1. PHY 101: Lecture 6 6.1 Work Done by a Constant Force
6.2 The Work-Energy Theorem and Kinetic Energy
6.3 Gravitational Potential Energy
6.4 Conservation Versus Nonconservative Forces
6.5 The Conservation of Mechanical Energy
6.6 Nonconservative Forces and the Work-Energy Theorem
6.7 Power

2. PHY 101: Lecture 6 Work and Energy
6.1
Work Done by a Constant Force

3. Work vs. Energy Energy is a property of matter
Mass and Inertia are other properties
Matter that has energy can do work by means of a force
Work transfers or changes energy

4. Work – Constant Force Work is a scalar
Work done on an object by a constant force F is:
W = (F cosq)d
F is the magnitude of the force
d is the magnitude of the displacement
q is angle between direction of force and direction of displacement
SI Unit of Work: Joule (J) d

5. Amount of Energy in Food
1 calorie = J
1 food calorie = 1000 calories = 4186 J
2000 food calorie diet = 8.37 x 106 J

6. Work – Example 1 You are carrying a backpack across campus
What is the work done by your vertical carrying force on the backpack?
Work done by vertical carrying force is zero
This force is perpendicular to the direction of motion so q = 900
cos90 = 0

7. Work – Example 2
A person pushes on an unmoving wall with a force of 10 newtons
How much work is he doing?
W = 0 because the force is not moving

8. Work – Example 3
A person does 50 J of work in moving a 30-kg box over a 10-m distance on a horizontal surface
What is the minimum force required?
cosq = 1
W = Fd
50 = F(10)
F = 50 /10 = 5.0 N

9. Work – Example 4
An object is being pulled along the ground by a 75-N force directed 280 above the horizontal
How much work does the force do in pulling the object 8.0 m?
W = Fdcosq = 75 N (8.0 m) cos28 = 530 J

10. Work – Example 5 A 5.0-kg box slides a 10-m distance on ice
The coefficient of kinetic friction is 0.20
What is the work done by the friction force
fk = mkN
W = -fkd
The sign is minus because the force is in the opposite direction from the motion, q = 1800
In vertical direction, we have N – mg = ma = 0
N = mg
W = -(0.20)(5 x 9.8)(10) = - 98 J

11. Work – Example 6a A 5-kg hammer moves at 8 m/sec
The hammer drives a nail 0.04 m into a piece of wood
Ability to do work comes from motion of hammer
What is the force applied to the nail?
vf2 = vi2 + 2a(xf – xi)
02 = a(0.04)
-64 = .08a
a = - 800m/s2
F = force on hammer = ma = 5 kg x (-800) = N
F = force on nail = N
Work done by hammer = Fd = 4000(0.040 = 160 J

12. Work – Example 6b A 5-kg hammer moves at 8 m/sec
The hammer drives a nail 0.01 m into a piece of wood
a = m/s2
F = n
F = n
W = Fd = x 0.01 = 160 J
Work comes from motion of hammer which is the same in parts a and b of the problem

13. PHY 101: Lecture 6 Work and Energy
6.2
The Work-Energy Theorem and Kinetic Energy

14. Kinetic Energy KE is the energy of a moving object
Work done by forces on a moving object
W = (SF)d = (ma)d
vf2 = vi2 + 2ad
W = (ma)d = m(1/2)(vf2 - vi2)
(1/2)mv2 is identified as Kinetic Energy

15. Work – Energy Theorem A net external force does work W on an object
The kinetic energy of the object changes from its initial value of KEi to a final value of KEf
The difference between the two values being equal to the work
W = KEf – KEi = ½ mvf2 – ½ mvi2

16. Work – Energy Theorem Example 1
A 1200-kg automobile travels at speed 25 m/s
(a) What is its kinetic energy?
(b) What is the net work that would be required to bring it to a stop?
(a)
KE = ½ mv2 = (0.5)(1200)(25)2 = 3.8 x 105 J
(b)
W = KEf – KEi = 0 – 3.8 x 105 = -3.8 x 105
Note: Work is negative because it is taking away KE

17. Work-Energy Theorem Example 2
A constant net force of 75 N acts on an object initially at rest through a parallel distance of 0.60 m
(a) What is the final kinetic energy of the object?
(b) If the object has a mass of 0.20 kg, what is its final speed?
(a)
W = Fd = KEf – KEi
W = 75 x 0.60 = KEf
KEf = 45 joules
(b)
KE = ½ mv2 = 45 joules
v = sqrt(2 x 45 / 0.20) = 21.2 m/s

