1. Physics

2. Session
Work, Power and Energy - 2

3. Session Objectives

4. Session Objective Gravitational Potential Energy
Potential Energy of an extended Spring
Conservation of Energy
Force and Potential Energy Relationship
Conditions for equilibrium

5. Gravitational Potential Energy
U (H) = mgH (at height H w.r.t : earth)
=-W
U is a properly of earth + block
At height h H h
U=mgh
KE=1/2 mv2
= mg(H-h)
U(h) = mgh
KE(h) = mv2 = mg (H-h)
E(h) = U(h)+K(h)

6. Potential Energy of an Extended Spring
At maximum extension (spring at rest)
Potential energy : spring
Kinetic energy : mass M X= M xo M x

7. Conservation of energy of isolated System
System : spring + mass
Spring energy E(x) = U(x) + K(x)
E,U,K K U x0
- x0
x=0

8. 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

9. Conservation of Energy
Three kinds of energy act on a system :
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

10. Force and Potential Energy Relationship

11. Condition of Equilibrium of a system under linear motion
Equilibrium along x (1)
Equilibrium along y (2)
Equilibrium along z (3) x q
q = 0 + - U
Spring - mass system
Linear equilibrium

12. Some useful relations
Work done by pressure on a fluid :

13. Power = Rate of Energy transfer
Rate of doing work is power
Power = Rate of Energy transfer
Nature : scalar
Units : Watt(W)
1 W = 1 J/s
1 hp = 746 W

14. Class Test

15. 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.
mgl (b) –mgl
(c) mgh (d) Zero l

16. 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)

17. Class Exercise - 2 A particle is moving in a region where
potential U is given by
The force acting on the particle is

18. Solution
U = K(x2 + y2 + z2)

19. 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
(a) at exists a stable equilibrium
(b) at exists an unstable equilibrium
(c) at exists a stable equilibrium
(d) at exists an unstable equilibrium

20. Solution For equilibrium: To check stability, Putting the value of r,
which is a position of stable equilibrium.
So at there is a stable equilibrium.
Hence answer is (c).

21. Class Exercise - 6 If the KE of a particle is doubled,
then its momentum will
remain unchanged
(b) be doubled
(c) be quadrupled
(d) increase by times

22. Solution
Hence answer is (d)

23. 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?
33 kW (b) 3.3 kW
(c) 0.33 kW (d) kW

24. Solution
Work done = 100 × 10 = 1000 J
Hence answer is (c)

25. 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 Dl?

26. Solution
At all time the tension T in the two springs is the same. The extension Dl = Dl1 + Dl2 … (i)
If we assume both springs are replaced by a single spring of stiffness k:
And the minimum work to be done on the system (equivalent spring) is

27. Class Exercise - 9
Find out whether the field of force is conservative, a being a positive constant

28. 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
= 0 + a = a
acting is not conservative.

29. 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.
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.

30. Solution
F = 0 Þ 4 – 2x = 0 or x = 2m
b. Total energy = K(x) + U(x) = 4 J
Þ K(x) + 4x – x2 = 4
Þ K(x) = x2 – 4x + 4

31. Thank you