Chapter 7
Kinetic Energy and Work

Kinetic Energy of a mass m is :
K  mv
1
2


2
Energy is a scalar quantity
The SI unit of energy is
1 joule  1 J  1 kg  m / s
2

2
This unit is named after an English scientist of
the 1800s, James Prescott Joule.
1
Sample Problem 7-1
In 1896 in Waco, Texas, William Crush of the “Katy” railroad
parked two locomotives at opposite ends of a 6.4-km-long track,
fired them up, tied their throttles open, and then allowed them to
crash head-on at full speed in front of 30,000 spectators.

Hundreds of people were
hurt by flying debris;
several were killed.
Assuming each locomotive
weighed 1.2 x 106 N and
its acceleration along the
track was a constant 0.26
m/s2, what was the total
kinetic energy of the two
locomotives just before the
collision?
2
SOLUTION:
v 2  v 0  2a (x  x 0 )
2
v 2  0  2 (0.26 m / s 2 ) (3.2 x103 m)
v  40.8 m / s
1.2 x 10 6 N
5
m

1
.
22
x
10
kg
2
9.8 m /s
K  2 ( 12 mv 2 )  (1.22 x 105 kg) (40.8 m / s) 2
 2.0 x 10 J
8
3
Work



Work W is positive if kinetic energy is
transferred to an object by an external force.
Work is negative if kinetic energy is transferred
from an object. In other words, negative work is
done if an object’s kinetic energy decreases.
Work is a scalar and the SI unit is Joule.
4
Work done by an External Force
W  F d cos 
 
W  F d

To calculate the work done on an object by a force
during a displacement, we use only the force component
along the object's displacement. The force component
5
perpendicular to the displacement does zero work.

A force does positive work when it has a vector
component in the same direction as the
displacement, and it does negative work when it
has a vector component in the opposite
direction. It does zero work when it has no such
vector component.
1 J  kg  m / s  1 N  m  0.738 ft  lb
2
2
6
 K  K f  Ki  W
 change in the kinetic

 energy of a particle
  net work done on 
  

  the particle

Kf  Ki  W
 kinetic energy after   kinetic energy   the net 

  
  

 the net work is done   before the net work   work done 
7
Sample Problem 7-2
Figure 7-4a shows two industrial spies sliding an initially

stationary 225 kg floor safe a displacement d of
magnitude 8.50 m, straight toward their truck.


The push F1 of Spy 001 is 12.0 N, directed atan angle
of 30° downward from the horizontal; the pull F2 of Spy
002 is 10.0 N, directed at 40° above the horizontal. The
magnitudes and directions of these forces do not
change as the safe moves, and the floor and safe make
frictionless contact.
8
(a) What
is thenet work done on the safe by


forces F1 and F2 during the displacement d ?
SOLUTION: 
Work done by F1 :
W1  F1 d cos 1  (12.0 N)(8.50 m)(cos 30 )
 88.33 J

Work done by F2 :
W2  F2 d cos  2  (10.0 N)(8.50 m)(cos 40 )
 65.11 J
Total work done :
W  W1  W2  88.33 J  65.11 J
 153.4 J  153 J
9
(b) During the displacement, what is the work Wg done on
the safe by the gravitational force Fg and what is the work

WN done on the safe by the normal force N from the floor?
SOLUTION:
Wg  mgd cos 90  0

WN  Nd cos 90  0

10
(c) The safe is initially stationary. What is its speed vf at
the end of the 8.50 m displacement?
SOLUTION:
Work done on object equals increase in kinetic energy :
W  K f  K i  mv f  mv i
1
2
2
1
2
2
2W
2 (153.4 J)
vf 

m
225 kg
 1.17 m / s
We have assumed no frictional forces exist.
11
Sample Problem 7-3

During a storm, a crate of crepe is sliding
 across a slick,
oily parking lot through a displacement d while a
steady wind pushes against the crate with a force

F  (20 N) î  ( 6.0 N) ĵ. The situation and coordinate axes
are shown in Fig. 7-5.
12
(a) How much work does this force from the wind
do on the crate during the displacement?
SOLUTION:
Work done by the wind force on crate :
 
W  F  d  (2.0 N) î  (  6.0 N) ĵ   ( 3.0 m) i 


 (2.0 N) ( 3.0 m) î  î  (  6.0 N) (  3.0 m) ĵ  î
 (  6.0 J) (1)  0   6.0 J
The wind force does negative work, i.e. kinetic energy is
taken out of the crate.
13
(b) If the crate has a kineticenergy of 10 J at the
d , what is its kinetic
beginning of displacement

energy at the end of d ?
SOLUTION:
K f  Ki  W  10 J  (  6.0 J)  4.0 J
14
Work done by Gravitational Force


Wg  m g d cos 

Fg (down)
The angle

=
is
between

and d . For the upward path,
.
  180
Wg (up path )  m g d cos180  m g d (1)   m g d


