1. WORK
In order for work to be done, three things are necessary:
There must be an applied force.
The force must act through a certain distance, called the displacement.
The force must have a component along the displacement.

2. Work is a scalar quantity equal to the product of the magnitudes of the displacement and the component of the force in the direction of the displacement.
W = F . x or W = F cos  x
 UNITS: N.m this unit is called a Joule (J)

3. As long as this person does not lift or lower the bag of groceries, he is doing no work on it. The force he exerts has no component in the direction of motion.
Work done by forces that oppose the direction of motion, such as friction, will be negative.

4. Centripetal forces do no work, as they are always perpendicular to the direction of motion.

5. If the force acting on an object varies in magnitude and/or direction during the object’s displacement, graphical analysis can be used to determine the work done. The Force is plotted on the y-axis and the distance through which the object moves is plotted on the x-axis. The work done is represented by the area under the curve.

7. Forces acting: Ff FA FG and FN FN does NO work.
5.1 A push of 200 N moves a 100 N block up a 30inclined plane. The coefficient of kinetic friction is 0.25 and the length of the plane is 12 m.
a. Find the work done by each force acting on the block. W
FA = 200 N
FG = 100 N
θ = 30˚
μ = 0.25
x = 12 m
Forces acting: Ff FA FG and FN
FN does NO work.

8. WFG = FG x = -FGx x = - FG sin30˚x = - 100 sin 30˚ (12) = - 600 J FGy
W FA = FA x
= 200 (12)
= 2400 J
Ff = μ FN
= μ FG cos 30˚
= 0.25 (100) cos 30˚ = 21.6 N
WFf = - Ff x
= (12)
= J FN FA
WFG = FG x = -FGx x = - FG sin30˚x
= sin 30˚ (12)
= J
FGy θ Ff FG
FGx

9. Net work: ΣW = 2400 - 259.2 - 600 = 1540.8 J FN The resultant force:
b. Show that the net work done by these forces is the same as the work of the resultant force.
Net work:
ΣW =
= J FN
The resultant force:
ΣFx = FA - Ff - FGx =
= N
WF = Fx . x
= (12)
= J FA
FGy θ Ff
FGx FG

10. ENERGY
Energy is that which can be converted into work. When something has energy, it is able to perform work or, in a general sense, to change some aspect of the physical world.

11. In mechanics we are concerned with two kinds of energy:
KINETIC ENERGY: K, energy possessed by a body by virtue of its motion.
Units: Joules (J)
POTENTIAL ENERGY: PE, energy possessed by a system by virtue of position or condition.
PE = m g h Units: Joules (J)

12. WORK-ENERGY PRINCIPLE:
The work of a resultant external force on a body is equal to the change in kinetic energy of the body.
W =  K Units: Joules (J)

13. W = PE

14. 5.2 What average force F is necessary to stop a 16 g bullet traveling at 260 m/s as it penetrates into wood at a distance of
12 cm? WE
W = ΔK
vf = 0 m/s
m = kg
vo = 260 m/s
x = 0.12 m
= N

15. CONSERVATIVE AND NON-CONSERVATIVE FORCES
The work done by a conservative force depends only on the initial and final position of the object acted upon. An example of a conservative force is gravity.
The work done equals the change in potential energy and depends only on the initial and final positions above the ground and NOT on the path taken.

16. Friction is a non-conservative force and the work done in moving an object against a non-conservative force depends on the path. For example, the work done in sliding a box of books against friction from one end of a room to the other depends on the path taken.

17. For mechanical systems involving conservative forces, the total mechanical energy equals the sum of the kinetic and potential energies of the objects that make
up the system and is always conserved.