18. PHY 101: Lecture 6 Work and Energy
6.3
Gravitational Potential Energy

19. Work Done by Force of Gravity
An object moves from initial height hi to final height hf
The force of gravity acts on the object
Wgravity = Fd
F = mg = weight
d = hi - hf
Wgravity = mg(hi – hf) = mghi - mghf

20. Gravitational Potential Energy
The gravitational potential energy, PE, is the energy that an object of mass m has because of its position relative to the surface of the earth
That position is measured by the height h of the object relative to an arbitrary zero level
PE = mgh

21. Potential Energy – Example 1
What is the gravitational potential energy, relative to the ground, of a 1.0-kg box at the top of a 50-m building?
PE = mgh = 1 x 9.8 x 50 = 490 J

22. Potential Energy – Example 2
How much more gravitational potential energy does a 1.0-kg hammer have when it is on a shelf 1.5 m high than when it is on a shelf 0.90 m high?
DPE = mg(hH – hL) = 1 x 9.8 x (1.5 – 0.9) = 5.88 J

23. PHY 101: Lecture 6 Work and Energy
6.4
Conservative Versus Nonconservative Force

24. Conservative Force Version 1 Version 2
A force is conservative when the work it does on a moving object is independent of the path between the object’s initial and final position
Version 2
A force is conservative when it does no net work on an object moving around a closed path starting and finishing at the same point

25. Conservative Forces Nonconservative Forces
Gravity
Spring force
Electrostatic force
Magnetic force
Nonconservative Forces
Friction
Air resistance

26. Demonstration Gravity is Conservative Force
A 10-kg mass rises 20 m
The 10-kg mass is then lowered 20 m to it’s starting position
How much work is done by gravity?
Raise
W = mg(hi – hf) = 10(9.8)(0 – 20) = J
Lower
W = mg(hi – hf) = 10(9.8)(20 – 0) = J
Total Work = = 0 J

27. Conservative Force Example
While testing a cannon, a 1.0-kg ball is fired straight up into the air
The cannon ball rises 22.5 m and then falls back to the height at which it was launched
What is amount of work done on ball by gravity?
Gravity is a conservative force
Work only depends on the end points of the path
The starting and ending points are the same
Work of gravity = 0

28. Work Done by Gravity Path Independence 1
Mount Everest is 9000 m high
How much work is done against gravity by a 70 kg person climbing straight up to the top of Mount Everest?
W = mg(hi – hf)=70(9.8)(-9000)=-6.17 x 106 N

29. Work Done by Gravity Path Independence 2
How much work is done against gravity by a 70 kg person climbing a ramp to the top of Mt. Everest?
Because of the angle q
Force along ramp = -mgsinq
Distance moved, d, is along ramp
Angle between F and d is 0 degrees
W = Fd = Fdsinq
h/d = sinq
W = -mg(h/d)d = -mgh
Same work as climbing straight up
Note: This shows that path doesn’t matter when gravity is the force

30. Demonstration Friction is Nonconservative Force
A 10-kg mass slides 20 m on a table
10-kg mass then slides back 20 m to it’s starting position
Coefficient of kinetic friction is 0.20
How much work is done by fricition?
Slide Forward
W = Fdcosq = mk(mg)dcos180
W = (0.2)(10)(9.8)(20)(-1) = -392 J
Slide Back
W = Fdcosq=mk(mg)dcos180
Total Work = = -784 J <> 0

31. PHY 101: Lecture 6 Work and Energy
6.5
The Conservation of Mechanical Energy

32. Work-Energy Conservative / Nonconservative Forces
W = Wc + Wnc
Wc + Wnc = ½ mvf2 – ½ mvi2
Only conservative force is gravity
mg(hi – hf) + Wnc = ½ mvf2 – ½ mvi2
Wnc = ½ mvf2 – ½ mvi2 + mg(hf – hi)
Wnc = (KEf – KEi) + (PEf – PEi)

33. Derivation Conservation of Mechanical Energy
Suppose that the net work Wnc done by external nonconservative forces is zero
Wnc = 0 J
0 = (KEf – KEi) + (PEf – PEi)
(KEf – KEi) = -(PEf – PEi)
(KEf + PEf) = (KEi + PEi)
½ mvf2 + mghf = ½ mvi2 + mghi
Mechanical Energy = E = ½ mv2 + mgh
Ef = Ei