The work done by the gravitational
force is negative for the upward
path.
The change of gravitational potential
energy of the object equals the
negative of the work done by the
gravitational force.
15

After the object has reached its maximum height and is
falling back down, the angle  = 0o. The work done is
Wg (down path )  m g d cos 0  m g d (1)   m g d
16
Work done on Lifting and Lowering an
Object against a Field at Constant Velocity

An applied force of magnitude = mg pointed up
can barely lift an object of weight mg. Assuming
no change in the velocity, the work done by the
applied force on the object in lifting it a distance
d is :
Wa  mgd   Wg (up path )

The work done by the applied force in lifting the
object is positive. It has the opposite sign as the
work done by the gravitational force.
17

The applied force of magnitude mg (pointed up) lowers
the object with no change in velocity. The work done by
the applied force on the object in lowering it a distance
d is :
Wa   mgd


The work done by the applied force in lowering the
object is negative.
The work done by the applied force equals the change
in the gravitational potential energy of the object.
18
Sample Problem 7-4
Let us return to the lifting feats of Andrey
Chemerkin and Paul Anderson.

(a) Chemerkin made his record-breaking lift with rigidly
connected objects (a barbell and disk weights) having a total
mass m = 260.0 kg; he lifted them a distance of 2.0 m. During
the lift, how much work
was done on the objects by the

gravitational force Fg acting on them?
SOLUTION:
The gravitational force points down and the displacement
d points up :
Wg (up path )  mgd cos   (2548 N) (2.0 m) (cos 180 )
  5100 J
The work done by the gravitational force is negative to
the work done by the applied force.
19
(b) How much work was done on the objects by
Chemerkin's force during the lift?
SOLUTION:
The work done by the applied force is :
WAC (up )   Wg (up )   5100 J
(c) While Chemerkin held the objects stationary
above his head, how much work was done on
them by his force?
Since the displacement d is zero, the work done is zero.
20
(d) How much work was done by the force Paul
Anderson applied to lift objects with a total weight
of 27 900 N, a distance of 1.0 cm?
SOLUTION:
The work done by the applied force of Paul Anderson is :
WPA (up path )   Wg (up path )
 (27900 N) (0.010 m)  280 J
21
Sample Problem 7-5

An initially stationary 15.0 kg
crate of cheese wheels is
pulled, via a cable, a distance
L = 5.70 m up a frictionless
ramp, to a height h of 2.50 m,
where it stops (Fig. 7-8a).
(a) How much work Wg is
done on the crate by
 the
gravitational force Fg during
the lift?
22
SOLUTION:
The work done by the gravitational force during the lift
is negative :
Wg (up path )   mg sin  d
Since d sin  = h, we have :
Wg (up path )   mgh
  (15.0 kg) (9.8 m / s 2 ) (2.50 m)
  368 J
23
(b) How much
work WT is done on the crate by

the force T from the cable during the lift?
SOLUTION:
Since the crate has zero velocity before and after the
lift, the work done by the applied force must be equal
and opposite to the work done by the gravitational
force.
WT   Wg (up path )  368 J
24
Sample Problem 7-6

An elevator cab of mass m
= 500 kg is descending with
speed vi = 4.0 m/s when its
supporting cable begins to
slip, allowing it to fall with
constant
  acceleration
a = g /5.

(a) During the fall through a
distance d = 12 m, what is
the work Wg done on the
cab by the gravitational
force Fg ?
25
SOLUTION:
During the fall, the work done on the cab by the
gravitational force is positive.
Wg (down path )  mgd cos 0  (500 kg) (9.8 m / s 2 ) (12 m) (1)
 5.88 x 10 4 J  59 kJ
26
(b) During the 12 m fall, what is the work WT done

on the cab by the upward pull T of the elevator
cable?
SOLUTION:
mg  T  ma
(positive direction is down)
T  m (g  a )  54 mg
This gives the magnitude of
T, the direction of T is up
The work done on the cab by T during the fall is negative.
WT (down path )   54 mg d
  4.70 x 10 4 J   47 kJ
27
(c) What is the net work W done on the cab
during the fall?
SOLUTION:
W  Wg  WT  5.88 x 10 4 J  4.70 x 10 4 J
 1.18 x 10 4 J  12 kJ
(d) What is the cab's kinetic energy at the end of
the 12 m fall?
SOLUTION:
K f  K i  W  mv i  W
1
2
2
 12 (500 kg) (4.0 m / s) 2  1.18 x 10 4 J
 1.58 x 10 4 J  16 kJ
28
The Spring Force


When a spring of length L0 is either
compressed or extended,
 a restoring
force acts opposite to d .
This restoring force is known as the
spring force and is given by :


FS   k d



The constant k is a scalar and is
called the spring constant or force
constant. The SI unit is N/m.
The negative sign indicates that the
spring force is always opposite in
direction to the displacement.
Hooke’s law is named after Robert
29
Hooke of the late 1600s.
Work done by the Spring Force

