Transcript Work and Kinetic Energy

Physics 111: Mechanics
Lecture 6
Bin Chen
NJIT Physics Department
Chapter 6 Work and Kinetic Energy

1.10 Scalar Product of Vectors

6.1 Work

6.2 Kinetic Energy and the Work-Energy Theorem

6.3 Work and Energy with Varying Forces

6.4 Power
Scalar Product of Two Vectors

The scalar product of
two vectors is written
as

It is also called the dot
product


q is the angle
between A and B
Scalar Product:
A Graphic Representation
The scalar, or “dot” product
says something about how
parallel two vectors are.
 The scalar product of two
vectors can be thought of as
the projection of one onto the
direction of the other.




q is the angle between the
vectors
Scalar product of any
perpendicular vectors = zero
Scalar product is maximum for
parallel vectors

B
( A cos q ) B

A
A( B cos q )
q
Scalar Product is a Scalar
 Not
a vector
 May be positive, negative, or zero
Scalar Product in Components



The scalar product of two
vectors can be thought of as
the projection of one onto the
direction of the other.
What is the scalar products of
unit vectors?
Components of a vector can be
regarded as the scalar product
of the vector and the unit
vectors
 
A  B  AB cos q

A  iˆ  A cos q  Ax
iˆ  ˆj  0; iˆ  kˆ  0; ˆj  kˆ  0
iˆ  iˆ  1; ˆj  ˆj  1; kˆ  kˆ  1

B
Projection is zero
p/2

A
Calculating Scalar Product Using
Components
 
A  B  Ax Bx  Ay By  Az Bz


ˆ
ˆ
ˆ
Start with A  Axi  Ay j  Az k

B  Bxiˆ  By ˆj  Bz kˆ

 
ˆ  Ay ˆj  Az kˆ)  ( Bxiˆ  By ˆj  Bz kˆ)
A

B

(
A
i
x
Then
 Axiˆ  ( Bxiˆ  By ˆj  Bz kˆ)  Ay ˆj  ( Bxiˆ  By ˆj  Bz kˆ)  Az kˆ  ( Bxiˆ  By ˆj  Bz kˆ)


But
So
iˆ  ˆj  0; iˆ  kˆ  0; ˆj  kˆ  0
iˆ  iˆ  1; ˆj  ˆj  1; kˆ  kˆ  1
 
A  B  Axiˆ  Bxiˆ  Ay ˆj  By ˆj  Az kˆ  Bz kˆ
 Ax Bx  Ay By  Az Bz
Scalar Product: An Example


A  2iˆ  3 ˆj and B  iˆ  2 ˆj
 The vectors
 
 Determine the scalar product A  B  ?
 
A  B  Ax Bx  Ay By  2  (-1)  3  2  -2  6  4
 Find the angle θ between these two vectors
A  Ax2  Ay2  2 2  32  13
 
A B
4
4
cos q 


AB
13 5
65
4
q  cos 1
 60.3
65
B  Bx2  B y2  (1) 2  2 2  5
Definition of Work W

The work, W, done by a constant force on an object is
defined as the scalar (dot) product of the component of
the force along the direction of displacement and the
magnitude of the displacement

𝑭 is the magnitude of the force
∆𝒙 is the the object’s displacement

q is the angle between 𝑭 and ∆𝒙

Unit of Work

This gives no information about


the time it took for the displacement to occur
the velocity or acceleration of the object
Work is a scalar quantity
 SI Unit


Newton • meter = Joule


N•m=J
J = ( kg • m / s2 ) • m

Scalar Product
and
Work

Steve apply a force F  3iˆ  3 ˆj N on a car and

makes it to move a displacement of s  3iˆ  5 ˆj
m. How much work (in J) does Steve do in this
case?
A) 9iˆ
B)  9iˆ
C) 6
D) 9
E) 15
 
W  F  s  Fx x  Fy y  3  (-3)  3  5  -9  15  6
Work: Positive or Negative?

Work can be positive, negative, or zero. The
sign of the work depends on the direction of
the force relative to the displacement


Work
Work
Work
Work

Work minimum if  = 180°


positive: if 0°<  < 90°
negative: if 90°<  <180°
zero: W = 0 if  = 90°
maximum if  = 0°
Example: When Work is Zero
A man carries a bucket of water
horizontally at constant velocity.
 The force does no work on the
bucket
 Displacement is horizontal
 Force is vertical
 cos 90° = 0

Example: Work Can Be
Positive or Negative
Is the work positive or negative
when lifting the box?
 Positive
 Is the work positive or negative
when lowering the box?
 Negative

Work Done by a Constant Force
1.
A.
The right figure shows four situations in which a force is
applied to an object. In all four cases, the force has the
same magnitude, and the displacement of the object is to
the right and of the same magnitude. Rank the situations in
order of the work done by the force on the object, from
most positive to most negative.

