Transcript Chapter 7

CHAPTER-7
Kinetic Energy and Work
Ch 7-2,3 Kinetic Energy
 Energy: a scalar quantity associated with state or
condition of one or more objects
 Kinetic Energy (K): energy associated with state of
motion of an object. The faster an object moves,
greater is its K
K=mv2/2
 Unit of energy (K or any type of energy) is joule (J)
1J=1 kg. m2/s2
Ch 7-4 Work
Work W is energy transferred to or from
an object by means of a force acting on
the object.
 If the object is accelerated by applying a force,
its kinetic energy K increases. Energy transferred
to the object is positive work +W.
 If you decelerate the object by applying a force,
you decrease its kinetic energy K. Energy
transferred from the object is negative work -W.
Ch 7-5 Work and Kinetic Energy
 Work done in accelerating a
bead through a distance d along
x-axis under a constant force F
acting at an angle  with
respect to x-axis
W= K= m(v2-v02)/2 but
axd=(v2-v02)/2
W= K=m axd=Fxd=Fcos  d=F.d
W= Fxd=Fcos  d=F.d
 Positive Work : Displacement
along force direction
 Negative Work: Displacement
opposite to force direction
Ch 7-5 Work and Kinetic Energy
Work-Kinetic Energy Theorem
W= K = Kf-Ki= m(v2-v02)/2
Kf = W+ Ki
v2 =2(W+Ki)/m
Ch 5-Check-Point-1
A particle moves along x-axis.
Does its kinetic energy
increase or decrease , or
remain the same if the
particle velocity changes
a) from -3 m/s to -2 m/s
b) -2 m/s to + 2 m/s
c) in each situation the work
done on the particle is
positive, negative or zero?
K= m(vf2-vi2)/2
a) K= m(vf2-vi2)/2
= m/2(4-9)=5m/2
K is negative
b) K= m*(vf2-vi2)/2
= m(4-4)/2=0
K is constant
c) W is negative
W is Zero
Ch 5-Check-Point-2
The figure show four situation in which a
box acts on a box while the box slides
rightward a distance d across a
frictionless floor. The magnitude of
the forces is identical: their
orientation are as shown. Rank the
situation according to the work done
on the box during the displacement,
most positive first
W = Fdcos
d) W= Fd
c) W=Fdcos
b) W= 0
a) W = - Fdcos
Ch 7-6 : Work Done by Gravitational Force
(Constant Force)
 A tomato, thrown upward, is slowed down
from initial velocity v0 to v under the effect
of gravitational force Fg:
 Work Wg done by the gravitational force Fg
in rising objects:

Wg=Fgd cos =mgd cos = mgd (-1)= mgd
 Work Wg done by the gravitational force Fg
in falling objects :

Wg=Fgd cos =mgd cos = mgd (+1)= +
mgd
Work done in lifting an object
 Work Wa done by an
applied force F in lifting /
lowering an object through
a distance d:
K = Kf-Ki = Wa+ Wg
 In lifting object is at rest
in initial and final position
K = Kf-Ki= 0 then

Wa= - Wg = -mg d cos
Lifting  = 180
Lowering  = 0
Ch 7-7: Work done by a spring Force
(Variable Force)
 Relaxed state of a spring (
Fig. a): Spring neither
compressed nor extended
 Spring Fore Fx (Restoring
Forces) acts to restore the
spring to its relaxed state
 Fx=-kx (Hooke’s Law)
 where k is spring constant (
force constant) and x is
compression or extension in the
relaxed length of the spring
 -ve Fx for +ve value of x
 +ve Fx for -ve value of x
Ch 7-7: Work done by a spring Force
(Variable Force)
Work done by a spring force Ws
xf
Ws=xi Fx dx =
xf
xi
(-kx) dx= k(xi2-xf2)/2
Ws= -kx2/2 (if xi=0 and xf=x)
Work done by an applied force Wa in
stretching/compressing a spring
K = Wa+Ws
If the block is initially and finally at rest
then K =0 and
Wa= - Ws
Ch 5-Check-Point-4
For three situations the
initial and final position
respectively , along x-axis
for the block is a) -3 cm,
2cm b) 2 cm, 3 cm c) -2
cm, 2 cm. In each situation
the work done by the spring
force on the block is
positive, negative , or zero.
Work done by spring force is
Ws=k(xi2-xf2)/2
a) Ws= k(9-4)/2=+5k/2 J
b) Ws= k(4-9)/2=-5k/2 J
c) Ws= k(4-4)/2=0 J
Ch 7-8: Work done by a General Variable Force
xf
 W= xi Fx dx
 Work done by a variable
force is equal to area
between F(x) curve and the
x-axis , between the limits
xi and xf
 Work-Kinetic Enegy
Theorem for a variable
force
xf

K= Kf-Ki=W= xi Fx dx
Ch 7-9 Power
 Power P
: Time
rate of doing work of a force
Average Power Pavg= W/t
 Instantaneous Power P= dW/dt = d/dt (Fcos  dx)
P = Fcos  v=F.v
 Unit of power : 1 watt= 1 W= 1J/s