### What is Physics?

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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)
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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
KпЂЅ
.
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 вЂњKatyвЂќ 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.
вЂў 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 пЂЅ v 0 пЂ« 2a (x пЂ­ x 0 )
2
v 2 пЂЅ 0 пЂ« 2 (0.26 m / s 2 ) (3.2 x103 m)
v пЂЅ 40.8 m / s
mпЂЅ
1.2 x 10 6 N
пЂЅ 1.22 x 10 5 kg
2
9.8 m /s
K пЂЅ 2 ( 12 mv 2 ) пЂЅ (1.22 x 105 kg) (40.8 m / s) 2
пЂЅ 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.
вЂў Energy transferred to the object is positive work,
вЂў 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 пЃ¦ to the wire is acting on the bead.
We apply Newton's second law: Fx пЂЅ 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 пЂ­ v02 пЂЅ 2ax d . We multiply both sides by m / 2 п‚®
m 2 m 2 m
m F
m
v пЂ­ v0 пЂЅ 2ax d пЂЅ 2 x d пЂЅ Fx d пЂЅ F cos пЃ¦ d .
K i пЂЅ v02
2
2
2
2 m
2
m 2
K f пЂЅ v п‚® The change in kinetic energy K f пЂ­ K i пЂЅ Fd cos пЃЄ.
2
Thus the work W done by the force on the bead is given by W пЂЅ Fx d пЂЅ Fd cos пЃЄ.
W пЂЅ Fd cos пЃЄ
FA
W пЂЅ F пѓ—d
W пЂЅ Fd cos пЃЄ
FC
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m
m
W пЂЅ F пѓ—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 пЂѕ 0 if 0 пЂј пЃ¦ пЂј 90п‚°, W пЂј 0 if 90п‚° пЂј пЃ¦ пЂј 180п‚°.
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 пЂЅW A пЂ«W B пЂ«W C .
Method 2: Calculate first Fnet пЂЅ FA пЂ« FB пЂ« FC ; then determine W net пЂЅ F пѓ— d .
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Properties of Work
вЂў Only the force component along the objectвЂ™s
displacement will contribute to work.
вЂў The force component perpendicular to the
displacement does zero work.
вЂў A force does positive work when it has a vector
component in the same direction displacement,
вЂў A force does negative work when it has a vector
component in the opposite direction.
вЂў Work is a scalar quantity.
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Work-Kinetic Energy Theorem
We have seen earlier that K f пЂ­ K i пЂЅ Wnet .
m
m
We define the change in kinetic energy as
пЃ„K пЂЅ K f пЂ­ K i . The equation above becomes
the work-kinetic energy theorem:
пЃ„K пЂЅ K f пЂ­ Ki пЂЅ Wnet
пѓЄenergy of a particle пѓє пЂЅ пѓЄ the particle
пѓє
пѓ«
пѓ» пѓ«
пѓ»
The work-kinetic energy theorem holds for both positive and negative values of Wnet .
If
Wnet пЂѕ 0 п‚® K f пЂ­ Ki пЂѕ 0 п‚® K f пЂѕ Ki
If
Wnet пЂј 0 п‚® K f пЂ­ Ki пЂј 0 п‚® K f пЂј 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 anпЃІ initially
stationary 225 kg floor safe a displacement d of
magnitude 8.50 m, straight toward their truck.
пЃІ
вЂў The push F1 of Spy 001 is 12.0 N, directed atпЃІan angle
of 30В° downward from the horizontal; the pull F2 of Spy
002 is 10.0 N, directed at 40В° 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 пЃІby
пЃІ
пЃІ
forces F1 and F2 during the displacement d ?
SOLUTION:
пЃІ
Work done by F1 :
W1 пЂЅ F1 d cos пЃ¦1 пЂЅ (12.0 N)(8.50 m)(cos 30пЃЇ )
пЂЅ 88.33 J
пЃІ
Work done by F2 :
W2 пЂЅ F2 d cos пЃ¦2 пЂЅ (10.0 N)(8.50 m)(cos 40пЃЇ )
пЂЅ 65.11 J
Total work done :
W пЂЅ W1 пЂ« W2 пЂЅ 88.33 J пЂ« 65.11 J
пЂЅ 153.4 J п‚» 153 J
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(b) During the displacement, whatпЃІ is the work W g done on
the safe by the gravitational force Fg and what is the work
пЃІ
WN done on the safe by the normal force N from the floor?