34. Principle Conservation of Mechanical Energy
Total mechanical energy (E = KE + PE) of an object remains constant as the object moves
Provided that the net work done by external nonconservative forces is zero, Wnc = 0 J

35. Kinetic Energy vs. Potential Energy
The sum of the kinetic and potential energies at any point is conserved
Kinetic and Potential Energies may be converted or transformed into one another

36. Conservation of Mechanical Energy Example 1a
A person standing on a bridge at a height of 115 m above a river drops a kg rock
(a) What is the rock’s mechanical energy at the time of release relative to the surface of the river?
Initially, at top of bridge, rock is not moving
KEi = 0
Ei = KEi + PEi = 0 + mgh
Ei = x 9.8 x 115 = joules

37. Conservation of Mechanical Energy Example 1b
A person standing on a bridge at a height of 115 m above a river drops a kg rock
(b) What are rock’s kinetic, potential, and mechanical energies after it has fallen 75.0 m?
After the rock falls 75.0 m, its height is 40 m and the total mechanical energy is the same
E = J
PE = mgh = 0.25 x 9.8 x 40 m = 98 J
E = KE + PE
KE = E – PE = – 98 = J

38. Conservation of Mechanical Energy Example 1c
A person standing on a bridge at a height of 115 m above a river drops a kg rock
(c) Just before the rock hits the water, what are its speed and total mechanical energy?
At bottom of fall, total mechanical energy, E, is 115 J
Potential energy is zero, because the height is now zero
E = KE + PE
= KE + 0
KE = = ½ mv2
v =sqrt( x 2 /0.25) = 47.5 m/s

39. Conservation of Mechanical Energy Example 2
A 60-kg stunt person runs off a cliff at 5.0 m/s and lands safely in the river 10.0 m below. What was his speed when he landed?
Mechanical energy is conserved in this problem
Ef =Ei
mghf + ½ mvf2 = mghi + ½ mvi2
Mass, m, cancels out of this problem
hi = 10.0 m
hf = 0 m
vi = 5.0 m/s
(9.8)(0) + ½ vf2 = (9.8)(10.0) + ½ (9.8)(5)2
vf = sqrt[2( )] = 29.7 m/s

40. Conservation of Mechanical Energy Example 3
A skier coasts down a very smooth, 10-m-high slope
The speed of the skier on the top of the slope is 5.0 m/s
What is his speed at the bottom of the slope?
Mechanical energy is conserved because there is no friction
Ef = Ei
½ mvf2 + mghf = ½ mvi2 + mghi
At the bottom of the hill hf = 0
Mass, m, cancels out of the problem
½ vf2 + 0 = ½ vi2 + ghi
vf2 = vi2 + 2ghi
vf = sqrt(vi2 + 2ghi) = sqrt( (9.8)(10) = 14.8 m/s

41. Conservation of Mechanical Energy Example 4
A 14.2 x 103 N auto is traveling at 26.7 m/s
Auto runs out of gas 16 x 103 m from a service station
Neglecting friction, if station is on a level 15.2 m above elevation where car stalled, how fast will the car be going when it rolls into the station, if in fact it gets there?
There is no friction. Mechanical energy is conserved.
Ef = Ei
½ mvf2 + mghf = ½ mvi2 + mghi
The initial height is taken as zero
½ mvf2 + mghf = ½ mvi2 + 0
½ vf2 + ghf = ½ vi2
vf2 + 2ghf = vi2
vf = sqrt(vi2 – 2ghf) = sqrt(26.72 – 2(9.8)(15.2))
vf = sqrt(414.97) = m/s

42. PHY 101: Lecture 6 Work and Energy
6.6
Nonconservative Forces and the Work-Energy Theorem
Skipped

43. PHY 101: Lecture 6 Work and Energy
6.7
Power

44. Average Power Average rate at which work W is done
It is obtained by dividing W by time required to perform the work
P = Work / time = W / t
SI Unit of Power: joules/s = watt (W)
P = W / t = Fd / t = F(d / t) = Fv

45. Average Power - Example
How much power does it take to raise an object weighing 100 N a distance of 20.0 m in 50.0 s
It begins at rest and ends at rest?
Because the object starts at rest and ends at rest, work only changes potential energy
W = mgh = 100(20) = 2000 J
Power = W/t = 2000/50.0 = 40 W