Transcript Lecture 6

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 mv2
(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 FNET = F1 + F2 and the
displacement is S.
The work done by each force
is:
W1 = F1 r
W2 = F2 r
WNET = W1 + W2
= F1 r + F2 r
= (F1 + F2 ) r
WNET = FNET  r
F1
FTOT
r
F2
Review: Constant Force...
W = F r
• No work done if  = 90o.
T
– No work done by T.
v
v
N
– No work done by N.
Work/Kinetic Energy Theorem:
{Net Work done on object}
=
{change in kinetic energy of object}
WF = K = 1/2mv22 - 1/2mv12
v1
v2
F
m
x
WF = Fx
Work done by gravity:
• Wg = F r = mg r cos 
= -mg y
(remember y = yf - yi)
m
mg
r 
y
Wg = -mg y
Depends only on y !
j
m
Work done by gravity...
• W NET = W1 + W2 + . . .+ Wn
= F r 1+ F r2 + . . . + F rn
= F (r1 + r 2+ . . .+ rn)
= F r
= F y
Wg = -mg y
m
r1
y
r3
rn
Depends only on y,
not on path taken!
r
mg
r2
j
Falling Objects
• 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
vf
Free Fall
(a) Vf > Vi > Vp
vi
Frictionless incline
(b) Vf > Vp > Vi
vp
Pendulum
(c) Vf = Vp = Vi
Solution
v=0
v=0
v=0
H
vf
Free Fall
vi
vp
Frictionless incline
Pendulum
Only gravity will do work: Wg = mgH = 1/2 mv22 - 1/2 mv12 = 1/2 mv22
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:
Wg
x
x
• For variable force, we find the area
by integrating:
– dW = F(x) dx.
F(x)
x2
W   F ( x )dx
x1
x1
dx
x2
Work/Kinetic Energy Theorem for a
Variable Force
x2
W   F dx
x1
x2
m
x1
v2
dv dx
dt
mv
v1
F  ma  m dv
dt
dv dx dv
dv
=
= v dx (chain rule)
dt
dt dx
dv
dx
dx
v2
 m  v dv
v1
1
1
1
 m (v22 v12 )  m v22  mv12  ΔKE
2
2
2
1-D Variable Force Example:
Spring
• For a spring, Hooke’s Law states: Fx = -kx.
F(x)
x1
x2
x
relaxed position
-kx
F = - k x1
F = - k x2
Spring...
• The work done by the spring Ws during a
displacement from x1 to x2 is the area
under the F(x) vs x plot between
x1 and x2.
F(x)
x
x
1
relaxed position
2
x
Ws
-kx
Spring...
• The work done by the spring Ws during a
displacement from x1 to x2 is the area
under the F(x) vs x plot between
x1 and x2.
x2
Ws   F ( x )dx
F(x)
x1
x1
x2
x2
  ( kx )dx
x
Ws
-kx
x1
1
  kx 2
2
Ws  
x2
x1
1
k x22  x12 
2
Work & Energy
• A box sliding on a horizontal frictionless
surface runs into a fixed spring,
compressing it a distance x1 from its
relaxed position while momentarily coming
to rest.
– If the initial speed of the box were doubled and
its mass were halved, how far x2 would the
spring compress ?
(a)
x2  x1
(b) x2  2 x1
x
(c)
x2  2 x1
Lecture 10, Act 2
Solution
• Again, use the fact that WNET = K.
WNET = WSPRING = -1/2 kx2
K = -1/2 mv2
In this case,
and
so
kx2 =
mv2
v1
In the case of x1
x1
m1
m1
x1  v1
m1
k
x v
Lecture 10, Act 2
Solution
m
k
So if v2 = 2v1 and m2 = m1/2
x 2  2v 1
m1 2
k
 v1
2m1
k
x2  2 x1
v2
x2
m2
m2