SOLUTION:
Wg пЂЅ mgd cos 90пЃЇ пЂЅ 0
WN пЂЅ Nd cos 90пЃЇ пЂЅ 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 пЂЅ K f пЂ­ K i пЂЅ 12 mv f пЂ­ 12 mv i
2
vf пЂЅ
2W
пЂЅ
m
2
2 (153.4 J)
225 kg
пЂЅ 1.17 m / s
We have assumed no frictional forces exist.
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Sample Problem 7-3
вЂў During a storm, a crate of crepe is sliding
пЃІ across a slick,
oily parking lot through a displacement d while a
steady wind pushes against the crate with a force
пЃІ
F пЂЅ (20 N) Л†i пЂ« (пЂ­ 6.0 N) Л†j. 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 :
пЃ›
пЃќ
пЃІ пЃІ
W пЂЅ F пѓ— d пЂЅ (2.0 N) Л†i пЂ« ( пЂ­ 6.0 N) Л†j пѓ— пЃ› (пЂ­ 3.0 m) i пЃќ
пЂЅ (2.0 N) (пЂ­ 3.0 m) Л†i пѓ— Л†i пЂ« ( пЂ­ 6.0 N) ( пЂ­ 3.0 m) Л†j пѓ— Л†i
пЂЅ ( пЂ­ 6.0 J) (1) пЂ« 0 пЂЅ пЂ­ 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 kineticпЃІenergy of 10 J at the
beginning of displacement d , what
is its kinetic
пЃІ
energy at the end of d ?
SOLUTION:
K f пЂЅ K i пЂ« W пЂЅ 10 J пЂ« ( пЂ­ 6.0 J) пЂЅ 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 пЂЁ A п‚® B пЂ© done by the gravitational force on the
tomato as it travels from point A to point B is
Wg пЂЁ A п‚® B пЂ© пЂЅ mgd cos180п‚° пЂЅ пЂ­ mgd .
The work Wg пЂЁ B п‚® A пЂ© done by the gravitational force on the
tomato as it travels from point B to point A is
Wg пЂЁ B п‚® A пЂ© пЂЅ mgd cos 0п‚° пЂЅ mgd .
B
A
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Work Done by a Force in Lifting an Object :
The work-kinetic energy theorem states that пЃ„K пЂЅ K f пЂ­ K i пЂЅW net .
We also have that K i пЂЅ K f п‚® пЃ„K пЂЅ 0 п‚®W net пЂЅ 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 пЂЅW a пЂЁ A п‚® B пЂ© пЂ«W g пЂЁ A п‚® B пЂ© пЂЅ 0 п‚®
W a пЂЁ A п‚® B пЂ© пЂЅ пЂ­W g пЂЁ A п‚® B пЂ© .
W g пЂЁ A п‚® B пЂ© пЂЅ mgd cos180п‚° пЂЅ -mgd п‚®W a пЂЁ A п‚® B пЂ© пЂЅ mgd .
Work Done by a Force in Lowering an Object :
In this case the object moves from B to A .
W g пЂЁ B п‚® A пЂ© пЂЅ mgd cos 0п‚° пЂЅ mgd
W a пЂЁ B п‚® A пЂ© пЂЅ пЂ­W g пЂЁ B п‚® A пЂ© = пЂ­ mgd
B
.
A m
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Sample Problem 7-4
Chemerkin and Paul Anderson.
вЂў (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
пЃІ
gravitational force Fg acting on them?
SOLUTION:
The gravitational force points down and the
displacement d points up :
Wg (up path) пЂЅ mgd cos пЃ¦ пЂЅ (2548 N) (2.0 m) (cos 180пЃЇ )
пЂЅ пЂ­ 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) пЂЅ пЂ­ Wg (up) пЂЅ пЂ« 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) пЂЅ пЂ­ Wg (up path)
пЂЅ (27900 N) (0.010 m) пЂЅ 280 J
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Sample Problem 7-5
вЂў 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
пЃІ 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) пЂЅ пЂ­ mg sin пЃ± d
Since d sin пЃ± = h, we have :
Wg (up path) пЂЅ пЂ­ mgh
пЂЅ пЂ­ (15.0 kg) (9.8 m / s 2 ) (2.50 m)
пЂЅ пЂ­ 368 J
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(b) HowпЃІ 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 пЂЅ пЂ­ Wg (up path) пЂЅ 368 J
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Sample Problem 7-6
вЂў 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
пЃІ пЃІ acceleration
a = g /5.
