is (-) and/or the
slope of d vs t is
getting (+) (i.e.,
slowing down north) or
more (-) (i.e., speeding
up south)
28
Newton's Laws
Newton's Laws are fundamental in that they explain the causes of motion of
(relatively) large, solid masses. These laws involve the relationship of forces
and motion, particularly (a) rest, (b) constant velocity, (c) constant acceleration.
For our purposes, forces in mechanics have only 4 sources:
 Fg : force of gravity (aka weight = mg)
 Ff : force of friction
 T : tension (in a wire, rope, etc.)
 Fapp : an applied force (by a person or inanimate object)
NEWTON'S 1st LAW: This is also known as the LAW OF INERTIA *
An object at rest remains at rest and an object in motion remains in motions
unless acted upon by an outside force.
* The Law of Inertia was first developed by Galileo! On the reference page that
follows, we'll see how he reasoned out the idea of inertia.
The statement that we are all familiar with, but it is too vague!! Specifically, "a
body in motion" refers to a mass that is moving with a constant velocity. Also,
"outside force" refers to an unbalanced force. (There are always outside forces
present!)
The law of inertia states that masses resist sudden changes in what they are
doing. In this way, masses are like people. We tend to resist sudden changes
in our hair styles, the types of clothes we wear, the music we listen to, or our
political beliefs. We do change, but it doesn't usually happen suddenly.
Inertia is directly related to mass. Which is harder to do - throw a baseball or
"throw" a bowling ball? Big masses resist changes in their motion more than
small masses do. It's common sense!!!
29
Reference - Galileo & Inertia
Galileo used inclined planes to develop his idea about inertia.
By
experimenting, he found that a ball that rolled down one ramp would roll up a
second ramp until it reached the same height that it began with. By making
the second ramp less and less inclined, the ball would roll further and further
along the second ramp before reaching the same height as the starting height.
(See first three illustrations below)
Here's where Galileo gets creative… he figured that if the second ramp was not
inclined at all (see 4th illustration below), the ball would "forever" be seeking
out its initial height. In other words, a body in motion will remain in motion….
Galileo's work will inclined planes also anticipated the law of Conservation of
Energy, which we'll see later this year.
30
EQUILIBRIUM:
Equilibrium is the mathematical representation of Newton's 1st Law.
Basically, it says that when an object is (a) at rest, or (b) moving with a
constant velocity (not a constant speed, which is different than velocity),
then the sum of the force vectors equals zero (F = 0).
Static Equilibrium
Static equilibrium occurs when an object is at rest. Although any number
of forces can be present, typically there are 2 or 3 present in a given
situation (like you sitting in your seat, for example). Visually, the force
vectors would appear as shown below:
2 Forces
3 Forces (Lab Situation)
3
+
=0
2
3
1
1
2
Translational Equilibrium
Translational equilibrium occurs when an object moves with a constant
velocity. We have observed this situation in the weird "frictionless" cases
of the air track and the "Hoovercraft", represented below.
v
FAPP
mg
31
Usually, friction is present. Friction depends on only 2 factors:
1. the smoothness of the surfaces (called the coefficient of friction, )
2. how much one surface "presses" into the other (called the normal
force, N)
Ff = N
2 Standard cases of Friction:
Ff
Fpull/push
level surface:
m
mg
incline:
N
Ff
m
mg sin
mg cos
N

Note - When the mass is moving, the only forces necessary to put into
the mathematical formula for Newton's 1st (or 2nd) law are those forces
which are parallel to the motion. Any other forces which are present are
assumed to balance each other (since the object doesn't move in the
direction of those other forces) and therefore add up to zero
mathematically.
32
NEWTON'S 2nd LAW: We know from the 1st law that when an object is
at rest or moving with a constant velocity, then the F = 0. The 2nd Law
deals with what happens when the forces don't balance (i.e. add up to
zero) Under these conditions, the object will accelerate, i.e. will (a) speed
up, (b) slow down, (c) change direction or (d) some combination of (a), (b)
and (c).
If the forces don't add up to zero, what do they add up to?
Well,
according to Newton, the net forces (i.e. F) will equal the product of the
mass of the object and its acceleration.
F = ma
Newton's 2nd Law
This expression makes sense, if you think about it, as stickman will be
happy to show you below:
Constant Force:
If stickman pushes a certain amount on a baby carriage, it will accelerate
a certain amount. If he applies the same push to his stalled stickmobile
(which has a much bigger mass), it will accelerate less..
right?
