Mechanics 105

Work done by a constant force

Scalar product

Work done by a varying force

Kinetic energy, Work-kinetic energy theorem

Nonisolated systems

Kinetic friction

Power

Mechanics 105
In discussion of work and energy, it is important that we are clear about the objects we are considering. A system is an object, group of objects or region of space with well defined boundaries.

Mechanics 105

Effect of force acting on an object over some distance.
W

F
 r cos


The component of the force in the direction of the displacement, multiplied by the magnitude of the displacement (or the component of the displacement in the direction of the force times the magnitude of the force.)

Result is a scalar quantity

Units: N·m

J b(Joule)

Mechanics 105
product of two vectors:
A

B

A B cos
 e.g.
W

F
 r cos
 

F
  r
 where
 is the angle between the two vectors
In terms of components:

A


B

A x
B x

A y
B y

A z
B z

Mechanics 105

Commutative

A


B


B


A

Distributive

A

(

B


C )


A


B


A


C for the unit vectors i
ˆ  i
ˆ  i
ˆ  j
ˆ  j
ˆ  j
ˆ  k
ˆ  k
ˆ 
1 j
ˆ  k
ˆ  k
ˆ  i
ˆ 
0

Mechanics 105

Increment of work: force applied over small displacement

W


F
  r


Total work: sum of increments
W
  
W
  i

F (
 r i
)
  r
 i

Taking the limit for infinitesimally small displacements
W
  
W

 lim r

0
 i

F (
 r i
)
  r
 i

 r i
 r
 f

F ( r

)
 d r


Mechanics 105
Example: work done pushing a block up a frictionless inclined plane at constant

F

N velocity
 m g

F

F x

F mg sin

 mg sin
 
0
 x W
  x f x i
Fdx
 mg sin
  x

F

Mechanics 105

What do Winnie the Pooh and Attila the
Hun have in common?

Mechanics 105

What do Winnie the Pooh and Attila the
Hun have in common?

Answer: Same middle name.

Mechanics 105

F x-direction will do an amount of work
W
 x x i f

F
 x
Using Newton’s
2 nd law
W
  f x x i
  dx
 m
 f x x i dv dt d x
 m
 f x x i dv dx dx d x
 dt m
 f v v i dx d v dt

1
2 mv
2 f

1
2 mv i
2

Mechanics 105
From this expression we define the kinetic energy
K

1
2 mv
2 and the expression for net work becomes
W

1
2 mv
2 f

1
2 mv i
2 
K f

K i
 
K
This is the work-kinetic energy theorem

Mechanics 105
A block slides down a frictionless plane – what is the velocity at the bottom?
 x
Forces perpendicular to the plane do no work

N

 m g 
W

F x
 mg
F
 x
 sin mg
 sin
  x sin
  x
 initial
W
 mgh height of block
 h

Mechanics 105

Mechanics 105

Mechanics 105
Example – work done by a spring
 Force exerted by spring (Hooke’s law)
F s
( x )
  kx where x is the displacement of the spring from its unstretched position
The work done by the spring is
W
  f x x i
F s
( x ) dx

1
2 kx i
2 
1
2 kx
2 f

Mechanics 105
Example – work done by a spring
A spring (k=100 N/m) is slowly stretched by 2 cm from its unstretched length. What is the work done by the force of the spring?
W

1
2 kx i
2 
1
2 kx
2 f

0

1
2
( 100 N / m )( 0 .
02 m )
2  
0 .
02 J
The result is negative since the force is applied in the opposite direction to the displacement.
What would be the work done by the spring if it were compressed by the same amount?
What is the work done by the force stretching or compressing the spring?

Mechanics 105
Example – speed of a block on a spring
The same spring in the last example is put in contact with a block
(mass m=1.00 kg), compressed 2.00 cm and then let go. How fast is the block moving when it loses contact with the spring?
v
Work done by the spring (now positive – spring is
W returning to unstretched length)

1
2 kx i
2

1
2 kx
2 f

1
2
( 100 .
N / m )( 0 .
0200 m )
2

0 .
00

2 .
00

10

2
J

Must equal the change in kinetic energy
K

1
2 mv
2 f

1
2 mv i
2 
1
2
( 1 .
00 kg ) v
2 
0 .
00 v

0 .
200 m / s

2 .
00

10

2
J

Mechanics 105
Example –block dropped onto a spring
Same spring, same mass, now the mass is dropped from a height 1.00 m above the uncompressed spring. How far down does the spring compress?
Work done by gravity on the block is
W g
 mg ( 1 .
00 m
 d )
1.00m + d
The work done on the block by the spring is
W s

1
2 kx
2 i

1
2 kx
2 f

0 .
00

1
2
( 100 .
N / m )( d )
2
Since the change in kinetic energy is zero, the total work done must also be zero
W
 g

W
9 .
80 J s


( mg
9 .
( 1 .
80 N
00
/ m m )
 d )
 d
1
( 100 .
N
2

50 .
0 ( N / m )
/ d m )( d
2 
0
)
2 d

0 .
551 m

0

Mechanics 105
Nonisolated systems

Work can be thought of as the transfer of energy between a system and its environment

Forms of energy of a system other than kinetic: internal
(thermal)

Ways of energy transfer other than work (mechanical waves, heat, matter transfer, electrical transmission,
EM radiation)

Mechanics 105
Kinetic friction
The new form of the work-KE theorem:

K
  f k
 x
 
W looks the same, but note that now the application point of the force (friction) is changing
For the large system of both objects
0
 
E int
 
K
 
E int
 f k
 x

E int
 f k
 x

Mechanics 105
Example
A block slides down an inclined plane with coefficient of kinetic friction
 k
– what is the velocity at the bottom?
 x f
 k

N
  m g


W
F x
 mg sin

F y


N
 mg
F x
 x

  f k cos


 mg (sin
 mg
0

 sin
 k
N


  k mg cos

N cos

)
 x
 
K

1
2 mv f

Mechanics 105
Power
 power = rate of energy transfer
 average power
P

W
 t
 instantaneous power
P
 dW dt


F
 d r
 dt


F
 v


 more general definition:
Units: Watt (=J/s)
P
 dE dt hp=550 ft·lb = 746 W

Mechanics 105
Example
How much power is needed to accelerate a 1.00

10 3 kg car from 0-60.0 mph in 5.00 seconds, ignoring air resistance and friction? How much is needed to keep it moving at 60.0 mph if friction and air resistance equal
1.00

10 2 N?
Assuming a constant acceleration (60 mph = 26.8 m/s)
P


W
 t



K t

1
2
 mv t
2
0 .
500 ( 1 .
00

10
3 kg )( 26 .
8 m / s )
2
5 .
00 s

7 .
18

10
4
W

Mechanics 105
To keep it moving at a constant velocity, the magnitude of the applied force must equal that from the air resistance and friction
P


F

 v

( 1 .
00

10
2
N )( 26 .
8 m / s )

2680 W