Transcript Work, Kinetic Energy

Physics 111: Mechanics
Lecture 6
Dale Gary
NJIT Physics Department
Energy





Energy and Mechanical Energy
Work
Kinetic Energy
Work and Kinetic Energy
The Scalar Product of Two Vectors
March 28, 2016
Why Energy?
Why do we need a concept of energy?
 The energy approach to describing motion is
particularly useful when Newton’s Laws are
difficult or impossible to use
 Energy is a scalar quantity. It does not have a
direction associated with it

March 28, 2016
What is Energy?
Energy is a property of the state of a system,
not a property of individual objects: we have to
broaden our view.
 Some forms of energy:


Mechanical:






Kinetic energy (associated with motion, within system)
Potential energy (associated with position, within system)
Chemical
Electromagnetic
Nuclear
Energy is conserved. It can be transferred from
one object to another or change in form, but
cannot be created or destroyed
March 28, 2016
Kinetic Energy
Kinetic Energy is energy associated with the
state of motion of an object
 For an object moving with a speed of v

1 2
KE  mv
2

SI unit: joule (J)
1 joule = 1 J = 1 kg m2/s2
March 28, 2016
Why
1 2
KE  mv
2
?
March 28, 2016
Work W
1 2 1
2
mv  mv0  Fx x
 Start with
2
2
Work “W”
Work provides a link between force and energy
 Work done on an object is transferred to/from it
 If W > 0, energy added: “transferred to the
object”
 If W < 0, energy taken away: “transferred from
the object”

March 28, 2016
Definition of Work W

The work, W, done by a constant force on an
object is defined as the product of the component
of the force along the direction of displacement
and the magnitude of the displacement
W  ( F cos q ) x



F is the magnitude of the force
Δ x is the magnitude of the
object’s displacement
q is the angle between F and x
March 28, 2016
Work Unit

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


1 2 1
2
mv  mv0  ( F cos q )x
2
2
N•m=J
J = kg • m2 / s2 = ( kg • m / s2 ) • m
W  ( F cos q ) x
March 28, 2016
Work: + or -?

Work can be positive, negative, or zero. The
sign of the work depends on the direction of
the force relative to the displacement
W  ( F cos q ) x





Work
Work
Work
Work
Work
positive: W > 0 if 90°> q > 0°
negative: W < 0 if 180°> q > 90°
zero: W = 0 if q = 90°
maximum if q = 0°
minimum if q = 180°
March 28, 2016
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

W  ( F cos q ) x
March 28, 2016
Example: Work Can Be
Positive or Negative
Work is positive when lifting
the box
 Work would be negative if
lowering the box


The force would still be upward,
but the displacement would be
downward
March 28, 2016
Work Done by a Constant Force

The work W done on 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
WII   Fr
WI  0

F

F
 
W  F  r  Fr cosq

r

r
III
WIII  Fr
IV
WIV  Fr cos q
March 28, 2016
Work and Force

An Eskimo pulls a sled as shown. The total mass
of the sled is 50.0 kg, and he exerts a force of
1.20 × 102 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 ?
W  ( F cos q )x
 (1.20 10 2 N )(cos 30 )(5.0m)
 5.2 10 2 J
March 28, 2016
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  Wby individual forces

Remember work is a scalar, so
this is the algebraic sum
Wnet  Wg  WN  WF  ( F cos q )r
March 28, 2016
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 ?
Fnet , y  N  mg  F sin q  0
N  mg  F sin q
W fric  ( f k cos180 )x   f k x
   k Nx    k (mg  F sin q )x
 (0.200)(50.0kg  9.8m / s 2
Wnet  WF  W fric  WN  Wg
 1.2 10 2 N sin 30 )(5.0m)
 5.2 10 2 J  4.3 10 2 J  0  0
 90.0 J
 4.3 10 2 J
March 28, 2016
Kinetic Energy

Kinetic energy associated with the motion of
an object
1 2
KE 
2
mv
Scalar quantity with the same unit as work
 Work is related to kinetic energy

1 2 1
2
mv  mv0  Fnet x
2
2
Wnet  KEf  KEi  KE
March 28, 2016
March 28, 2016
Work-Kinetic Energy Theorem