F
A
a
F
M
33
Make sense,
Constant Mass
If stickman lobs a baseball, it doesn't gain much speed.
If he throws the
ball as hard as he can, it gains much more speed. Makes sense, no?
v
F
v
F
p.s. I HATE STICKMEN!
MEANING of the 2nd LAW
Newton's 2nd law involves (a) force vectors, (b) acceleration and (c) mass.
(a) The net force (F) determines the direction of the acceleration, not any
individual force. If a dry ice puck is acted upon by 2 tension forces of
equal magnitude but different directions. The puck accelerates in the
same direction as the vector sum of these individual forces.
34
(b) The net force is directly related to the acceleration, not the velocity (a
common Aristotelian misconception). We can see this by examining
the computer-generated graphs for force, acceleration and velocity for
a cart which is acceleration constantly in a line. The force graph is
similar to the acceleration graph, not the velocity graph.
F
v
a
(c) Suppose the same force as that used above is used to accelerate a
larger mass. Use the 2nd law (and common sense) to predict what will
happen to the acceleration graph
35
Reference: Standard Situation of Newton's 1st and 2nd Laws
When the situation represented below correspond to an object's being at
Rest or moving with a constant velocity, then the 1st Law is in effect. That
is, F = 0.
If the situations correspond to an object's speeding up or slowing down,
then the 2nd Law is in effect. That is, F = ma.
Remember, the only forces that are necessary to use when the object is in
motion are those that are parallel to the direction of motion of the object.
36
2nd Law Example - The Elevator Problem
Stickman represents a man in an elevator.
For each of the following cases, predict (using 1st or 2nd law) what the
scale will read relative to stickman's weight.
(a) Elevator is at rest.
(b) Elevator initially speeds up.
(c) Elevator reaches a constant upward speed.
(d) Elevator slows down when reaching the 10th floor.
(e) Elevator speeds up as it leaves the 10th floor to return to the 1st floor.
(f) Elevator slows down as it approaches the 1st floor.
37
Reference - The Acceleration Bridge
Acceleration links the subject of kinematics to the subject of dynamics.
It is possible to:
(a) use Newton's 2nd Law to find acceleration, and then "cross the bridge"
to find a kinematics value such as distance, time and speed.
(b) Use distance, time and/or speed to find acceleration, and then "cross
the bridge" to find a force value or .
Acceleration has dual citizenship in both "Kinematics land" and
"Dynamics Land"!
Acceleration Bridge
"Kinematics" Land
"Dynamics" Land
Residents: vf, vI, d, t &
Residents: mg, T, Ff, Fapp, &
acceleration
acceleration
Formulas: vf = vI + at
Vf
2
Formulas: F = ma
= vi 2 = 2ad
D = vit + (1/2)at2
38
Newton's Laws Problems
1. An elevator weighs 8000 N. If it accelerated downward for 6 seconds
through 20 meters, starting from rest, find the cable tension.
2. A 600 kg elevator accelerates upward from rest for 10 seconds
through 40 meters. Find the cable tension.
3. Bruno, the "Human Bicep", can dead lift 220 kg by accelerating it from
rest through 1 meter in 0.20 seconds. What force does Bruno exert on
the barbells during this time?
39
4. A certain physics teacher, in another life, used to push start his car
nearly every day. His car weighted 810 kg. If the coefficient of friction
was 0.10 and he pushed the car with a force of 900 N, how fast would
the car be moving after 5 seconds. Assume the car started from rest
and that the road was level.
5. A trotter pulls a rider and carriage weighting 85 kg along a level racing
track with coefficient of friction 0.40. With what force does the horse
pull if the rider reaches a speed of 8 m/s after 4 seconds, starting
from rest?
6. A car pulls a trailer weighing 320 kg along a level road. If the car pulls
with a force of 2000 N, the trailer attains a speed of 9 m/s after
moving 30 meters, starting from rest. Find the coefficient of friction
between the trailer and the road.
40
7. A certain physics teacher, in that other life, sometimes used to park his
810 kg car on a hill so that gravity would push start the car for him.
The hill was inclined at 10º, and the car would roll 25 meters in 8
seconds, starting from rest. What was the coefficient of friction for
this hill?
8. A 23 kg child sleds down a hill that is inclined 15º. The hill is 100
meters long and has a coefficient of friction of 0.10. If the child starts
from rest, how fast will she be moving when she reaches the bottom of
the hill?
