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Chapter 7. Kinetic Energy and Work
7.1. What is Physics?
7.2. What Is Energy?
7.3. Kinetic Energy
7.4. Work
7.5. Work and Kinetic Energy
7.6. Work Done by the Gravitational Force
7.7. Work Done by a Spring Force
7.8. Work Done by a General Variable Force
7.9. Power
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What is Physics?
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Chapter 7
Kinetic Energy and Work
In this chapter we will introduce the following concepts:
Kinetic energy of a moving object
Work done by a force
Power
In addition, we will develop the work-kinetic energy theorem and
apply it to solve a variety of problems.
This approach is an alternative approach to mechanics. It uses
scalars such as work and kinetic energy rather than vectors such as
velocity and acceleration. Therefore it is simpler to apply.
(7-1)
3
Kinetic Energy:
We define a new physical parameter to describe the state of
motion of an object of mass m and speed v.
We define its kinetic energy K as
mv 2
K
.
2
Unit for Kinetic energy is:
Kinetic energy is a scalar quantity.
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(7-2)
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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?
5
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
m
1.2 x 10 6 N
 1.22 x 10 5 kg
2
9.8 m /s
K  2 ( 12 mv 2 )  (1.22 x 105 kg) (40.8 m / s) 2
 2.0 x 108 J
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Work
Work W is energy transferred to or from an object by
means of a force acting on the object.
• Energy transferred to the object is positive work,
• Energy transferred from the object is negative work.
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m
m
Work: (symbol W)
If a force F is applied to an object of mass m it can accelerate it and
increase its speed v and kinetic energy K. Similarly F can decelerate m
and decrease its kinetic energy.
We account for these changes in K by saying that F has transferred energy
W to or from the object. If energy is transferred to m (its K increases) we
say that work was done by F on the object (W > 0). If on the other hand,
energy is transferred from the object (its K decreases) we say that work
was done by m (W < 0).
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m
m
Finding an Expression for Work :
Consider a bead of mass m that can move
without friction along a straight wire along
the x-axis. A constant force F applied at an
angle  to the wire is acting on the bead.
We apply Newton's second law: Fx  max . We assume that the bead had an initial
velocity v0 and after it has traveled a distance d its velocity is v . We apply the
third equation of kinematics: v 2  v02  2ax d . We multiply both sides by m / 2 
m 2 m 2 m
m F
m
v  v0  2ax d  2 x d  Fx d  F cos  d .
K i  v02
2
2
2
2 m
2
m 2
K f  v  The change in kinetic energy K f  K i  Fd cos .
2
Thus the work W done by the force on the bead is given by W  Fx d  Fd cos .
W  Fd cos 
FA
W  F d
W  Fd cos 
FC
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m
m
W  F d
FB
The unit of W is the same as that of K , i.e., joules.
Note 1:The expressions for work we have developed apply when F is constant.
Note 2:We have made the implicit assumption that the moving object is point-like.
Note 3: W  0 if 0    90, W  0 if 90    180.
Net Work: If we have several forces acting on a body (say three as in the picture)
there are two methods that can be used to calculate the net work W net .
Method 1: First calculate the work done by each force: W A by force FA ,
W B by force FB , and W C by force FC . Then determine W net W A W B W C .
Method 2: Calculate first Fnet  FA  FB  FC ; then determine W net  F  d .
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Properties of Work
• Only the force component along the object’s
displacement will contribute to work.
• The force component perpendicular to the
displacement does zero work.
• A force does positive work when it has a vector
component in the same direction displacement,
• A force does negative work when it has a vector
component in the opposite direction.
• Work is a scalar quantity.
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Work-Kinetic Energy Theorem
We have seen earlier that K f  K i  Wnet .
m
m
We define the change in kinetic energy as
K  K f  K i . The equation above becomes
the work-kinetic energy theorem:
K  K f  Ki  Wnet
Change in the kinetic  net work done on 
energy of a particle    the particle


 