When work is done by a net force on an object
and the only change in the object is its speed,
the work done is equal to the change in the
object’s kinetic energy


Speed will increase if work is positive
Speed will decrease if work is negative
Wnet  KEf  KEi  KE
Wnet
1 2 1
2
 mv  mv0
2
2
March 28, 2016
Work and Kinetic Energy

The driver of a 1.00103 kg car traveling on the interstate
at 35.0 m/s slam on his brakes to avoid hitting a second
vehicle in front of him, which had come to rest because of
congestion ahead. After the breaks are applied, a constant
friction force of 8.00103 N acts on the car. Ignore air
resistance. (a) At what minimum distance should the brakes
be applied to avoid a collision with the other vehicle? (b) If
the distance between the vehicles is initially only 30.0 m, at
what speed would the collisions occur?
March 28, 2016
Work and Kinetic Energy


3
3
(a) We know v0  35.0m / s, v  0, m  1.00 10 kg, f k  8.00 10 N
Find the minimum necessary stopping distance
Wnet
1 2 1 2
 W fric  Wg  WN  W fric  mv f  mvi
2
2
1 2
 f k x  0  mv0
2
1
 (8.00 10 N )x   (1.00 103 kg)(35.0m / s ) 2
2
3
x  76.6m
March 28, 2016
Work and Kinetic Energy



(b) We know x  30.0m, v0  35.0m / s, m  1.00 103 kg, f k  8.00 103 N
Find the speed at impact.
Write down the work-energy theorem:
1
1
Wnet  W fric   f k x  mv 2f  mvi2
2
2
2
v 2f  v02  f k x
m
2
3
2
2
v 2f  (35m / s) 2  (
)(
8
.
00

10
N
)(
30
m
)

745
m
/
s
1.00 103 kg
v f  27.3m / s
v0  35.0m / s, v  0, m  1.00 103 kg, f k  8.00 103 N
March 28, 2016
Work Done By a Spring
 Spring
force
Fx  kx
March 28, 2016
Spring at Equilibrium
F
=0
March 28, 2016
Spring Compressed
March 28, 2016
xf
lim  Fx x   Fx dx
x 0
xi
xf
xi
xf
xf
xi
xi
0
kx dx  12 kx 2
W   Fx dx   kx dx

 xmax
xf
W   kx dx  12 kxi2  12 kx 2f
xi
Work done by
spring on block
Fig. 7.9, p. 173
Measuring Spring Constant
Start with spring at its
natural equilibrium length.
 Hang a mass on spring and
let it hang to distance d
(stationary)
 From Fx  kx  mg  0

mg
k
d
so can get spring constant.
March 28, 2016
Scalar (Dot) Product of 2 Vectors

The scalar product of
two vectors is written
as A  B


A  B  A B cos q


It is also called the dot
product
q is the angle
between A and B
Applied to work, this
means
W  F r cosq  F  r
March 28, 2016
Dot Product


The dot product says
something about how parallel
two vectors are.
The dot product (scalar
product) of two vectors can be
thought of as the projection of
one onto the direction of the
other.
 
A  B  AB cos q

A  iˆ  A cos q  Ax

Components
 
A  B  Ax Bx  Ay By  Az Bz

B
( A cos q ) B

A
A( B cos q )
q
March 28, 2016
Projection of a Vector: Dot Product


The dot product says
something about how parallel
two vectors are.
The dot product (scalar
product) of two vectors can be
thought of as the projection of
one onto the direction of the
other.
 
A  B  AB cos q

A  iˆ  A cos q  Ax

Components
 
A  B  Ax Bx  Ay By  Az Bz
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
March 28, 2016
Derivation

 
A
How do we show that  B  Ax Bx  Ay By  Az Bz ?

Start with A  A iˆ  A ˆj  A kˆ
x
y
z

B  Bxiˆ  By ˆj  Bz kˆ

 
Then A  B  ( A iˆ  A ˆj  A kˆ)  ( B iˆ  B ˆj  B kˆ)
x
y
z
x
y
z

 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
March 28, 2016
Scalar Product


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
March 28, 2016