1. Chapter 6: Work and Energy
Work Energy
 Work done by a constant force
(scalar product)
 Work done by a varying force
(scalar product & integrals)
Kinetic Energy
Work-Energy Theorem
Work and Energy

2. Work and Energy

3. Work by a Baseball Pitcher
A baseball pitcher is doing work on
the ball as he exerts the force over
a displacement.
v1 = 0
v2 = 44 m/s
Work and Energy

4. Work Done by a Constant Force (I)
Work (W)
 How effective is the force in moving a body ?
 Both magnitude (F) and directions (q ) must be taken into account.
W [Joule] = ( F cos q ) d
Work and Energy

5. Work Done by a Constant Force (II)
Example: Work done on the bag by the person..
 Special case: W = 0 J
a) WP = FP d cos ( 90o )
b) Wg = m g d cos ( 90o )
 Nothing to do with the motion
Work and Energy

6. Example 1A A 50.0-kg crate is pulled 40.0 m by a
constant force exerted (FP = 100 N and
q = 37.0o) by a person. A friction force Ff =
50.0 N is exerted to the crate. Determine
the work done by each force acting on the
crate.
Work and Energy

7. Example 1A (cont’d) F.B.D. WP = FP d cos ( 37o )
Wf = Ff d cos ( 180o )
Wg = m g d cos ( 90o )
WN = FN d cos ( 90o )
180o d
90o
Work and Energy

8. Example 1A (cont’d) WP = 3195 [J] Wf = -2000 [J] (< 0) Wg = 0 [J]
WN = 0 [J]
180o
Work and Energy

9. Example 1A (cont’d) The body’s speed increases. Wnet = SWi
= 1195 [J] (> 0)
The body’s speed increases.
Work and Energy

10. Work-Energy Theorem Wnet = Fnet d = ( m a ) d
= m [ (v2 2 – v1 2 ) / 2d ] d
= (1/2) m v2 2 – (1/2) m v1 2
= K2 – K1
Work and Energy

11. Example 2 A car traveling 60.0 km/h to can brake to
a stop within a distance of 20.0 m. If the car
is going twice as fast, 120 km/h, what is its
stopping distance ?
(a)
(b)
Work and Energy

12. Example 2 (cont’d) (1) Wnet = F d(a) cos 180o
= - F d(a) = 0 – m v(a)2 / 2
 - F x (20.0 m) = - m (16.7 m/s)2 / 2
(2) Wnet = F d(b) cos 180o
= - F d(b) = 0 – m v(b)2 / 2
 - F x (? m) = - m (33.3 m/s)2 / 2
(3) F & m are common. Thus, ? = 80.0 m
Work and Energy

13. Forces on a hammerhead
Forces
Work and Energy

14. Work and Energy

15. Work and Energy

16. Spring Force (Hooke’s Law)FS
Spring Force
(Restoring Force):
The spring exerts its force in the
direction opposite
the displacement. FP
x > 0
Natural Length
x < 0
FS(x) = - k x
Work and Energy

17. Work Done to Stretch a SpringFS FP
FS(x) = - k x
Natural Length x2
W = FP(x) dx x1 W
Work and Energy

18. Work and Energy

19. Work Done by a Varying Forcelb
W = F|| dl la
Dl  0
Work and Energy

20. Example 1A A person pulls on the spring, stretching it
3.0 cm, which requires a maximum force
of 75 N. How much work does the person
do ? If, instead, the
person compresses
the spring cm,
how much work
does the person do ?
Work and Energy

21. Example 1A (cont’d) x2 = 0.030 m (a) Find the spring constant k
k = Fmax / xmax
= (75 N) / (0.030 m) = 2.5 x 103 N/m
(b) Then, the work done by the person is
WP = (1/2) k xmax2 = 1.1 J
(c)
x2 = m
WP = FP(x) d x = 1.1 J
x1 = 0
Work and Energy

22. Example 1B A person pulls on the spring, stretching it
3.0 cm, which requires a maximum force
of 75 N. How much work does the spring
do ? If, instead, the
person compresses
the spring cm,
how much work
does the spring do ?
Work and Energy

23. Example 1B (cont’d) x2 = -0.030 m (a) Find the spring constant k
k = Fmax / xmax
= (75 N) / (0.030 m) = 2.5 x 103 N/m
(b) Then, the work done by the spring is
(c) x2 = m  WS = -1.1 J
x2 = m
WS = FS(x) d x = -1.1 J
x1 = 0
Work and Energy

24. Example 2 A 1.50-kg block is pushed against a spring
(k = 250 N/m), compressing it m, and
released. What will be the speed of the
block when it separates from the spring at
x = 0? Assume mk =
0.300.
FS = - k x
(i) F.B.D. first !
(ii) x < 0
Work and Energy

25. Example 2 (cont’d) x2 = 0 m (a) The work done by the spring is
(b) Wf = - mk FN (x2 – x1) = ( )
(c) Wnet = WS + Wf = x 0.200
(d) Work-Energy Theorem: Wnet = K2 – K1
 4.12 = (1/2) m v2 – 0
 v = 2.34 m/s
x2 = 0 m
WS = FS(x) d x = J
x1 = m
Work and Energy

26. Work and Energy