This document covers Newton's laws of motion, work, energy, impulse, and momentum. It begins by defining Newton's first law of inertia and describes balanced and unbalanced forces. It then defines force, discusses action-reaction pairs, and covers frictional forces. Several examples of Newton's second law are provided. The document also defines work and energy, distinguishing between kinetic and potential energy. It presents the work-energy theorem. Impulse and momentum are then defined, and the impulse-momentum theorem is stated. Several example problems are worked through to demonstrate applications of these concepts.

1. UNIT-V
Presented by
Dr.G.Elangovan Dean(i/c)
University College Of Engineering
Dindugul

2. Topics To Be Covered
Newton’s Law of Motion
Work & Energy Method
Impulse & Momentum Relation

3. Newton’s Law Of Motion

4. Every body continues to be in its state of rest or of
uniform motion in a straight line unless and until it is
acted upon some external force to change the state.
It is push or pull, which either changes or tends to change
the state of rest or of uniform motion of a body.
In the absence of an external force, no body can change
on it’s own , it’s state of rest or the state of uniform
motion along a state line.

5. Balanced Force
Equal forces in opposite
directions produce no motion

6. Unbalanced Forces
Unequal opposing forces
produce an unbalanced force
causing motion

7. The rate of change of momentum of a moving body is directly
proportional to the impressed force & take place in the
direction of the force applied.
Force = Mass x Acceleration
Force is measured in Newtons

8. To every action, there is always an equal &
opposite reaction.

9. • When two bodies are in contact with one
another, the property of two bodies by
virtue of which a force is exerted b/w them
at their point of contact to prevent one
body from sliding on the other, is called
‘Frictional force’. Its denoted by F.
• The maximum frictional force developed at
the contact area is called as ‘Limiting
friction’. Its denoted by the Fm

10. Contd…..
• The ratio of limiting friction to the normal
reaction is known as ‘Co-efficient of
friction’, denoted by the symbol ‘µ’.
Co-efficient of friction = Limiting Friction
Normal reaction
µ = Fm
NR
Fm = µ x NR

11. Case I: Body moving on rough horizontal
surface
Consider a body moving on a rough horizontal surface towards
right as shown in fig.
IF
Co-efficient of friction = µ
Fm = µ x NR
But NR = Weight ‘W’
Fm = µ x W

12. Case II: Body pulled up on an inclined surface
When a body is pulled up on
an inclined force as shown
In fig. then the frictional
Force Fm acts in the opposite
(downward) direction.
NR –W COS θ = 0
NR = W COS θ
Fm = µ x W COS θ

13. Case III: Body sliding downwards
When a body is sliding downwards as shown in fig. then the
frictional force Fm acts in the upward direction. Resolving the
forces normal to the plane & equated to zero.
NR –W COS θ = 0
NR = W COS θ
Fm = µ x W COS θ

14. D’Alembert’s Principle
D’Alembert’s Principle is an application of Newton’s 2nd
law.For the static equilibrium of forces in plane,we have seen
the equations to be satisfied are ∑H=0; ∑V=0 and ∑M=0.
P= m a
P = External force
m = Mass of the moving body
a = Acceleration of the body

15. CaseI:-
CaseII:-
P=ma
∑H=0
P-ma=0

16. Case III:- Case IV
∑F=ma
∑F-ma=0

17. Case V
When a body is subjected to two forces P1 and P2 as shown in fig.
then the resultant force is P1 - P2 ( assuming P1 > P2 ).
Here,P – ma= 0 equation may be written as ∑F=0
ie, P1 - P2 – ma = 0
a = P1 - P2
m

18. Hence, D’Alembert’s principle States that
” The system of forces acting on a body in
motion is in dynamic equilibrium, with the
inertia force of the body”.

20. WORK
Work is defined as the product of force &
displacement of the body.
Case I
Work done by the force = Force x Distance
Moved
= P X S
S
P

21. Case II
 UNIT OF WORK
In SI system of units, force in Newton & the
distance in meters.
Unit of work = 1 Nm= 1 joule

22. ENERGY
The capacity of doing work is known as ‘Energy’.
Unit of energy is same as that of work.
Mechanical energy is classified into 2types:
i. Potential energy
ii. Kinetic energy

23. It is the capacity to do work by virtue of position of the body. It is denoted
by P.E.
The force of attraction or the weight
of the body=W=mg
Work done by the body = Force X Distance
= mg x h
P.E = mgh

24. It is the capacity to do work by virtue of motion of the body.
It is denoted by k.E.
Kinetic Energy = ½ mv2

25. Work-Energy Equation
Work done = Final Kinetic Energy – Initial Kinetic Energy
P.S =Work done by the moving body = Force X Distance
(W/2g)v2 = Final Kinetic Energy
(W/2g)u2 = Initial Kinetic Energy
∑Fx . S = W/2g ( v2 - u2 )

26. When a large force acts for a short period of time, that force
is called an impulsive force. It is denoted by the symbol I. It
is a vector quantity.
t
2
I = ʃ F dt
t
1
Linear Impulse = Force x Time

27. It is defined as the total quantity of motion contained
in a body and is measured as the product of the mass
of the body and it’s velocity.
(momentum) M= m x v
m =mass of the body (Kg)
v =velocity (m/s)
M = Linear momentum (Kg.m/s)

28. The Impulse- Momentum equation is also derived from the
Newtons second law,
t
2
ʃ F dt = m (v-u)
t
1
t
2
ʃ F dt = Impulse m (v-u) = Change of momentum
t
1
Ie, Final momentum – Initial Momentum

29. Contd…..
Therefore Impulse = m(v-u) or W/g(v—u)
The impulse of the force acting on a particle is equal to
the change in the linear momentum of the particle.
ie., Impulse = Final momentum – Initial momentum
Final momentum = Initial momentum + Impulse

30. Problems
1. A block of weight 2000 N is in contact
with a plane inclined at 30 ˚ to the
horizontal. A force ‘P’ parallel to the plane
& acting up the plane is applied to the
body as shown in fig. The coefficient of
friction between the contact surfaces is
0.20.Find (i) The value of P just cause the
motion to impend up the plane (ii) The
value of P just prevent the motion down
the plane.
2. Figure shows two blocks of weight 60N &

31. 3). A 500N block is in contact with a level
plane, the co-efficient of friction between
two contact surfaces being 0.25. if the
block is acted upon by a horizontal force
of 1300N, what time it will elapse before
the block reaches a velocity of 24 m/s.
4)An aeroplane of mass 8t is flying at a rate
of 250kmph, at a height of 2km above the
ground level. Calculate the total energy
possessed by the aeroplane.