вЂў (a) During the fall through a
distance d = 12 m, what is
the work W g done on the
cab byпЃІ 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) пЂЅ mgd cos 0пЃЇ пЂЅ (500 kg) (9.8 m / s 2 ) (12 m) (1)
пЂЅ 5.88 x 104 J п‚» 59 kJ
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(b) During the 12 m fall, what is the work WT done
пЃІ
on the cab by the upward pullT of the elevator
cable?
SOLUTION:
mg пЂ­ T пЂЅ ma
(positive direction is down)
T пЂЅ m (g пЂ­ a ) пЂЅ 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) пЂЅ пЂ­ 54 mg d
пЂЅ пЂ­ 4.70 x 10 4 J п‚» пЂ­ 47 kJ
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(c) What is the net work W done on the cab
during the fall?
SOLUTION:
W пЂЅ Wg пЂ« WT пЂЅ 5.88 x 10 4 J пЂ­ 4.70 x 10 4 J
пЂЅ 1.18 x 10 4 J п‚» 12 kJ
(d) What is the cab's kinetic energy at the end of the
12 m fall?
SOLUTION:
K f пЂЅ K i пЂ« W пЂЅ 12 mv i пЂ« W
2
пЂЅ 12 (500 kg) (4.0 m / s) 2 пЂ« 1.18 x 10 4 J
пЂЅ 1.58 x 10 4 J п‚» 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 пЂЅ пЂ­kx. 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 пЂЅ пЂ­kx
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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 пЂЅ
xf
xf
xf
xf
xi
xi
xi
пѓІ F ( x) dx пЂЅ
пѓІ пЂ­kxdx пЂЅ пЂ­k пѓІ xdx.
kx 2 kx
Ws пЂЅ пЂ­k пѓЄ пѓє пЂЅ i пЂ­
2
2
пѓ« 2 пѓ» xi
2
f
spring (xi пЂЅ 0) and we either stretch or compress the spring by an
amount x ( x f пЂЅ п‚± x). In this case Ws пЂЅ пЂ­
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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
вЂў 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 пЂЅ пЂ­ 12 kd 2
Kinetic energy change of the canister is :
k f пЂ­ k i пЂЅ пЂ­ 12 mv 2
Therefore,
пЂ­ 12 kd 2 пЂЅ пЂ­ 12 mv 2
dпЂЅv
m
0.40 kg
пЂЅ (0.50 m / s)
k
750 N / m
пЂЅ 1.2 x 10 пЂ­ 2 m пЂЅ 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 пЂЁ xi , x f
пЂ© into N "elements" of length
пЃ„x each, as is shown in fig. b. The work done by F in the jth
interval is пЃ„W j пЂЅ Fj ,avg пЃ„x, where Fj ,avg is the average value of F
N
over the j -th element. W пЂЅ пѓҐ Fj ,avg пЃ„x. We then take the limit of
j пЂЅ1
the sum as пЃ„x п‚® 0, (or equivalently N п‚® п‚Ґ).
N
xf
j пЂЅ1
xi
W пЂЅ lim пѓҐ Fj ,avg пЃ„x пЂЅ
пѓІ F ( x)dx.
Geometrically, W is the area
between the F ( x) curve and the x-axis, between xi and x f
xf
WпЂЅ
пѓІ 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.
Л†
The force has the form F пЂЅ F пЂЁ x, y, z пЂ© Л†i пЂ« F пЂЁ x, y, z пЂ© Л†j пЂ« F пЂЁ x, y, z пЂ© k.
x
y
z
Points A and B have coordinates пЂЁ xi , yi , zi пЂ© and пЂЁ x f , y f , z f пЂ© , respectively.
dW пЂЅ F пѓ— dr пЂЅ Fx dx пЂ« Fy dy пЂ« Fz dz
B
xf
yf
A
xi
yi
W пЂЅ пѓІ dW пЂЅ
пѓІ Fx dx пЂ«
xf
WпЂЅ
пѓІ
zf
Fy dy пЂ« пѓІ Fz dz
zi
yf
z
zf
пѓІ F dx пЂ« пѓІ F dy пЂ« пѓІ 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 пЂЅ xi )
dv
. We t hen
dt
dv
multiply both sides of the last equation with dx and get Fdx пЂЅ m dx.