The work done by the spring force as the block is
moved from position xi to xf is :
xf
WS   F dx
xi
xf
xf
xi
xi
WS   ( kx ) dx   k  x dx
 ( 12 k )[ x 2 ]xx if  ( 12 k ) ( x f  x i )
2

2
We have assumed that all the mass is at the block and
the spring itself is massless.
30


The work done by the spring force on the block is
negative if xf > xi.
If we set at L0 , xi = 0, then the work done by the spring
force on the block for extension x is
WS   k x
1
2

2
Change in potential energy of spring = - work done by
the spring force.
P.E. change  kx
1
2

2
The work done by the spring force is to increase P.E.
and to decrease K.E.
31
Work Done on the Spring by
an Applied Force

If a block is stationary before and after a displacement,
the work done on the block by an applied force is
negative to the work done on it by the spring force.

Therefore, for an extension x, the work done by an
applied force on the block is :
WA  k x
1
2


2
The work done by the applied force equals to elastic
potential energy of the spring.
The work done by the applied force is to increase the
P.E. of the spring.
32
Sample Problem 7-7
A package of spicy Cajun pralines lies on a frictionless
floor, attached to the free end of a spring in the
arrangement of Fig. 7-10a. An applied force of magnitude
Fa = 4.9 N would be needed to hold the package
stationary at x1 = 12 mm.
(a) How much work does the spring force do on the
package if the package is pulled rightward from x0 = 0 to
x2 = 17 mm?
33
SOLUTION:
FS   k x
FS
 4.9 N
k 
 408 N / m
3
x1
12 x 10 m
Work done by the spring is :
3
Ws   kx 2   (408 N / m) (17 x 10 m)
1
2
2
1
2
2
  0.059 J
34
(b) Next, the package is moved leftward to x3 = -12 mm.
How much work does the spring force do on the package
during this displacement? Explain the sign of this work.
SOLUTION:
Ws   12 kx f  12 kx i
2
2

 12 (408 N / m)  ( 12 x 10 3 m) 2  (17 x 10 3 m) 2

 0.030 J  30 mJ
The spring force does positive work as the block moves
from +17mm to its relaxed position and negative work as
the block moves from the relaxed position to -12mm. The
former work is larger resulting in WS being positive.
35
Sample Problem 7-8

In Fig. 7-11, a cumin canister of mass m = 0.40 kg
slides across a horizontal frictionless counter with
speed v = 0.50 m/s. It then runs into and compresses a
spring of spring constant k = 750 N/m. When the
canister is momentarily stopped by the spring, by what
distance d is the spring compressed?
36
SOLUTION:
We assume the spring is massless. Work done by the
spring on the canister is negative. This work is :
WS   12 kd 2
Kinetic energy change of the canister is :
k f  k i   12 mv 2
Therefore,
 12 kd 2   12 mv 2
m
0.40 kg
dv
 (0.50 m / s)
k
750 N / m
 1.2 x 10  2 m  1.2 cm
37
Work done by a General Variable Force
W
 
F  d r ( work by var iable force )
38
Power

The instantaneous power P is
dW
P
dt

Power is a scalar and the SI unit is J / s which is known
as the Watt (W), named after James Watt.
1 watt  1 W  1 J / s  0.738 ft  lb / s
1 horsepower  1 hp  550 ft  lb / s  746 W
1 kiwowatt  hour  1kW  h  (103 W) (3600 s)
 3.60 x 10 6 J  3.60 MJ

Note that the kilowatt-hr is a unit of work
39
 dx 
d W F cos  d x

P

 F cos  
dt
dt
 dt 
P  F v cos 
The angle  is between the force and velocity.
Therefore,
 
P  F v
40
Sample Problem 7-10



Figure 7-14 shows constant forces F1 and F2
acting on a box as the box slides rightward
across a frictionless floor. Force
F1 is horizontal,

with magnitude 2.0 N; force F2 is angled upward
by 60° to the floor and has magnitude 4.0 N.
The speed v of the box at a certain instant is 3.0
m/s.
41
(a) What is the power due to each force acting on the box
at that instant, and what is the net power? Is the net power
changing at that instant?
SOLUTION:
P1  F1 v cos 180  (2.0 N) (3.0 m / s) cos 180


  6.0 W
P2  F2 v cos 60   (4.0 N) (3.0 m / s) cos 60 
 6.0 W
Pnet  P1  P2
0
The kinetic energy of the box is not changing. The speed of
the box remains at 3 m/s. The net power does not change.
42

(b) If the magnitude of F2 is, instead, 6.0 N, what
now is the net power, and is it changing?
SOLUTION:
P2  F2 v cos 60   (6.0 N) (3.0 m / s) cos 60 
 9.0 W
Pnet  P1  P2   6.0 W  9.0 W
 3.0 W
There is a net rate of transfer of energy to the box. The
kinetic energy of the box increases. The net power also
increases.
43
Homework (due Oct 18)









5P
7E
11P
17P
19P
21E
23P
31E
33P
44