Download 13. Work Power & Energy - 2

Physics
Session
Work, Power and Energy - 2
Session Objectives
Session Objective
1.
Gravitational Potential Energy
2.
Potential Energy of an extended Spring
3.
Conservation of Energy
4.
Force and Potential Energy Relationship
5.
Conditions for equilibrium
Gravitational Potential Energy
U (H) = mgH (at height H w.r.t : earth)
=-W
U is a properly of earth + block
At height h
U(h) = mgh
1
KE(h) = mv2 = mg (H-h)
2
E(h) = U(h)+K(h)
H
U=mgh
KE=1/2 mv2
= mg(H-h)
h
Potential Energy of an Extended
Spring
1 2
U(x)  W  kx0
2
At maximum extension (spring at rest)
1 2
At x : U(x)
 kx
Potential
energy
: spring
2
1 2
 2 kx 


1

2
Kinetic energy : mass
1 2 mv 
E  U(x)  k(x)  kx02U(x0 ) 
2
M
M
1
1
2
K (x)  mv  k x20  x2
2
2


X= 0
x
xo
M
Conservation of energy of isolated
System
System : spring + mass
Spring energy E(x) = U(x) + K(x)
1 2 1
 kx  k(x20  x2 )
2
2
1 2 E,U,K
E  kx0
2
K
U
- x0
x=0
x0
Conservation of energy of isolated
System
General Rule
For an isolated system, in the
absence of non – conservative
forces , the total mechanical
energy remains a constant
Ufinal  k final  Uinitial  kinitial
 U   k  0
 E  0
Conservation of Energy
Three kinds of energy act on a system :
1. Mutual forces : internal and
conservative  work = - PE
2. Mutual forces : internal and
non conservative  work
changes mechanical energy
3. External forces : Work changes
total energy
Force and Potential Energy
Relationship
Fx = -
U
U
U
,Fy = ,Fz = x
y
z
f
U    Fxdx  Fydy  Fzdz 
i
Condition of Equilibrium of a
system under linear motion
F0
U

 0 Equilibrium along x
x
U
 0 Equilibrium along y
y
U
 0 Equilibrium along z
z
U U  U



0
x y z
Linear equilibrium
(1)
(2)
(3)
U
q
-
q0
+
Spring - mass system
x
Some useful relations
Work done by pressure on a fluid :
W  V
Kinetic energy 
Efficiency 
linear momentum
2.mass
Output power
Input power
2
Power
Rate of doing work is power
P
W
t
Power = Rate of Energy transfer
W dW

t 0 t
dt
dW
ds
P
 F.
 F.v
dt
dt
P  lim
Nature : scalar
Units : Watt(W)
1 W = 1 J/s
1 hp = 746 W
Class Test
Class Exercise - 1
A body of mass m was slowly pulled
up the hill by a force F which at each
point was directed along the tangent
of the trajectory. All surfaces are
smooth. Find the work performed by
this force.
(a) mgl
(c) mgh
(b) –mgl
(d) Zero
l
Solution
Surfaces are smooth, so no friction
exists. The force F is always
tangential. Work is done against
gravity is conservative. So work
done is path independent and
equal to increase in potential
energy : W = mgh
Hence answer is (c)
Class Exercise - 2
A particle is moving in a region where
potential U is given by U  K x2  y2  z2


The force acting on the particle is

ˆ
ˆ  zk
(a) K xˆi  yj


ˆ
ˆ  zk
(c)  2K xˆi  yj

ˆ
ˆ  zk
(b)  K xˆi  yj

(d) Zero

Solution
U = K(x2 + y2 + z2)
 Fx  –
U
 –2Kx
x
U
Fy  –
 –2Ky
y
Fz  –
U
 –2Kz
z
Class Exercise - 3
The potential energy of a particle in
a conservative field has the form
a b
U
– , where a and b are
2
r
r
positive constants, r is the distance
from the centre of the field. Then
(a) at
(b) at
(c) at
(d) at
1
r  exists a stable equilibrium
b
2a
exists an unstable equilibrium
r
b
2a
exists a stable equilibrium
r
b
a
exists an unstable equilibrium
r
b
Solution
U
a
r2
–
b
r
For equilibrium:
dU
2a b
dU
–

0
0 
3
2
dr
dr
r
r
2a
r 
b
To check stability,
Putting the value of r,
d2U
6a 2b

–
4
dr
r
r3
3
d2U 2b3  3a  b
 2b  b 

– b 
3  2a
3  2 
dr

8a
8a
which is a position of stable equilibrium.
2a
So at r 
there is a stable equilibrium.
b
Hence answer is (c).
Class Exercise - 6
If the KE of a particle is doubled,
then its momentum will
(a) remain unchanged
(b) be doubled
(c) be quadrupled
(d) increase by 2 times
Solution
p12
K1 
2m
p22
K2 
2m
K 2 p22

2
2
K1 p
1
p2

 2  p2  2p1
p1
Hence answer is (d)
Class Exercise - 7
An engine pumps up 100 kg of water
through a height of 10 m in 5 s.
Given that the efficiency of engine is
60%, what is the power of the
engine?
(a) 33 kW
(b) 3.3 kW
(c) 0.33 kW
(d) 0.033 kW
Solution
Work done = 100 × 10 = 1000 J
Utilised power 
w 1000

 200 W
t
5
Utilised power
Efficiency  60% 
Power of engine
 0.6
So power 
200 W
 0.33 kW
0.6
Hence answer is (c)
Class Exercise - 8
A system has two light springs with
stiffness k1 and k2 joined in series and
hanging from a rigid support. What is the
minimum work that needs to be done to
stretch the system by a length l?
Solution
At all time the tension T in the two
springs is the same. The extension
l = l1 + l2 … (i)
If we assume both springs are
replaced by a single spring of
stiffness k:

kk
T
T
T

 k  1 2
k1 k 2 k
k1  k 2
K
1
 l
And the minimum work to be done on
the system (equivalent spring) is
1
2 1  k1k 2 
2
W  k(  )  
 ( )
2
2  k1  k 2 
K
2
 l
 l2
1
Class Exercise - 9
Find out whether the field of force
F  ayiˆ is conservative, a being
a positive constant
y
3
(0, 1)
4
(1, 1)
2
x
0 (0, 0)
1
(1, 0)
Solution
Consider a particle at origin.
It is taken along the perimeter of
unit square as shown in the figure.
Total work done on the particle:
W = W1 + W2 + W3 + W4
1
1
0
0
0
0
1
1
 ayiˆ .dxiˆ   ayiˆ .dyjˆ    ayiˆ .  dxiˆ    ayiˆ .  dyjˆ 
=0+a+0+0=a
 F acting is not conservative.
Class Exercise - 10
A particle moves along the X-axis
through a region in which the
potential energy U(x) varies as
U(x) = 4x – x2.
(i) Find the position of the particle when
the force on it is zero
(ii) The particle has a constant mechanical energy
of 4.0 J. Find the kinetic energy as a function of x.
Solution
a.
U  4x  x 2
dU
F
 4  2x
dx
F = 0  4 – 2x = 0 or x = 2m
b.
Total energy = K(x) + U(x) = 4 J
 K(x) + 4x – x2 = 4
 K(x) = x2 – 4x + 4
Thank you