Download 4.1 The Concepts of Force and Mass

Chapter 6
Work and Energy
• Work done by a constant force
• Work-energy theorem, kinetic energy
• Gravitational potential energy
• Conservation of mechanical energy
• Conservative and non-conservative forces
• Work-energy theorem and
non-conservative forces
• Power
• Work done by a variable force
IMPORTANT:
YOU NEED TO KNOW HOW TO APPLY THESE
1)
v  vo  at
2)
1 2
x  xo  vot  at
2
3)
1
x  xo  v vo t
2
4)
2a x  xo   v 2 vo2
F 
2ax  v vo  2 x
m 
1
1
2
 Fx  mv  mv o2
2
2
2
2
Work-Energy theorem
1
Kinetic Energy : EK  mv 2
2
Work done = change in kinetic energy
1
1
2
Fx  mv  mv o2
2
2
Fx = “work” done by the force = force times displacement
The work changes the speed of the object thereby increasing its
kinetic energy.
Initial kinetic energy:
1
KE o  mv o2
2
Final kinetic energy:
1
KE  mv 2
2
Work done:
W  KE  KEo  KE
The unit of work and energy: Joule (J)
1 J  1 N m
Work Done by a Constant Force
The mechanical work done is equal to the force times the distance
something is moved:
W  Fs
1 N  m  1 joule J 
Work
Only the component of the force in the direction of the displacement
Counts:
Fx F cos 
a

m
m
1
1
 F cos  
2
2
2
2
So, 2
x  v vo  Fx cos   mv  mv o
2
2
 m 
Work done: W
 Fx cos   KE
WORK-ENERGY THEOREM
Work done: (force in direction of displacement)  (displacement)
A friction force fk = 70 N acts on the skis.
Initial speed, v0 = 3.6 m/s
Find final speed, vf =?, after skiing 57 m down the slope
Work-Energy Theorem: W = Fnet  (displacement) = KE

F

m
g
sin
25
 fk  net force down the slope

W  Fnet d  KEf  KEo  KE

F

m
g
sin
25
 fk

 58 kg  g sin 25   70 N  170 N
Displacement d = 57 m
So, W
 170 N  57 m  9690 J
1
1
2
2


KE o  mv o  58 3.6  375 .8 J
2
2
So,W
 9690 J  KEf  375.8 J  KEf  10066 J
1
2
and KEf  (58)vf  10066 J
 vf  18.6 m/s
2
Check:
2




vf  vo  2 Fnet / m d  3.6  2  57 170 / 58
2
2
vf  18.6 m/s
Serway, Physics for Scientists and Engineers
Problem 6.10
A 55-kg box is being pushed a distance of 7.0 m across the
floor by a force whose magnitude is 150 N. The force is
parallel to the displacement of the box. The coefficient of
kinetic friction is 0.25.
Determine the work done on the box by each of the four
forces that act on the box. Be sure to include the proper
plus or minus sign for the work done by each force.

FP  150 N

FN
m
fk ,x  0.25FN ,y

w
7m

FP  150 N

FN
m
fk ,x  0.25FN ,y

w
7m
No work is done in holding an object
at rest:
• no displacement, no work
• also no net work in lifting an object
up, then returning it to its starting
point as the net displacement is
zero
Work done in lifting an object
A force F lifts the mass at constant speed
through a height h.
The displacement is h
The applied force in the direction of the
displacement is:
F  mg
(no acceleration)
The work done by the force F is:
W  Fh  mgh
But the kinetic energy has not changed – the gravity force mg has
done an equal amount of negative work so that the net work done on
the mass is zero.
Work done in lifting an object
Alternative view:
define a different form of energyGravitational potential energy:
PE  mgy
Define:
Mechanical energy = kinetic energy + potential energy
Mechanical energy:
Then:
1
E  mv 2  mgy
2
Work done by applied force, F, is (change in KE) + (change in PE)
So,
W  Fh  KE  PE
Check, using forces and acceleration
Net upward force on the mass is F  mg
Apply Newton’s second law to find the acceleration:
F  mg  ma
So,
a
F  mg
m
From equation of kinematics number 4:
v 2  vo2  2ah
h
 vo  2F  mg 
m
2
So,
1
1
2
mv  mv o2  Fh  mgh
2
2
W  Fh  KE  PE
Two ways to look at the problem
1) A net upward force (F - mg) does work and
moves the mass upward and changes its kinetic
energy:
(F  mg)h  KE
Work - Energy Theorem 
2) An applied force F does work and changes
the kinetic and potential energies of the mass:
W  Fh  KE  mgh  KE  PE
The second is more powerful as it can be turned into a general
principle that:
Work done by an applied force = change in mechanical energy
If no work is done, then the mechanical energy is conserved!
W  Fh  KE  PE
If there is no net external force, W = 0 and
so that
0  KE  PE
KE  PE
As the mass falls and loses potential energy, it gains an equal
amount of kinetic energy.
Potential energy is converted into kinetic energy.
Energy is conserved overall.
Conservation of Mechanical Energy
In the absence of applied forces and friction:
The work done by a net zero applied force is zero
So, 0  KE  PE
And
KE + PE = E = total energy
is constant
Total Energy is Conserved
in a Closed System
Other kinds of potential energy:
• elastic (stretched spring)
• electrostatic (charge moving in an electric field)
No applied (i.e. external) forces
E = KE + PE = constant
KE = mv2/2
PE = mgy
So E = mv2/2 + mgy = constant, until the ball hits the ground
Check:
Problem 6.40
A particle, starting from point A in the drawing, is
projected down the curved runway. Upon leaving the
runway at point B, the particle is traveling straight
upward and reaches a height of 4.00 m above the
floor before falling back down.
Ignoring friction and air resistance, find the speed of
the particle at point A.
Conservative Forces
Gravitational potential energy depends only on height
The difference in PE, mg(ho - hf) is independent of path taken
Gravity is a “conservative force”
Conservative Forces
Alternative definitions of conservative forces:
• The work done by a conservative force in moving an object is
independent of the path taken.
(Compare pushing a crate across the floor – the most direct path
requires the least work. Friction is not a conservative force).
• A force is conservative when it does no net work in moving an
object around a closed path, ending up where it started.
In either case, the potential energy due to a conservative force
depends only on position.
Examples of conservative forces:
• Gravity, elastic spring force, electric force
vo  0
Skier starts from rest
A
v
B
Problem 6.42:
What must be the height h if the skier just loses contact
with the snow at the crest of the second hill at B?
FN
v
B
mg
r  36 m