Download Kinetic energy

Dr- Sonia Reda
Dr- Sonia Reda
chapter 7
Kinetic energy and Work
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
Power
Outline Chapter 7
Work and Kinetic energy
Work done by a net force results in
kinetic energy
Some examples: gravity, spring, friction
What is Energy?
The term energy is so broad that a
clear definition is difficult to write.
Technically,
Energy is a scalar quantity associated
with the state (or condition) of one or
more objects.
However, this definition is too vague
to be of help to us now.
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Kinetic Energy
Kinetic energy K is energy associated
with the state of motion of an object.
For an object of mass m whose speed v is well
below the speed of light, Kinetic energy K is:
Unit for Kinetic energy is:
Kinetic energy is a scalar quantity.
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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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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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Finding an Expression for Work
we can use Eq. 2-16 to write, for components along the x axis,
v2 =vo2 + 2axd
By multiplying the above Eq with ½ m
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Finding an Expression for Work
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Kinetic Energy
Work-Kinetic Energy Theorem
Change in KE work done by all forces
DK  Dw
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Work-Kinetic Energy Theorem
Vector sum of all forces
acting on the body
SF
w  xx F .dx
f
i
 xx ma.dx
f
i

x
x i f
dv
m .dx 
dt
x
x i f
dx
m .dv
dt
xi
xf
x
 m  v .dv  m[1 / 2v ]
vf
vi
2 vf
vi
= 1/2mvf2 – 1/2mvi2
= Kf - Ki
=
DK
Work done by net force
= change in KE
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Checkpoint 1
A particle moves along an x axis. Does the
kinetic energy of the particle increase,
decrease, or remain the same if the
particle’s velocity changes
(a) from −3 m/s to −2 m/s and
(b) from −2 m/s to 2 m/s?
(c) In each situation, is the work done on the
particle positive, negative, or zero?
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Example 7-3
During a storm, a crate of crepe is sliding across
a slick, oily parking lot through a displacement
while a steady wind pushes against
the crate with a force
. The
situation and coordinate axes are shown in Fig.
7-5.
.
(a) How much work does this force do on the
crate during the displacement?
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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) 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.
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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  Ki  W  10 J  (  6.0 J)  4.0 J
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Gravitation and work
Work done by me (take down as +ve)
h
F
mg
Lift mass m with
constant velocity
= F.(-h) = -mg(-h)
= mgh
Work done by gravity
= mg.(-h)
= -mgh
________
Total work by ALL forces (DW) =
0
=DK
Work done by ALL forces = change in KE
DW = DK
What happens if I let go?
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Work Done by a Spring Force
The spring force given by
Hooke’s Law:
spring
x
F
 k Dx
The work done by spring
force:
1
1
W spring  ( kx22  kx12 )
2
2
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Compressing a spring
Compress a spring by an amount x
F -kx
x
Work done by me Fdx = kxdx = 1/2kx2
Work done by spring
-kxdx =-1/2kx2
Total work done (DW)
=
0
=DK
What happens if I let go?
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Moving a block against friction
at constant velocity
f
F
d
Work done by me
= F.d
Work done by friction = -f.d = -F.d
Total work done
What happens if I let go?
=
0
NOTHING!!
Gravity and spring forces are Conservative
Friction is NOT!!
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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
m
0.40 kg
dv
 (0.50 m / s)
k
750 N / m
 1.2 x 10  2 m  1.2 cm
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Power
The rate at which work is done by a force
is called the power.


The average power due to the work done by a force
during that time interval as
We define the instantaneous power P as the
instantaneous rate of doing work, so that
W = F . Δx
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The units of power
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Sample Problem 7-10



Figure 7-14 shows constant forces F1 and F2
acting on a box as the box slides rightward
F
across a frictionless floor. Force
1 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  F1v 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.
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
(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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