work, work-energy theorem and kinetic energy

1. Presented by:
Karen A. Adelan
BSE 3
Classical Mechanics

2. 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

3. 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

4. Kinetic Energy
 Kinetic Energy is energy associated with the
state of motion of an object
 For an object moving with a speed of v
KE
1 2
mv
2
 SI unit: joule (J)
1 joule = 1 J = 1 kg m2/s2

6. Work W
1 2 1
2
Fx x
 Start with 2 mv 2 mv0
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”

7. 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
1 2 1
2
 SI Unit
mv
mv0 ( F cos
2
2
 Newton • meter = Joule
N • m = J
 J = kg • m2 / s2 = ( kg • m / s2 ) • m

W
( F cos ) x
) x

8. 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 ) x
Work positive: W > 0 if 90°> > 0°
Work negative: W < 0 if 180°> > 90°
Work zero: W = 0 if = 90°
Work maximum if = 0°
Work minimum if = 180°

9. 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 ) x

10. 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

11. 
F
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:
W

F

r

F

r

r
II
I

0F
WI
WII
F r

F

r

r
F r cos
III
WIII
F r
IV
WIV
F r cos
February 11, 2014

12. 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
Wnetnet
W

Wby by individual forces
Windividual forces
Remember work is a scalar, so
this is the algebraic sum
Wnet Wg WN WF
Wnet Wg WN WF
( F cos ) r
( F cos ) r
February 11, 2014

13. 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 ?
( f k cos180 ) x
W fric
k
N x
k
fk x
(mg F sin ) x
(0.200)(50.0kg 9.8m / s 2
1.2 102 N sin 30 )(5.0m)
4.3 102 J
Wnet
WF W fric WN Wg
5.2 102 J
90.0 J
4.3 102 J
0 0

14. Kinetic Energy
object is the energy which it possesses due to
its motion.[1] It is defined as the work needed to accelerate
a body of a given mass from rest to its stated velocity.
Having gained this energy during its acceleration, the body
maintains this kinetic energy unless its speed changes. The
same amount of work is done by the body in decelerating
from its current speed to a state of rest.
In classical mechanics, the kinetic energy of a non-rotating
object of mass m traveling at a speed v is ½ mv².
In relativistic mechanics, this is only a good approximation
when v is much less than the speed of light.

15. Work-Kinetic Energy Theorem
Is the net work done on an object is
equal to the change in the kinetic
energy of the object.
Wnet = ∆KE
Net work is equal to kinetic energy

16. • Kinetic energy depends on speed and mass:
KE = ½mv2
Kinetic energy = ½ x mass x (speed)2
KE is a scalar quantity, SI unit (Joule)

17. • TRY TO SOLVE: Ex. A 7.00 kg bowling ball
moves at 3.00 m/s. how much kinetic
energy does the bowling ball have? how
fast must a 2.24 g table-tennis ball move in
order to have the same kinetic energy as
the bowling ball? Is the speed reasonable
for a table-tennis ball?
Given:
the subscript b and t indicate the
bowling ball and the table-tennis ball,
respectively.
Mb = 7.00 kg Mt = 2.24g Vb = 3.00m/s
Unknown: KEb = ? Vt = ?

18. 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
1
mv 2
2
1
2
mv0
2

19. Work-Energy Theorem
W= KE
W=KEf-KEi
“In the case in which work is done on a system
and the only change in the system is in its
speed, the work done by the net force equals
the change in kinetic energy of the system.”

20. • The Work-Kinetic Energy Theorem
can be applied to non isolated
systems
• A non isolated system is one that is
influenced by its environment
(external forces act on the system)

21. 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?

22. Work and Kinetic Energy
3
v 0 35/,s m s m 1. m. 10 10 kg3 f 8 f k8 10 310 10 3
v
8 N
v0 0 35 .00.mms.0, v /0,,0, m 0,00001.00 3 ,10k, kg, .00 .00.00 3 N N
kg f k
35 / v
1
 (a) We know v

Find the minimum necessary stopping distance
1 2 1 2
Wnet W fric Wg WN W fric
mv
mv
1 22 f 2 i
f k x 0 1 mv0
2
f k x 0 2 mv0
2
1
1 103 kg)(35.0m / s) 2
(8.00 108.N ) 103 N ) x (1.001.00 103 kg)(35.0m / s) 2
( 00 x
(
2
2
3
x 76.6.6m
x 76m
February 11, 2014

23. Work and Kinetic Energy



x
30 .0m, v0
35 .0m / s, m 1.00 10 3 kg, f k
(b) We know
Find the speed at impact.
Write down the work-energy theorem:
1 2
Wnet W fric
fk x
mv f
2
2
2
v 2 v0
fk x
f
m
v2
f
vf
1 2
mvi
2
2
(35 m / s) 2 (
)(8.00 10 3 N )(30 m)
1.00 10 3 kg
27.3m / s
8.00 10 3 N
745 m 2 / s 2