The work-kinetic energy theorem holds for both positive and negative values of Wnet .
If
Wnet  0  K f  Ki  0  K f  Ki
If
Wnet  0  K f  Ki  0  K f  Ki
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(a) decrease; (b) same; (c) negative, zero
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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
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frictionless contact.
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(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
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(b) During the displacement, what is the work W g 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
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(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  12 mv f  12 mv i
2
vf 
2W

m
2
2 (153.4 J)
225 kg
 1.17 m / s
We have assumed no frictional forces exist.
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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) ˆi  ( 6.0 N) ˆj. The situation and coordinate axes
are shown in Fig. 7-5.
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(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) ˆi  (  6.0 N) ˆj   ( 3.0 m) i 
 (2.0 N) ( 3.0 m) ˆi  ˆi  (  6.0 N) (  3.0 m) ˆj  ˆi
 (  6.0 J) (1)  0   6.0 J
The wind force does negative work, i.e.
kinetic energy is taken out of the crate.
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(b) If the crate has a kineticenergy of 10 J at the
beginning of displacement d , what
is its kinetic

energy at the end of d ?
SOLUTION:
K f  K i  W  10 J  (  6.0 J)  4.0 J
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Work Done by the Gravitational Force :
Consider a tomato of mass m that is thrown upward at point A
with initial speed v0 . As the tomato rises, it slows down by the
gravitational force Fg so that at point B it has a smaller speed v.
The work Wg  A  B  done by the gravitational force on the
tomato as it travels from point A to point B is
Wg  A  B   mgd cos180   mgd .
The work Wg  B  A  done by the gravitational force on the
tomato as it travels from point B to point A is
Wg  B  A   mgd cos 0  mgd .
B
A
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Work Done by a Force in Lifting an Object :
The work-kinetic energy theorem states that K  K f  K i W net .
We also have that K i  K f  K  0 W net  0. There are two forces
acting on the object: The gravitat ional force Fg and the applied force F
that lifts the object. W net W a  A  B  W g  A  B   0 
W a  A  B   W g  A  B  .
W g  A  B   mgd cos180  -mgd W a  A  B   mgd .
Work Done by a Force in Lowering an Object :
In this case the object moves from B to A .
W g  B  A   mgd cos 0  mgd
W a  B  A   W g  B  A  =  mgd
B
.
A m
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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
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to the work done by the applied force.
(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.
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(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
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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 W g is
done on the crate by
 the
gravitational force Fg during
the lift?
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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
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(b) How much work WT is done on the crate by
the force
from the cable during the lift?
T
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
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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 W g done on the
cab by the gravitational
force Fg ?
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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 104 J  59 kJ
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(b) During the 12 m fall, what is the work WT done

on the cab by the upward pullT 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
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(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  12 mv i  W
2
 12 (500 kg) (4.0 m / s) 2  1.18 x 10 4 J
 1.58 x 10 4 J  16 kJ
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The Spring Force :
Fig. a shows a spring in its relaxed state. In fig. b
we pull one end of the spring and stretch it by an
amount d . The spring resists by exerting a force F
on our hand in the opposi te direction.
In fig. c we push one end of the spring and compress
it by an amount d . Again the spring resists by
exerting a force F on our hand in the opposite
direction.
The force F exerted by the spring on whatever agent (in the picture it is our
hand) is trying to change its natural length either by extending or by
compressing it is given by the equation F  kx. Here x is the amount by
which the spring has been extended or compressed. This equation is known
as "Hooke's law" and k is known as the "spring constant" F  kx
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x
O
xi
(a)
x
O
(b)
xf
x
O
(c)
Work Done by a Spring Force :
Consider the relaxed spring of spring constant k shown in (a).
By applying an external force we change the spring's
length from xi (see b) to x f (see c). We will
calculate the work Ws done by the spring on the external agent
(in this case our hand) that changed the spring length. We
assume that the spring is massless and that it obeys Hooke's law.
We will use the expression Ws 
xf
xf
xf
xf
xi
xi
xi
 F ( x) dx 
 kxdx  k  xdx.
 x2 
kx 2 kx
Ws  k    i 
. Quite often we start with a relaxed
2
2
 2  xi
2
f
spring (xi  0) and we either stretch or compress the spring by an
amount x ( x f   x). In this case Ws  
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kx 2
.
2
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(a) positive; (b) negative; (c) zero
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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?
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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
dv
m
0.40 kg
 (0.50 m / s)
k
750 N / m
 1.2 x 10  2 m  1.2 cm
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Work Done by a Variable Force F( x ) Acting Along the x - Axis :
A force F that is not constant but instead varies as a function of x
is shown in fig. a. We wish to calculate the work W that F does
on an object it moves from position xi to position x f .
We partition the interval  xi , x f
 into N "elements" of length
x each, as is shown in fig. b. The work done by F in the jth
interval is W j  Fj ,avg x, where Fj ,avg is the average value of F
N
over the j -th element. W   Fj ,avg x. We then take the limit of
j 1
the sum as x  0, (or equivalently N  ).
N
xf
j 1
xi
W  lim  Fj ,avg x 
 F ( x)dx.
Geometrically, W is the area
between the F ( x) curve and the x-axis, between xi and x f
(shaded blue in fig. d).
xf
W
 F ( x)dx
xi
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Three - Dimensional Analysis :
In the general case the force F acts in three-dimensional space and moves an object
on a three-dimensional path from an initial point A to a final point B.
ˆ
The force has the form F  F  x, y, z  ˆi  F  x, y, z  ˆj  F  x, y, z  k.
x
y
z
Points A and B have coordinates  xi , yi , zi  and  x f , y f , z f  , respectively.
dW  F  dr  Fx dx  Fy dy  Fz dz
B
xf
yf
A
xi
yi
W   dW 
 Fx dx 
xf
W