9. A cyclist, just for kicks, decides to roll down a 400 meter hill, starting
from rest and not pedaling or using his brakes. The cyclist and bike
together weigh 82 kg. If the coefficient of friction is 0.13 and the hill
is inclined at an angle of 8º, find the cyclist's speed at the bottom of
the hill.
41
Additional Problems 1. Starting from rest, a ball rolls down an incline at a constant
acceleration of 2.00m/s2 . (a) What is the velocity of the ball after 8.5 s?
(b) How far does the ball roll in 10.0 s?
2. A car traveling at 88 km/h undergoes a constant deceleration of 8.0
m/s2. (a) How long does it take the car to come to a stop? (b) How far
does the car move after the brakes are applied?
3. A ball is thrown vertically upward with a velocity of 115 m/s. (a) To
what height will it rise? (b) How long will it take for the ball to fall back
to earth?
42
4. A ball drops from rest and attains a velocity of 53 m/s. How much
time has lapsed?
5. A ball initially at rest rolls down a hill with an acceleration 3.3 m/s2.
(a) If it accelerated for 7.5 s, how far will it move? (b) How far would it
have moved if it had started at 4.0 m/s rather than from rest?
6. A penny dropped from the roof of a building takes 6.3 seconds to
strike the ground. How high is the roof?
7. A physics student throws a softball straight up into the air. Her friend
has a stopwatch and determines that the ball was in the air for a total of
3.56 s. (a) With what speed was it thrown? (b) How high did it rise?
43
8. A boy sliding down a hill accelerated at 1.40 m/s2. If he started from
rest, in what distance would he reach a speed of 7.00 m/s?
9.A car weighs 1.50 x 104 N. What force is required to accelerate the car
at 4 m/s? (neglect friction and air resistance)
10. What force is needed to give a mass of 25 kg an acceleration of 20.0
m/s2?
11. A force of 30.0 N acting on an object gives it an acceleration of 5.00
m/s2. (a) What is the mass of the object? (b) What is its weight?
44
12. A car weighs 19,600N. (a) What is the mass of the car? (b) If a
braking force of 1250 N is needed to stop the car, what is its
acceleration?
13. An 1800 Kg car starting from rest accelerates constantly during the
first 9.50 sec of its motion. If the force acting on the car is 2150 N, what
is the car's speed at the end of this time?
14. During a throw, a baseball pitcher exerts a force on the ball through a
distance of 3.20 m before releasing it. If the 0.12 kg ball leaves the
pitcher's hand with a speed of 35.0 m/s, how much of a force did the
pitcher exert on the ball? A catcher stops this ball by pulling back 25.0
cm on the mitt during the catch. How much force did the catcher exert
on the ball?
45
NEWTON'S 3rd LAW: For every action, there's an equal but opposite
reaction.
Like the statement of the Law of Inertia, the 3rd law is to vague!!
Specifically, "action" and "reaction" refer to forces. That means that 2
masses must somehow be involved in an interaction. The force that m1
exerts on m2 is equal to but opposite in direction to the force that m2
exerts on m1. Both masses feel the same force, and it only takes one of
the masses to "initiate" the interaction. But as soon as it does,
automatically and simultaneously, the other mass exerts a force on the
"instigator" without even trying!!
Stated mathematically, the 3rd Law says:
F1 - 2 = - F2 - 1
m1a1 = m2a2
This formula shows that if one of the 2 masses is very large, we wouldn't
see much of a reaction (because of its small acceleration) In other words,
a large mass would speed up or slow down very slowly, while a small
mass would speed up or slow down very quickly. This is consistent with
what would happen due to the inertia of the large mass. If the masses
are the same, they will both speed up or slow down at about the same
rate (but in an opposite sense!).
Examples: Describe & Explain the action and reaction for each of the following
situations as specifically as you can, using the 3rd Law
1.
2.
3.
4.
5.
6.
Swimming
A basketball is hit using a baseball bat
A rotating lawn sprinkler
A football player is tackled head-on by another player
A small car collides with a large truck
Punch a piece of paper as hard as you can!
3rd Law Class Demonstrations
46
For each of these demonstrations, predict and explain what you think will happen. After
each demonstration, revise your prediction if necessary.
1.
2.
3.
4.
5.
6.
7.
8.
Collision between two carts of equal mass, one of which is at rest.
Explosion between 2 carts of unequal mass. (m1 = 4m2)
A ball thrown against a free standing piece of cardboard.
A ball thrown against a cement wall.
A student, on skates, pulls on a rope attached to the sink.
A student, on skates, pulls on a rope held by a 2nd student on skates.
A bucket of water Is whirled in a vertical circle.