19. A roller-coaster car moving without friction illustrates the conservation of mechanical energy.

21. In real life applications, some of the mechanical energy is lost due to friction. The work due to non-conservative forces is given by:
  WNC = ΔPE + ΔK or
WNC = Ef - Eo

23. 5.3 A ballistic pendulum apparatus has a 40-g ball that is caught by a 500-g suspended mass. After impact, the two masses rise a vertical distance of 45 mm. Find the velocity of the combined masses just after impact.
COE
m1 = 0.04 kg
m2 = kg
h = m
K0 = PEf

24. K0 = PEf PEf = (m1+m2) ghf = (0.04+0.500)(9.8)(0.045) = 0.24 J
m1 = 0.04 kg
m2 = kg
h = m
K0 = PEf
PEf = (m1+m2) ghf
= ( )(9.8)(0.045)
= 0.24 J
K0 = PEf = 0.24 J
= 0.94 m/s

25. A At point A: PEA + KA At point B: KB PEA + KA= KB B = 42.9 m/s
5.4 The tallest and fastest roller coaster in the world is the Steel Dragon in Japan. The ride includes a vertical drop of 93.5 m. The coaster has a speed of 3 m/s at the top of the drop.
a. Neglect friction and find the speed of the riders at the bottom.?
COE A
At point A: PEA + KA
At point B: KB
PEA + KA= KB
vA = 3 m/s
hA = 93.5 m
hB = 0 m B
= 42.9 m/s
(about 96 mi/h)

26. WNC = Ef - E0 = KB - (PEA + KA) = - 4416.5 J
b. Find the work done by non-conservative forces on a 55 kg rider during the descent if the actual velocity at the bottom is 41 m/s.
vA = 3 m/s
vB = 41 m/s
hA = 93.5 m
hB = 0 m
m = 55 kg
WNC = Ef - E0
= KB - (PEA + KA)
= J

27. 5.5 A 20-kg sled rests at the top of a 30˚ slope 80 m in length.
If μk= 0.2, what is the velocity at the bottom of the incline?
COE
WNC = Ef - Eo
= Kf - PE0
m = 20 kg
θ = 30°
r = 80 m
μk= 0.2

28. PE0 = mgh0 = 20(9.8)(40) = 7840 J Ff = μkFN = μk Fgy = μk Fgcos30°
m = 20 kg
θ = 30°
x = 80 m
μk= 0.2
h = x sin θ
h = 80 sin 30°
= 40 m x h
PE0 = mgh0
= 20(9.8)(40)
= 7840 J
Ff = μkFN = μk Fgy
= μk Fgcos30°
= (0.2)(20)(9.8)cos30°
= 34 N
WNC = - Ff r
= - 34 (80)
= J
WNC= Kf - PE0
Kf = PE0 + WNC
= = 5120 J

29. = 22.6 m/s

30. POWER
Is the rate at which work is performed. P = work/time
= Watt
UNITS:
The difference between walking and running up these stairs is power.
The change in gravitational potential energy is the same.

31. = 23 m/s The average velocity is: 23/2 = 11.5 m/s P = Fv
5.6 A 1100-kg car starting from rest, accelerates for 5.0 s. The magnitude of the acceleration is 4.6 m/s2. What power must the motor produce to cause this acceleration?
m = 1100 kg
vo = 0 m/s
t = 5 s
a = 4.6 m/s2
F = ma
= (1100)(4.6)
= 5060 N
vf = vo + at
= (5)
= 23 m/s The average velocity is: 23/2 = 11.5 m/s
P = Fv
= (5060)(11.5)
= 5.82x104 W

32. ELASTIC FORCE
The force Fs applied to a spring to stretch it or to compress it an amount x is directly proportional to x.
Fs = - k x Units: Newtons (N)
Where k is a constant called the spring constant and is
a measure of the stiffness of the particular spring. The spring itself exerts a force in the opposite direction:

33. This force is sometimes called restoring force because the spring exerts its force in the direction opposite to the displacement. This equation is known as the spring equation or Hooke’s Law.

37. The elastic potential energy is given by:
PEs = ½ kx2 Units: Joules (J)

38. 5.7 A dart of mass kg is pressed against the spring of a toy dart gun. The spring (k = 250 N/m) is compressed 6.0 cm and released. If the dart detaches from the spring when the spring reaches its normal length, what speed does the dart acquire?
m = 0.1 kg
k = 250 N/m
x = 0.06 m
PEs = K
½ kx2 = ½ mv2
= 3 m/s

39. page 108
Problem 4