zf
Fy dy   Fz dz
zi
yf
z
zf
 F dx   F dy   F dz
x
xi
y
yi
B
z
O
x
zi
path
y
A
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Work - Kinetic Energy Theorem with a Variable Force :
Consider a variable force F ( x) that moves an object of mass m from point A ( x  xi )
dv
. We t hen
dt
dv
multiply both sides of the last equation with dx and get Fdx  m dx.
dt
to point B ( x  x f ). We apply Newton's second law: F  ma  m
We integrate both sides over dx from xi to x f
xf
xf
xi
xi
dv
 Fdx   m dt dx .
dv dv dx
dv
dv dx

 dx 
dx  vdv. Thus the integral becomes
dt dx dt
dt
dx dt
xf
W  m  vdv 
xi
2
m 2 x f mv f mvi2
v  

 K f  K i  K .
xi
2
2
2
Note: The work-kinetic energy theorem has exactly the same form as in the case
when F is constant!
W  K f  Ki  K
O
m F(x)
.A
x
20
dx
.
B x-axis
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Power
We define "power" P as the rate at which work is done by a force F.
If F does work W in a time interval t then we define the average power as
Pavg 
W
t
The instantaneous power is defined as
P
dW
dt
Unit of P : The SI unit of power is the watt. It is defined as the power
of an engine that does work W = 1 J in a time t = 1 second.
A commonly used non-SI power unit is the horsepower (hp), defined as
1 hp = 746 W.
The kilowatt-hour The kilowatt-hour (kWh) is a unit of work. It is defined
as the work performed by an engine of power P = 1000 W in a time t = 1 hour ,
W  Pt  1000  3600  3.60 10 6 J . The kWh is used by electrical utility
companies (check your latest electric bill).
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Consider a force F acting on a particle at an angle  to the motion. The rate
at which F does work is given by P 
dW F cos  dx
dx

 F cos 
 Fv cos .
dt
dt
dt
P  Fv cos   F  v
v
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zero
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Sample Problem 7-11


• 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.
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(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. The49net
power does not change.

(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.
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Homework # 3 (due March 28)
‫ ربيع ثاني‬23 ‫تسليم الواجب والكويز االثنين‬
• Question 7
 Question 8
 10
 20
 27
• 37
• 39
• 43
• 67
• 71
51
Q. #
Ans.
7
--------------
8
(a) 3 m; (b) 3 m; (c) 0 and 6 m; (d) negative x
10
5.0 103 J
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(a) +1.31 J. (b) 0.935 m/s.
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a) 7.2 J; (b) 7.2 J; (c) 0; (d) - 25 J
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5.3×102 J
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(a) 42 J; (b) 30 J; (c) 12 J; (d) 6.5m/s, positive x direction; (e) 5.5m/s,
positive x direction;
43
4.9×102 W
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(a) 1.20 J; (b) 1.10 m/s
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(a) 23 mm; (b) 45 N
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