1. Kinetics of a Particle:
CHAPTER 14
Kinetics of a Particle:
Work and Energy

2. CHAPTER 14

3. In the Newton’s method (force and acceleration method);
SF = ma = m
In the Work and Energy methods;
SF = ma = m
The work-energy method combines the principles of kinematics
with Newton’s second law to directly relate the position and speed
of a body

4. SF = ma = m

5. - what forces act on the crate
- in what direction the forces act
- how far the crate move
independent
- involve same integration

6. Work of a force
[N.m]
[J]
U = Pcosq d
U = P dcosq

7. Work of a variable force

8. Work of a constant force moving along a straight line

9. Work of a weight

10. Work of a spring force Fs = k (l-lo) Fs = ks Fs
When connected to a particle; S1 S2

11. the direction of Fs acting on the particle
negative works
the direction of Fs acting on the particle
the direction of displacement of the particle

12. Example 1
The 10 kg block rests on the smooth incline. If the spring is originally
stretched 0.5m, determine the total work done by all the forces
acting on the block when a horizontal force P = 400 N pushes
the block up the plane s = 2 m

13. Horizontal force P (constant);
Up = Pcosq (s2-s1) = 400 cos30o (2)
= [J]

14. Spring force Fs;
s1 = 0.5 [m]
s2 = = 2.5 [m]
Negative/positive work?
Us = k(s2-s1) = (2.5 – 2) = -90 [J] 1 2
Weight W;
UW = - (98.1 sin30o)(2) = [J]
Normal force NB;
Perpendicular to the displacement  no work

15. Horizontal force P
Spring force Fs
Weight W
Normal force NB
: [J]
: [J]
: [J]
: 0
Total work
UT = – 90 – 98.1 = 505 [J]

16. Principle of Work and Energy
Kinetic energy
SU1-2 = mv mv12 1 2
T1 + SU1-2 = T2
system of particles
ST1 + SU1-2 = ST2

17. Example 2
The 17.5 kN ( ≈1750 kg) car is traveling down the 10o inclined
road at a speed of 6 m/s. If the driver jams on the brakes, causing
his wheels to lock, determine how far s his tires skid on the road.
mk = 0.5

18. Solution I : Principle of Work and Energy
T1 + SU1-2 = T2
To determine FA;
SFn=0;
NA – cos10o = NA = [N]
FA = 0.5NA = [N]
T1 + SU1-2 = T2
(6)2 + {(17500sin10o) s – s} = 0
s = 5.75 [m]

19. Solution II : Newton’s method
Equations of Motion :
SFs=mas ;
17500 sin10o – = (17500/9.81) a
a = [m/s2]
Kinematics :
Since a is constant,
( ) v2 = vo2 + 2ac (s – so)
(0)2 = (6)2 + 2(-3.13)(s – 0)
s = 5.75 [m]