CO2 is released from a canister that is free to rotate.
Do Action-Reaction Forces Cancel Out?
Consider this…. A friend who hears about Newton's 3rd Law says that it doesn't make
any sense. Here's the reason why: If you were to kick a football, the 3rd law says that the
force of the kick is equal and opposite to the force of the ball against the foot. As a result
the forces would cancel. If the forces cancel out, then the football would remain at rest
(according to the conditions for equilibrium). But we know, from our experience, that
this certainly does not happen. So, your friend claims that the 3rd law is nonsense.
Of course, with your current background as a 3rd law expert, you can easily set your
friend straight. Explain how you would resolve this apparent conflict between the 1st
Law, the 3rd Law and your experience.
47
Uniform Circular Motion
Uniform Circular Motion (UCM) is any circular motion for which the speed
remains constant.
Note that the velocity does not remain constant
because the direction is changing. Therefore, there is an acceleration and
Newton's 2nd Law is in effect. And that means that there is an unbalanced
force present which produces the circular motion.
Circular motion should not be unfamiliar to you. Amusement park riders,
athletic events (like the discus, figure skating and swinging a baseball
bat), rounding a turn (in a car or on a bike) and orbital motion are all
circular motions. What distinguishes these examples is the unbalanced
force that produces the circular motion. In the case of orbital motion, the
force is gravity; for the discus, the force is the tension in the arm; for a
car rounding a turn, friction is present; and for an amusement park ride
like the Rotor, an applied force (by the wall) enables a rider to move in a
circle. The 4 forces mentioned in these examples constitute our force
menu!!
The vectors present in UCM are illustrated below.
There are only 3:
velocity, force and acceleration.
F, a
F, a
v is tangent
a, F are centripetal
48
UCM FORMULA:
v = 2R/T
ac = 42R/T2
ac = v2/R
Fc = mv2/R = 42Rm/T2
T = 1/f
** In these formulas, T represents the period, that is, the time that is
needed to complete one complete revolution. Tension forces will
still be represented with a vector arrow above the T.
SPECIFIC SITUATIONS:
49
Reference: Rounding Turns at High Speeds
If you're like most people, you know intuitively that it is difficult to round
a turn at a high speed. Newton's 2nd Law for UCM explains why. In
Fc = mv2/R
general, for any UCM,
In particular, for rounding a turn (which represents a "temporary" UCM),
Fc = Ffriction = mg
mv2/R = mg
Therefore,
Notice that the expression for friction represents a constant value (for
any given car on a particular road surface). But the centripetal force
expression is not constant… as v increases, the centripetal force
increases.
So… friction is the force that is available and it is constant. The
centripetal force, which is the amount of force needed to make a turn of
radius R at a speed v, depends on v. When the speed gets to the point
where the needed force (mv2/R) is greater than the available force
(friction), guess what happens? Yup, the car (or whatever) goes off the
road because there's not enough available force to supply the required
centripetal force!! (By the way, this is like life in general… if what you
need is more than what you have, you are in trouble; think it!)
v
R
50
UCM Problems
1. The Rotor at Great Adventure has a radius of 2.5 meters and a period
of 2 seconds. Find the force, acceleration, speed and frequency of a
60 kg rider.
2. A discus weighs 1 kg. If a discus thrower turns one complete
revolution in 0.35 seconds with his arm extended 1 meter from the
axis of rotation, find a) the speed with which the discus is released
b) the force he feels in his arm
3. The earth weighs 6 x 1024 kg. It is 1.5 x 1011 meters from the sun.
The period of the earth around the sun is 3.15 x 107 sec (i.e., 1 year).
Find the a) speed of earth in its orbit b) its acceleration
c) the centripetal force that the sun exerts on the earth
51
4. A ball on a string is whirled in a horizontal circle with a force of 20 N.
Its acceleration is 12 m/s2 and its speed is 4 m/s. Find the
a) Mass of the ball
b) Radius of the ball
c) Period of the ball
d) Frequency of the ball
5. A car rounds a turn of radius 10 meters at a speed of 15 m/s. Find the
coefficient of friction necessary to keep the car on the road.
6. A car rounds a turn of radius 25 meters at a speed of 20 m/s. Find the
coefficient of friction necessary to keep the car on the road.
