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
KE(h) = mv2 = mg (H-h)
E(h) = U(h)+K(h)
12 H
h
U=mghKE=1/2 mv2
= mg(H-h)
U (H) = mgH (at height H w.r.t : earth)=-W U is a properly of earth + block
At height h
Potential Energy of an Extended Spring
20
1U(x) W kx
2
At maximum extension (spring at rest)
Potential energy : spring
Kinetic energy : mass
21kx
2
21mv
2 M
X= 0
M
xM
xo
21At x : U(x) kx
2
2 2 20
1 1K (x) mv k x x
2 2
20 0
1E U(x) k(x) kx U(x )
2
Conservation of energy of isolated System
Spring energy E(x) = U(x) + K(x)
System : spring + mass
2 2 20
1 1kx k(x x )
2 2
E,U,K
K
U
x0- x0 x=0
20
1E kx
2
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
final final initial initialU k U k
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
x y z
U U UF = - ,F = - ,F = -
x y z
f
x y zi
U F dx F dy F dz
Condition of Equilibrium of a system under linear motion
F 0
U U U0
x y z
Linear equilibrium
U0
xU
0y
U0
z
Equilibrium along x (1)
Equilibrium along y (2)
Equilibrium along z (3)
x
+-
U
Spring - mass system
Some useful relations
Work done by pressure on a fluid :
W V
2linear momentum
Kinetic energy2.mass
Output powerEfficiency
Input power
Power
Rate of doing work is power
WP
t
v.Fdtds
.FdtdW
P
dtdW
tW
limP0t
Nature : scalarUnits : Watt(W)1 W = 1 J/s1 hp = 746 W
Power = Rate of Energy transfer
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 (b) –mgl(c) mgh (d) Zero
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 2 2 2U K x y z
The force acting on the particle is
ˆ ˆˆ ˆ ˆ ˆ(a) K xi yj zk (b) K xi yj zk
ˆˆ ˆ(c) 2K xi yj zk (d) Zero
Solution
U = K(x2 + y2 + z2)
xU
F – –2Kxx
yU
F – –2Kyy
zU
F – –2Kzz
Class Exercise - 3
The potential energy of a particle in a conservative field has the form
, where a and b are
positive constants, r is the distance from the centre of the field. Then
2
a bU –
rr
(a) at exists a stable equilibrium
(b) at exists an unstable equilibrium
(c) at exists a stable equilibrium
(d) at exists an unstable equilibrium
1r
b
2ar
b
2ar
b
ar
b
Solution
2a b
U –rr
Hence answer is (c).
For equilibrium:
dU0
dr 3 2
dU 2a b– 0
dr r r
2ar
b To check stability,
2
4 3d U 6a 2b
–dr r r
2 3 3
3 3d U 2b 3a b 2b b
– bdr 2a 28a 8a
which is a position of stable equilibrium.
Putting the value of r,
So at there is a stable equilibrium.2a
rb
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 times2
Solution2
11
pK
2m
22
2p
K2m
Hence answer is (d)
22 1
1
p2 p 2p
p
22 2
21 1
K p2
K p
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
Hence answer is (c)
w 1000Utilised power 200 W
t 5
Utilised powerEfficiency 60%
Power of engine
0.6
200 WSo power 0.33 kW
0.6
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 ?
Solution
K1
K2
1
2
At all time the tension T in the two springs is the same. The extension = 1 + 2 … (i)
If we assume both springs are replaced by a single spring of stiffness k:
1 2
1 2 1 2
k kT T Tk
k k k k k
And the minimum work to be done on the system (equivalent spring) is
2 21 2
1 2
k k1 1W k( ) ( )
2 2 k k
Class Exercise - 9
Find out whether the field of force is conservative, a being
a positive constant
ˆF ayi
y
(0, 1)3
(1, 1)
2
x0 (0, 0) 1 (1, 0)
4
SolutionConsider 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
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆayi . dxi ayi . dyj ayi . dxi ayi . dyj
= 0 + a + 0 + 0 = a
acting is not conservative.F
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
2a. U 4x x
dUF 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
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