Physics 2011
Chapter 6: Work and Kinetic
Energy

Work

The Physics of Work
• By strict definition, in order for work to be performed, a Net Force must be applied to a body, resulting in the Displacement of that body.
Work = Force * Displacement
= Newtons * Meters
= Joules (Energy)

Calculating Work from Vectors
• Consider the Idiot pushing his girlfriend’s car in one direction while she steers in another:
• The useful work is:
• Thus the Scalar, Work, is a DOT PRODUCT:

Work has a Sign
• Work is calculated by finding the component of
Force acting along the line of Displacement, but they may be in opposite directions.
• ALSO, Work is W and Weight is w …..OK?

Work is ENERGY
• Work is the product of a Net Force and an accompanying displacement
• A body under the influence of a Net Force is accelerating (F = ma)
• An accelerating body is said to have increasing Kinetic Energy

Kinetic Energy
• A body with Mass, m , moving at velocity, v , has some ability to perform Work
(For example, a bowling ball rolling down the alley can knock over pins)
• This ability of a moving body to do work
(Work is Energy) is quantified as:
Kinetic Energy,
K = ½ mv 2 (Joules)

Work-Energy
• Positive Work on a Body INCREASES its
Kinetic Energy
• Negative Work on a Body DECREASES its Kinetic Energy
• A body that gains K must increase in speed and a body that loses K must decrease in speed.

gotta have POWER!!!!
Power is the RATE of Work: i.e. Power is the change in work over some unit of time
P = ΔW / Δt
(Average Power)
P = dW/dt (Instantaneous Power)
Power is Joules/Seconds or Watts

Review: Sum of Constant
Forces
Suppose F
NET
= F
1
+ F
2 and the displacement is S .
The work done by each force is:
W
W
1
2
= F
1

 r
= F
2

 r
W
NET
W
NET
= W
1
= F
1

+ W
2
 r + F
2

= (
= F
F
1
+ F
2
NET

)
 r

 r
 r
F
1
F
TOT
 r
F
2

Review: Constant Force...
W = F

 r
• No work done if 
= 90 o .
– No work done by T .
T v v
N
– No work done by N .

Work/Kinetic Energy Theorem:
{ Net Work done on object }
=
{ change in kinetic energy of object } v
1
W
F
=

K = 1 /
2 mv
2
2 1 /
2 mv
1
2 v
2
F m
W
F
= F
 x
 x

Work done by gravity:
• W g
= F

 r = mg  r cos

= mg
 y
(remember
 y = y f
- y i
) m
 r

 y mg
W g
= mg
 y m
Depends only on
 y !
j

Work done by gravity...
•
W
NET
= W
1
+ W
2
+ . . .+ W n
= F 
 r
1
+ F 
 r
2
= F  (
 r
1
= F 
 r
+
 r
2
+ . . . +
+ . . .+
 r
F n
)

 r n
= F
 y
W g
= mg
 y
 y  r
3
 r
 r
2
 r
1 m mg j
 r n
Depends only on
 y, not on path taken !

• Three objects of mass m begin at height h with velocity 0 .
One falls straight down, one slides down a frictionless inclined plane, and one swings on the end of a pendulum. What is the relationship between their velocities when they have fallen to height 0? v=0 v=0 v=0
H v f
Free Fall
(a) V f
> V i
> V p v i v p
Frictionless incline Pendulum
(b) V f
> V p
> V i
(c) V f
= V p
= V i

v = 0
v = 0 v = 0 v f
Free Fall
H v i
Frictionless incline v p
Pendulum
Only gravity will do work: W g
= mgH = 1 /
2 mv
2
2 1 /
2 mv
1
2 = 1 /
2 mv
2
2 v f
 v i
 v p

2 gH does not depend on path !!

Work done by Variable Force:
(1D)
• When the force was constant, we wrote W = F
 x
F
– area under F vs. x plot: W g x
 x
• For variable force, we find the area by integrating:
– dW = F(x) dx .
F(x)
W
 x

2
F x
1
( x ) dx x
1 dx x
2

Work/Kinetic Energy Theorem for a
Variable Force
W

 x

2
F dx x
1 m x

2 x
1 dv dt dx
 m v
2
 v
1 v dv dx dx
F
 ma
 m dv dt dv dt
= dx dt dv dx
= v dv dx
(chain rule)
 m v
2
 v dv v
1
 m
1
2
( v
2
2
 v
1
2 )

1
2 m v
2
2

1
2 m v
1
2
 ΔKE

1-D Variable Force Example:
Spring
• For a spring, Hooke’s Law states: F x
= -kx .
F(x) x
1 x
2 x relaxed position
-kx
F = - k x
1
F = - k x
2

Spring...
• The work done by the spring W s during a displacement from x
1 to x
2 is the area under the F(x) vs x plot between x
1 and x
2
.
F(x) x
1 x
2 relaxed position x
W s
-kx

Spring...
F(x)
• The work done by the spring W s during a displacement from x
1 to x
2 is the area under the F(x) vs x plot between x
1 and x
2
.
W s
 x
 2 x
1
F ( x ) dx x
1 x
2
 x

2 x
1
(
 kx ) dx x x
2
W s
 
1
2 kx 2
-kx x
1
W s
 
1
2 k
 x 2
2
 x
1
2


• A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x
1 from its relaxed position while momentarily coming to rest.
– If the initial speed of the box were doubled and its mass were halved spring compress ?
, how far x
2 would the
(a) x

2 x
1 x
2

2 x
1
(c) x

2
2 x
1 x

• Again, use the fact that W
NET
In this case, W
NET and
= W

K = 1 /
SPRING
2 mv 2
= 1 /
2 kx 2
=

K.
so kx 2 = mv 2 In the case of x
1 x
1
 v
1 m
1 k x
1 m
1 v
1 m
1

x
 v m
k
So if v
2
= 2v
1 and m
2
= m
1
/2 x
2

2 v
1 m
1 k
2
 v
1
2m
1 k x

2
2 x
1 x
2 m
2 v
2 m
2