F
I, IV, III, II

F
II, I, IV, III
C. III, II, IV, I
D. I, IV, II, III
E. III, IV, I, II
B.
I
III

F
II

F
IV
Work Done by a Constant Force

The work W done a system by
an agent exerting a constant
force on the system is the
product of the magnitude F of
the force, the magnitude Δr of
the displacement of the point
of application of the force, and
cosθ, where θ is the angle
between
the
force
and
displacement vectors:

F

F

r

r
II
I

F

F

r

r
III
IV
Work and Force

An Eskimo returning pulls a sled as shown. The
total mass of the sled is 50.0 kg, and he exerts a
force of 120 N on the sled by pulling on the rope.
How much work does he do on the sled if θ =
30° and he pulls the sled 5.0 m ?
Work Done by Multiple Forces

If more than one force acts on an object, then
the total work is equal to the algebraic sum of
the work done by the individual forces
Wnet = åWby individual forces

Remember work is a scalar, so
this is the algebraic sum
Wnet  Wg  WN  WF  ( F cos q )r
Work and Multiple Forces

Suppose µk = 0.200, How much work done on
the sled by friction, and the net work if θ = 30°
and he pulls the sled 5.0 m ?
F = 120 N
m = 50 kg
Work Done by a Constant Force

A (gravitational) force exerted by the Sun on the
Earth holds the Earth in an orbit around the Sun. Let
us assume that the orbit is perfectly circular. The
work done by the force during a short time interval in
which the Earth moves through a displacement in its
orbital path is
zero
B. positive
C. negative
D. impossible to determine
A.
Work and Motion
Kinetic Energy

For an object m moving with a
speed of v
Kinetic Energy is energy
associated with the state of
motion of an object
 SI unit: joule (J)
1 joule = 1 J = 1 kg m2/s2

Common Exam #1 Results
 Grades


posted on Moodle
Grades are not negotiable
Check your grade and let me know if there is
any problem by today
 70%
(B) and above: 14
 50%–70% (D–C): 10
 Below 50% (F): 15
How to improve your grades?


In fact, ALL exam questions were from homework
assignments or examples/quizzes discussed in
class. If you want a better grade, pay attention to them!
I want to help you!



iClicker quizzes will be presented more like exam questions.
More discussions on solving problems (which, in turn, means
going more quickly on concepts)
But you also need to help yourself!




Do your homework.
Go to tutoring sessions and my office hours.
Do NOT simply satisfied by getting an answer. Try to understand
EVERY STEP leading to your answer.
Review example problems and quiz questions after each lecture.
Test yourself on these problems.
Important Changes on Homework
 Homework
is now due Saturday nights
at 11:59 pm
 Adaptive follow-ups changed to 1 set of
questions, due 1 day after the parent
assignments (Sunday nights at 11:59
pm)
Friction Homework Review
e are updated periodically. Statistics were last updated October 12, 2016 at 4:38 am.
a 141-N block resting on a rough
horizontal table is pulled by a
a 141­ block resting on a rough horizontal table is pulled by a horizontal wire. The pull gradually increases until the block begins to
after. The figure shows a graph of the friction force on this block as a
horizontal wire. The pull
gradually increases until the
block starts to move and
continues to increase thereafter.
ment on friction, a w­N block resting on a rough horizontal table is pulled by a horizontal wire. The pull gradually increases until the

s to increase thereafter. The figure shows a graph of the...
Find the coefficients of static and
kinetic friction between the block
and the table.
 Why does the graph slant upward in
the first part but then level out?
 What would the graph look like if a
141 N brick were placed on the box,
and what would be the coefficients
where static and kinetic friction occur.
of friction be in that case?

0 to 75.0 , static for 75.0 .
Minimum friction
Two masses M1 = 1 kg and M2 = 10 kg are
attached by a wire as shown. The system is
at rest. Find the minimum coefficient of
static friction between the mass M2 and the
surface.
A. 0.2
B. 0.1
C. 0.3
D. 1.5
E. 0.05
Minimum Friction: Follow Up
Two masses M1 and M2 are attached by a
wire as shown. M2 weights 100 N. The
system is initially at rest. The coefficient of
static friction between the mass M2 and
the surface is 0.1.


What is the minimum mass
of M1 for it to drag M2 down?
What would the minimum M1
be if M2 is on a 30 degree slope?
Definition of Work W

The work, W, done by a constant force on an object is
defined as the scalar (dot) product of the component of
the force along the direction of displacement and the
magnitude of the displacement

𝑭 is the magnitude of the force
∆𝒙 is the the object’s displacement

q is the angle between 𝑭 and ∆𝒙

Work in pulling a truck

A 2000-kg truck is towed by Billy
along a horizontal road at a
constant speed by a 1500 N
force acting 15° above the
horizontal. How much work is
done by Billy as the object moves
6.0 m?
A. 2.1 kJ; B. 1.5 kJ;
C. 8.7 kJ; D. 11 kJ;
E. 5.2 kJ
W
F s (constant force, straight-line disp
Work-Energy Theorem
K 12 m 2 (definition of kinetic ener
 When work is done by a net force on an
object and the only change in the object is its
The work done by the net force on a particle equals the chan
speed, the work done is equal to the change
kinetic energy:
in the object’s kinetic energy
Wtot


