1) The document discusses the work-energy principle, which states that the work done on an object equals its change in kinetic energy. 2) It provides various examples of calculating work done by different types of forces like inclined, gravitational, frictional, and spring forces. 3) Key concepts covered include the definitions of work and work done, sign conventions for different forces, and using the work-energy principle to solve problems by setting the work equal to the change in kinetic energy.

1. Work- Energy Principle
Sub- Engg. Mechanics
By: Mr. Ghumare S. M.
Sanjivani COE, Kopargaon
Work- When particle undergoes a displacement along
the line of action of force then work done by force is
the product of force and displacement(S ).
Work done = Force x Displacement= F X S
Unit: N-M or Joule

2. Work or Work Done
Sign Convension: W. D. denoted by letter U
 W. D. will be + ve when displacement takes place in the direction
of force
• W. D. will be - ve when force and displacement are in the opposite
in the direction
• Note: After the application of force if there is no displacement
then W.D will be zero. Or work is not done
W.D = F X S
1 2. . UW D F dx  

3. W.D.by Inclined Force
W.D = Force x Displacement = F X S
If Force F is inclined with some angle
Then, W. D. in Hz Dir. by Horizontal Compo. = F cos X S
W.D in perpendicular direction is Zero. W. D. = F Sin X S = 0,
As S=0 there is no displacement in Y-direction


 
1 2. . U xW D F dx F dx   

4. W. D. by Gravity Force
Consider motion if the block from S1to S2 and Viseversa.
• W. D. + ve when motion is towards gravity from S2 to S1
• W. D. -ve when motion is against gravity S1 to S2
• Displacement in
• Dir the direction of Gravity
2
1
2 1 2 1
. . ,
. . ( ) ( )
. .
y
y
W D W dy
W D W y y y y h
W D Wh mgh

   
 


5. W. D. by Frictional Force
Work done by Frictional force-
Frictional force = F =
W.D. by Frictional force = -F . S
• Sign Conventions
W. D. by frictional force is always –ve because frictional
force always acts opposite to the direction of motion.
N S 
N

6. W. D. by Spring Force
When the spring is stretched or compressed it does the work.
If the spring is stretched from
Work done by Spring force
1 2tox x
 2 1
2
2 1
2 2
1
2
1
2
K x where x x x
K x x
  
 F
Where, K = Stiffness of spring or Spring constant or modulus of spring
K= Force required for unit displacement = N / M
Sign Convension
 W. D. by spring force is –ve when spring is stretched or compressed ,
in this state spring force and applied force acts in opposite direction
 W. D. by spring force is + ve when spring is returning to the original
position

7. Work- Energy Principle
W-E Principle derived form Newton’s Second Law
2
1
2 2
2 1
2 1
1 1
( )
2 2
( ) . . .
. . .
As we know that
S v
S u
dv
F m a a
dt
dv dv dv ds
F m Where can be written as X
dt dt ds dt
ds dv
Put v then F m v
dt ds
F ds m v dv
F S S mv mu
F S S Final K E Initial K E Change in K E
F S W D Change in K E
 
 
 

  
   
 



  




8. Work- Energy Principle
To Solve problem Following W-E Principle is used
2 2
2 1
2 2
2 2
1 1
( )
2 2
. . .
1 1
2 2
1 1
2 2
. . .
F S S mv mu
F S W D Final K E Initial K E
F S mv mu
mu F S mv
Initial K E W D Final K E
  
  
 
 
 






9. Work- Energy Principle
Example 1. A women having mass 70 Kg stands on elevator
which has downward acceleration 4 m/s2 starting from the
rest. Determine the work done by her weight and work done
by normal force which the floor exerts on her when elevator
descends by 6m.
Given, Mass of Women=70 Kg, Downward Accl a = 4 m/s2
Elevator descends by 6m, i.e. h = 6m
Work done by her weight = M g h
= 70 x 9.81 x 6
W. D. = 4120.2 N. M.

10. Example 1 Continue
To find Normal Force,= N
sin
70 9.81 70 4
Re 506.7
506.7 6 2440.2 . .
. .
Work Done by Normal Force
W.D
nd
y y
y y
y
U g Newton II law F m a
Motion is Down W N
F m a
W N m a
X N X
Normal action N N
N X h X N M
W D is ve becaue nomal rection act Opposite to motion


 
 

     





11. Example-2
Prob. The spring is upstretched when x = 0. If the block moves
from the initial position x1 = 100mm to the final position x2 =
200mm. Determine 1) W.D. spring on the body 2) W. D. by weight.
Take spring Constant K=4 KN/m
Given, Initially spring is upstretched x = 0.
Then stretched from 100mm to 200mm i. e. 0.1 m to 0.2 m
x1=0.1m and x2=0.2m K=4 KN/m=4000 N/m
W.D. by Spring
= - 60 N. M. or Joule
 2 1
2 2
2 2
1
2
1
4000(0.2 0.1 )
2
K x x  
  

12. Prob. 2 Continues
Consider F.B.D. of Block
Comp. of Wt. along the Plane:
= 7 x 9.81 x sin 20
= 23.48 N
W.D. By weight on the body:
W.D.= Force x Displacement
Here F=23.48N
Disp. S= 0.2-0.1 = 0.1M =Spring Disp.
W. D. = 23.48 X 0.1
= 2.35 N.M or Joule

13. Example-3
Prob. The car having mass 2000 Kg is originally travelling with 2
m/s. Determine the distance it must be travelled by force 4 KN in
order to attain a speed of 5m/s. Neglect friction.
Given Mass of Car M=2000Kg ,
W=Mg=2000 x 9.81 N
Initial Velocity = u =2 m/s
Final Velocity = v =5 m/s
Force F=4 KN= 4000N
Let be distance travelled S= ?
Solution
Using W-E Principle 2 21 1
2 2
= Final.K.E-Initial K.E.
F S mv mu 

14. Prob. 3 Conti… 2 21 1
2 2
F S mv mu  To find F
4000 20 2000 9.81 sin10
351.79
Resultant Force F Cos X X
F N
  



2 2
2 2
1 1
2 2
1 1
351.79 2000 5 2000 2
2 2
Using W-E Principle
Solving S= 59.69 m
F S mv mu put all the terms
X S X X X X
 
 


15. Example-4
Prob.4 A package is projected 10m up the inclined plane so that it
reaches at the top with zero velocity. Determine the initial velocity
of the package at A. Take Coe. of Kine. friction is 0.12.
Given
Displacement S =10 m
Final Velocity = v =0 m/s As
it reaches at zero velocity
Initial Velocity = u = ? To find
Assume Mass of package as m
Hence, W= mg
To solve using
Using W-E Principle
2 21 1
2 2
F S mv mu 

16. Prob. 4 Conti…
0
0
0
0.12 0.12 cos15
Re . cos15
0.12 cos15 sin15
(0.12cos15 sin15)
Frictional Force Fs
Here,
Resultant Force
k N N x mg
Normal c N m g
F
mg mg
mg


  


  
  

2 2
2
1 1
2 2
1 1
(0.12 cos15 sin15) ( 0 )
2 2
8.57
Using W-E Principle
Solving Initial velocity = m / s
F S mv mu
m g g X S m X X u
u
 
   


17. Thank You