7. A bucket of water weighing 3 kg is whirled in a vertical circle of radius
1 meter by a physics teacher entertaining guests at a wild party. If the
speed of the bucket is 4 m/s at the top and 8 m/s at the bottom, how
much tension does the teacher feel in his arm at the
a) highest point
b) lowest point
52
8. In fast pitch softball, a pitcher can wind up and release the ball in 0.2
seconds. The ball weighs 300 grams and the pitcher's arm is 0.66 m
long. Assuming that the speed of the ball remains constant
throughout the windup, find
a) the speed of the ball
b) the tension exerted on the pitcher's shoulder at the release point
(i.e. the bottom of the circle).
9. A loop-the loop ride at an amusement park has a radius of 5 meters.
At the highest point the rider moves at 8 m/s. At the lowest point, the
rider moves at 13 m/s. Find what you would feel like you weigh at
those points. This force represents the force of the seat against your
rear end; its Fapp and it takes the place of tension for a ball on string.
53
Work & Energy
Energy is the bottom line so far as our global and domestic economies are concerned.
The types of energy used, how abundant they are, how much they cost, and how
much they pollute has an effect on almost every aspect of our daily lives (not only
economically, but also personally). The cars we drive, the houses we live in, and any
commercial product we use are, in some way, energy dependent.
From the standpoint of science, the Law of Conservation of Energy is probably the
single most important of all the laws. This law governs every biological, chemical,
and physical process.
We use energy so that work can be done. Mowing the lawn, playing sports, driving
cars, and listening to music are examples of activities that require energy so that
work can be done.
This brings us back to the beginning of the year, when we looked at wave motion.
Energy is the ability to do work. And work is done whenever a force is applied to
move an object through a distance.
RELATIONSHIP of WORK & ENERGY:
This is sort of like the chicken and the egg problem. Energy represents the ability to
apply a force through a distance. This work is then applied to some object, changing
the energy of that object! That object might lose energy (in which case it is wasted/
dissipated as friction/heat) or it might gain energy (in which case it can be used to do
work on yet another object). The particular types of energy that we'll examine are
Kinetic Energy (the energy of motion) and Potential Energy (stored energy; this is
energy that has the potential to be converted into Kinetic Energy and/or work).
This sort of cycle is shown below (examples are on the following page).
WORK
ENERGY
HEAT
ENERGY
HEAT
WORK
54
Work & Energy Formulas
(A) Work can be determined 3 ways:
(1) It is done when it is used to change the energy of an object.
W = E
(2) It is done when a force is applied through a distance so that at least part of
the force {i.e., some component of the force) is parallel to the direction of
motion; this criteria enables the object to gain or lose energy.
W = Fdcos
note -  is the angle between F and the direction of motion; the cos of this angle gives
the component of F that is parallel to the motion.
note -the unit for work & energy is Joules (J).
(3) Work can be found graphically by the area under a force vs distance graph.
This is similar, mathematically, to what we did in the Linear Motion section
when we found distance from the area under a speed vs time graph.
(Remember d =vt & d = area under v vs t, we now have that W = Fd and W =
area under F vs d; See the mathematical similarity? Of course you do!)
Work = Area under F vs d
note - It's possible to apply forces and yet do no work! How?
55
{a) When there's a push or pull but no distance is covered. This happens, for
example, when you are simply standing still... you're applying a force to the
floor, but the floor doesn't move.
(b) When the angle between the force and the direction of motion is 90°.
Examples of this are orbital motion or carrying some object while walking
along.
(B) Power is the rate of doing work:
P = W/t
Since W = Fd, and v =d/t, we can also express power as
P = Fv
The unit Watts represents 1 Joule/sec. If you lift 1/4 lb (which is about 1 N) a
distance of 1 meter in 1 second, you've used 1 Watt of power.
Don't confuse Watts with horsepower (AKA hp). 1 hp = 746 Watts!!! You need to lift
185 lb a distance of 1 meter in 1 second, in order to exert the power of an average
horse (but only for a second!!!)
Challenge: 10 extra credit points for anybody who can generate 2 hp in running up
the stairs at the end of the hall. We will be completing the "Work Lab" if there is
anyone who would care to try?
56
(C) Energy comes in two forms: kinetic and potential. The only potential
energy that we'll deal with is gravitational potential energy (there are other types)
KE = (1/2)mv2
PE g = mgh*
* The "h" value in mgh is a relative value! It represents the height measured from the
lowest point in a given situation. This will become clear as we do some problems.
(D) Law of Conservation of Energy: Energy is neither created nor destroyed. It
is transformed (via work) and/ or wasted in the form of heat (usually friction).
This law is stated mathematically as follows
PEi + KEi = PEf + KEf
In using the Law of Conservation of Energy, pick a convenient initial and final point.
