1. Work and Energy
Physics Chapter 5

2. Chapter 5 Objectives
Recognize the difference between the scientific and ordinary definitions of works
Define work by relating it to force and displacement
Calculate the net work done with many forces are applied to an object

3. Work
How do you define work?

4. Definition of Work
Work (W) is done on an object when a force (F) causes a displacement (d) of the object
Work is done on an object only if it moves in the direction of the force
W = F d
Force units (N)  distance units (m)
N•m are also called joules (J)
W =  F ds

5. Work 2 categories of Work Work done against another force
Lift something – exerted against force of gravity
Work done to change the speed of something
Stopping a car or speeding it up
How would you calculate the work in this case?
What is the component of F in the direction of d?
Only the component of the force that is in the direction of the objects displacement does work!
F cos 
W = F cos  d

6. Energy Indentify several forms of energy
Calculate kinetic energy for an object
Apply the work-kinetic energy theorem to solve problems
Distinguish between kinetic & potential energy
Classify different types of potential energy
Calculate the potential energy associated with an object’s position

7. Energy Energy is “something” that enables an object to do work
What is that something?
Energy is associated with heat, light, electricity, mechanical motion, sound, and the nature of a chemical reaction
Mechanical energy - Energy due to the position or the movement of something
Mechanical energy has 2 forms
Kinetic Energy (KE) – energy of motion
Potential Energy (PE) – Stored energy

8. Kinetic Energy Kinetic Energy depends on speed and mass
Units – Joules (used for all forms of energy)
kg•m2/s2 or N•m or J

9. Kinetic Example Problem
A 100. Kg Linebacker moves at 8.90 m/s (runs 100m in 11.2 secs). How much KE does the linebacker have? How fast does a 54.5 Kg wide receiver have to run to have the same KE as the linebacker?
Given:
Linebacker
VL = 8.90 m/s
ML = 100. kg
KEL = ?
Wide Receiver
Vw = ?
Mw = 54.5 kg
KEW = KEL
½ MwVw2 = KEL
Vw2 = 2KEL/ Mw
Vw2 = 2 ( 3.96x103 J) / (54.5 kg)
Vw = 12.1 m/s
Soln:
KEL = ½ ML VL2
= ½ (100. Kg) (8.90 m/s)2
KEL = 3.96 x 103 J
note: wide receiver would have to run 100m in 8.26 secs

10. Work-Kinetic Energy Theorem
The net work (Wnet) done on an object is equal to the change in the KE of the object
Wnet =  KE = ½ mvf2 - ½ mvi2

11. Example problem P. 168 #2 Wnet =  Kef -  Kei W=Fnetd
Ff = 950 N
Given:
m = 2.0 x 103 Kg vi = 0
Vf = 2.0 m/s
d = ?
FA= 1140 N
Fnetd = ½ mVf2
d = mVf2
2Fnet
d = (2.0 x 103 Kg) (2.0 m/s)2
2 (1140N – 950N)
d = 21 m
Soln:
Wnet =  Kef -  Kei
Wnet = ½ mvf2 - ½ mvi2
W=Fnetd
FNetd = ½ mVf2 - ½ mVi2

12. gravitational PE = mass  free-fall acceleration  height
Potential Energy
Potential Energy is the energy associated with an object because of the position, shape, or condition of the object.
Stored energy
Chemical energy
Object due to its position (Gravitational, ex: rock on hilltop)
Elastic potential energy
Gravitational potential energy is the potential energy stored in the gravitational fields of interacting bodies.
depends on height from a zero level (zero level is arbitrary)
PEg = mgh
gravitational PE = mass  free-fall acceleration  height

13. Potential Energy
Elastic potential energy is the energy available for use when a deformed elastic object returns to its original configuration.
The symbol k is called the spring constant, a parameter that measures the spring’s resistance to being compressed or stretched.

14. Potential Energy Elastic Potential energy
The energy available for use in deformed elastic objects
Rubber bands, springs in trampolines, pole-vault poles, muscles
For springs, the distance compressed or stretched = x

15. Sample Problem: Elastic potential energy
A 70.0 kg stuntman is attached to a bungee cord with an unstretched length of 15.0 m. He jumps off a bridge spanning a river from a height of 50.0 m. When he finally stops, the cord has stretched a length of 44.0 m. Disregard the weight of the bungee cord. Assuming the spring constant of the bungee cord is 71.8 N/m, what is the total potential energy relative to the water when the man stops falling?
Given:
m = 70.0 kg
k = 71.8 N/m
g = 9.81 m/s2
h = 50.0 m – 44.0 m = 6.0 m
x = 44.0 m – 15.0 m = 29.0 m
PE = 0 J at river level
PEtot = ?

16. Sample Problem: Elastic potential energy
The zero level for PEg is chosen to be at the surface of the water.
h = 50.0 m – 44.0 m = 6.0 m
h= 6.0 m
The total Potential Energy is the sum of PEg and PEelastic
PEtot = PEg + PEelastic
PEg = mgh
= (70.0 kg)(9.81m/s2)(6.0m) = 4.1x103 J
PEelastic = ½ kx2
= ½ (71.8 N/m)(29.0m)2 = 3.02x104J
PEtot = 4.1x103J x104J
PEtot = 3.43x104J

17. Conservation of Energy
Energy cannot be created or destroyed, but only changed from one form to another. It transforms without net loss or gain
When we say that some is conserved, we mean that it remains constant
2 types of Energy
Mechanical Energy –kinetic, gravitational Potential, elastic potential, and chemical potential energy
2. Nonmechanical energy – Nuclear, chemical, internal, and electrical

18. Mechanical Energy MEi = MEf ME = KE + ∑PE
Mechanical Energy is the sum of KE and PE present in a situation
ME = KE + ∑PE
Mechanical Energy is also conserved
MEi = MEf
initial mechanical energy = final mechanical energy
(in the absence of friction)

19. Mechanical Energy example p. 177 #2
Initial
hi = 10.0m
KEi = 0
PEi = mghi
Midpoint
hm = 5.0m
KEm = ½ mVm2
PEm = mghm
Final
hf = 0
KEf = ½ mVf2
PEf = 0
Given:
Soln:
@ midpoint (hm = 5.0)
MEi = MEm
PEi + KEi = PEm + KEm
mghi = mghm + ½ mVm2
Vm2 = 2ghi – 2ghm
= 2g(hi-hm)
Vm = 9.9 m/s
@ Final
MEi = MEf
PEi + KEi = PEf + KEf
mghi = ½ mVf2
Vf2 = 2ghi
Vf = 14.0 m/s

20. power = work ÷ time interval
Power is a quantity that measures the rate at which work is done or energy is transformed
How fast you do the job
Rate of energy transfer
P = W/∆t
power = work ÷ time interval
Units for power – Watts (W) = J/s
Horsepower (hp) another unit for power
1hp = 746 watts

21. Power
Machines with different power ratings do the same amount of work in different time intervals
An alternate equation for power in terms of force and speed is
P = Fv
power = force  speed