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