Think about whether speed (meaning KE) and/or height (meaning PEg) exist at those
points.
A nice example of the Conservation of Energy is a simple pendulum. As it swings
back and forth, there is a continual transformation of PE and KE, yet the total amount
of energy never changes. This is illustrated on the following page.
57
Reference- Conservation of Energy & The Simple Pendulum
The illustration below shows the idea that although energy may change
type; the amount remains constant (it's conserved)
(a) At each endpoint, there is only PEg.
(b) As the pendulum swings from the endpoint to the midpoint, PEg
decreases while KE increases. In other words, some of the PE from the
end point is transformed into KE.
(c) At the midpoint, there is no PE. It has all been transformed into KE.
At all of these positions, the total amount of energy is constant!!!
(a) PE only
(b) PE & KE
(c) KE only
58
Work & Energy Problems
1. A girl pulls a sled with a force of 10 N over a distance of 200 meters.
If the rope she's pulling with makes an angle of 37° with the ground, how
much work does she do?
2. A woman does 160,000 J of work in mowing a lawn. If she applies a
force of 100 N to the handle over a distance of 4000 meters, find the
angle made between the mower's handle and the ground.
3. Chris carries a carton of milk, weighing 10 N, along a level hall to the
kitchen, a distance of 5 meters. How much work did he do? Explain.
4. A force is applied to a 2-kg object as described by the graph below.
The object's initial KE is 10 J.
(a) find the work done over 8 meters
20
(b) find the KE after 8 meters
10
2
(c) find the speed after 4 meters
59
-10
4
6
8 10
5. A 0.5-kg mass with an initial speed of 3 m/s has a force applied to it as
described by the graph below.
(a) find the work done after 3 meters
4
2
b) find the speed after 6 meters
-2
2
4
6
8 10
-4
6. A force is applied to a 2 kg mass as shown by the graph below, If the
mass had an initial KE of 7 J, find its speed after it has moved through a
distance of 8 meters.
10
5
2
4
6
8 10
7. A 1-kg mass has an initial speed of 2 m/s when it is acted upon by a
force as described by the graph below. Find the speed of the mass after
moving 5 meters.
10
5
2
60
4
6
8 10
8. A force of 300 N is used to push a 145-kg mass 30 meters horizontally
in 3 sec.
(a) Find the work done on the mass
(b) Find the power exerted by the force
9. A lawn mower is pushed by a force of 115 N along the direction of the
handle, which is 22.5° above the horizontal. If 64.6 W are power are
exerted each 90 seconds, over what distance is the mower pushed?
10. Will pulls a 305N sled along a snowy path using a rope that makes a
45° angle with the ground. Will pulls with a force of 42.3 N. The sled
moves 16 meters in 3 seconds. What power did Will develop?
61
11. A ball rolling horizontally approaches a hill with a speed of 10 m/s.
Assuming no friction find:
(a) how far the ball rolls up the hill before stopping
(b) how fast the ball is moving halfway before it stops?
h
12. A 30-kg boy decides to go on a water slide that is 30 meters long.
The top of the slide is 15 meters above a pool at the bottom of the slide.
(a) How fast will the boy be moving when he enters the pool?
(b) How far down the slide is he when his speed is 10 m/s?
13. A 0.01-kg marble is thrown straight up in the air with a speed of 20
m/s. Find:
(a) How high the marble rises
(b) The speed of the marble at the instant it has risen 10 meters
62
14. A 0.25-kg rock is thrown straight down with a speed of 5 m/s from
the edge of a 100-m high cliff.
(a) How fast is it traveling when it hits the ground?
(b) How high is the rock when its speed is 40 m/s?
15. A baseball has a mass of about 0.5-kg. A pitcher releases the ball
with a speed of 35 m/s (about 80 mph). If the pitcher's arm moved
through a distance of 1.5 m in the act of throwing the ball, find the force
that the pitcher applies to the ball.
16. A weightlifter does 3440 J of work in lifting a 180-kg barbell. How
high did he lift the barbell?
63
17. A force of 410 N is applied in a direction straight up to a stone that
weighs 32 N. If the force is applied through a distance of 2 m,
(a) how fast will the will stone be moving at the release point?
(b) how high will the stone rise above the release point?
18. A steel ball of mass 0.5-kg is at rest 2 meters from the edge of a level
table that is 1.5 meters above the floor. A force of 10 N acts on the ball
until it reaches the edge of the table.
(a) How fast does the ball leave the table?
(b) How fast is the ball traveling when it hits the floor?
64