dt
to point B ( x пЂЅ x f ). We apply Newton's second law: F пЂЅ ma пЂЅ m
We integrate both sides over dx from xi to x f
xf
xf
xi
xi
dv
пѓІ Fdx пЂЅ пѓІ m dt dx .
dv dv dx
dv
dv dx
пЂЅ
п‚® dx пЂЅ
dx пЂЅ vdv. Thus the integral becomes
dt dx dt
dt
dx dt
xf
W пЂЅ m пѓІ vdv пЂЅ
xi
2
m 2 x f mv f mvi2
пЂ­
пЂЅ K f пЂ­ K i пЂЅ пЃ„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 пЂЅ K f пЂ­ Ki пЂЅ пЃ„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 пЃ„t then we define the average power as
Pavg пЂЅ
W
пЃ„t
The instantaneous power is defined as
PпЂЅ
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 пЂЅ Pt пЂЅ 1000 п‚ґ 3600 пЂЅ 3.60 п‚ґ10 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 пЃ¦ to the motion. The rate
at which F does work is given by P пЂЅ
dW F cos пЃ¦ dx
dx
пЂЅ
пЂЅ F cos пЃ¦
пЂЅ Fv cos пЃ¦.
dt
dt
dt
P пЂЅ Fv cos пЃ¦ пЂЅ F пѓ— v
v
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zero
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Sample Problem 7-11
пЃІ
пЃІ
вЂў Figure 7-14 shows constant forces F1 and F2
acting on a box as the box slides пЃІrightward
across a frictionless floor. Force
F1 is horizontal,
пЃІ
with magnitude 2.0 N; force F2 is angled upward
by 60В° 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 пЂЅ F1 v cos 180 пЃЇ пЂЅ (2.0 N) (3.0 m / s) cos 180 пЃЇ
пЂЅ пЂ­ 6.0 W
P2 пЂЅ F2 v cos 60 пЃЇ пЂЅ (4.0 N) (3.0 m / s) cos 60 пЃЇ
пЂЅ 6.0 W
Pnet пЂЅ P1 пЂ« P2
пЂЅ0
The kinetic energy of the box is not changing.
The speed of the box remains at 3 m/s. The49net
power does not change.
пЃІ
(b) If the magnitude of F2 is, instead, 6.0 N, what
now is the net power, and is it changing?
SOLUTION:
P2 пЂЅ F2 v cos 60 пЃЇ пЂЅ (6.0 N) (3.0 m / s) cos 60 пЃЇ
пЂЅ 9.0 W
Pnet пЂЅ P1 пЂ« P2 пЂЅ пЂ­ 6.0 W пЂ« 9.0 W
пЂЅ 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.
25
50
12/04/32
Homework # 3 (due March 28)
вЂ« Ш±ШЁЩЉШ№ Ш«Ш§Щ†ЩЉвЂ¬23 вЂ«ШЄШіЩ„ЩЉЩ… Ш§Щ„Щ€Ш§Ш¬ШЁ Щ€Ш§Щ„ЩѓЩ€ЩЉШІ Ш§Ш§Щ„Ш«Щ†ЩЉЩ†вЂ¬
вЂў Question 7
пЃ± Question 8
пЃ± 10
пЃ± 20
пЃ± 27
вЂў 37
вЂў 39
вЂў 43
вЂў 67
вЂў 71
51
Q. #
Ans.
7
--------------
8
(a) 3 m; (b) 3 m; (c) 0 and 6 m; (d) negative x
10
5.0 п‚ґ103 J
20
(a) +1.31 J. (b) 0.935 m/s.
27
a) 7.2 J; (b) 7.2 J; (c) 0; (d) - 25 J
37
5.3Г—102 J
39
(a) 42 J; (b) 30 J; (c) 12 J; (d) 6.5m/s, positive x direction; (e) 5.5m/s,
positive x direction;
43
4.9Г—102 W
67
(a) 1.20 J; (b) 1.10 m/s
71
(a) 23 mm; (b) 45 N
52
26
```