K2 K1
K (work energy theorem
Speed will increase if work is positive
x2 decrease if work is negative
Speed will
W
W
x1
Fx dx (varying x-component of force, straight-l
1
1
2
𝑊tot = 𝑚𝑣2 − 𝑚𝑣12
2
2
P2
P1
F cos dl
P2
P1
F||dl
P2
P1
F d
(work done o
Work in Skydiving
 Let’s
watch a video first:
https://www.youtube.com/watch?v=Rb
cbjMhvjEs
 After some time, the 80 kg skydiver was
flying at a constant speed of 60 m/s
(i.e., terminal speed). How much work
is done by the force the air exerts on
the skydiver as he fell 100 m in height?
A. 47 kJ; B. -47 kJ; C. 78 kJ; D. -78 kJ
Announcement
 Common
Exam 1 Version B question 15
(on average velocity of a glider sliding
down a beam) has an error on the key
card.
 If you got the correct answer, C (0.02
m/s), send me an email. I’ll correct your
grade.
Work with Varying Forces



On a graph of force as a function of
position, the total work done by
the force is represented by the
area under the curve between the
initial and the final position
Note there could be negative work!
Straight-line motion
W  Fax xa  Fbx xb  ......
x2
W   Fx dx
x1

Motion along a curve

P
P
P 
W   F cos dl   F||dl   F  dl
2
P1
2
P1
2
P1
Work-Energy with Varying Forces

Work-energy theorem Wtol = K holds for
varying forces as well as for constant ones
ax 
dvx dv x dx
dv

 vx x
dt
dx dt
dx
x2
x2
x2
x1
x1
x1
Wtot   Fx dx   max dx  
dvx
mvx
dx
dx
v2
Wtot   mvx dv x
v1
1 2 1 2
Wtot  mv2  mv1  K
2
2
Spring Force: a Varying Force
Involves the spring constant, k
 Hooke’s Law gives the force

𝐹𝑥 = 𝑘𝑥



Where 𝐹𝑥 is the force exerted on the spring in the
same direction of x
The force exerted by the spring is 𝐹𝑠 = −𝐹𝑥 = −𝑘𝑥
k depends on how the spring is made of. Unit: N/m.
Work done on a Spring
To stretch a spring, we must do
work
 We apply equal and opposite
forces to the ends of the spring
and gradually increase the forces
 The work we must do to stretch
the spring from x1 to x2

x2
x2
x1
x1
W   Fx dx   kxdx 
1 2 1 2
kx2  kx1
2
2
W
Fscos
(constant force, straight-line d
Homework
Review
I
W F s#6
(constant
force, straight-line
disp
A
5.00-kg box slides 7.002 m across the
1
K
kinetic ener
floor before coming to2 mrest.(definition
What isofthe
coefficient of kinetic friction between the
Thefloor
workand
donethe
by the
netifforce
a particle
equals
the chan
box
the on
box
had an
initial
kinetic
energy:
speed
of 3.00 m/s?
Wtot
K2 K1
K (work energy theorem
𝑊 = 𝑭 ∙ 𝒔 = 𝐹𝑠 cos𝜃
W
x2
x1
𝜇𝑘 𝑁 x-component of force, straight-l
Fx dx𝑓𝑘 =
(varying
Homework #6 Review II

The force on a 3.00-kg object
as a function of position is
shown in the figure. If an
object is moving at 2.50 m/s
when it is located at x = 2.00
m, what will its speed be
when it reaches x = 8.00 m?
(Assume that the numbers on
the graph are accurate to 3
significant figures.)
Homework
#6 Review
IIIstraight-line d
W Fscos
(constant force,

If a fish is attached to a vertical spring and
W itsFequilibrium
s (constantposition,
force, straight-line
disp
slowly lowered to
it is
found to stretch the spring by an amount d.
K𝑭 = 120m 2 (definition of kinetic ene
 If the same fish is attached to the end of the
unstretched spring and then allowed to fall from
The work done by the net force on a particle equals the chan
rest, through what maximum distance does it
kinetic energy:
stretch the spring?
𝑭net =
Wtot
x2
K2 K1
K (work energy theorem
Power
Work does not depend on time interval
 The rate at which energy is transferred is
important in the design and use of practical
device
 The time rate of energy transfer is called power
 The average power is given by


when the method of energy transfer is work
Instantaneous Power
Power is the time rate of energy transfer. Power
is valid for any means of energy transfer
 Other expression


A more general definition of instantaneous
power
Units of Power
 The

SI unit of power is called the watt
1 watt = 1 joule / second = 1 kg . m2 / s3
A
unit of power in the US Customary
system is horsepower

1 hp = 550 ft . lb/s = 746 W
 Units
of power can also be used to
express units of work or energy

1 kWh = (1000 W)(3600 s) = 3.6 x106 J