1. Mass training of trainers General Physics 1
May 20-22, 2017
The Law Conservation of Mechanical Energy & Power
Prof. MARLON FLORES SACEDON
Physics Instructor
Visayas State University
Baybay City, Leyte

2. Objectives To study and calculate WorkDone by the system.
To study and calculate Potential Energy of the system.
To study and calculate Kinetic Energy by the system.
To derive and apply the Work-Energy theorem.
To derive and apply The Law of Conservation of Mechanical Energy.
To study and calculate Average Power.

3. Conservation of Energy Energy is conserve.
The total energy for every point of travel are the same.
p.1
𝐸 1
p.2
p.3
𝐸 2
𝐸 3
p.4
𝐸 4
𝐸 1 = 𝐸 2 = 𝐸 3 = 𝐸 4

4. What is Work?
In physics and engineering, work is accomplished only when a force acts on a body and this force is able to move the body.

5. What is Work?
In physics and engineering, work is accomplished only when a force acts on a body and this force is able to move the body.
Work Done is defined is either of two ways:
1. It is the product of the force and the component of the displacement in the direction of the force.
2. The product of the displacement and the component of the force in the direction of the displacement. 𝑦 𝑥 𝑦 𝑥 𝐹 m 𝜃 𝐹 𝑠 m 𝜃 𝑠
Def’n 1: 𝑊=(𝐹)(𝑠 𝑐𝑜𝑠𝜃)
Def’n 2: 𝑊=(s)(𝐹𝑐𝑜𝑠𝜃)
𝑊=𝐹𝑠𝑐𝑜𝑠𝜃
𝑊= 𝐹 ∙ 𝑠
Note: Work (𝑊) is scalar physical quantity.

6. Units of Work What is Work?
In physics and engineering, work is accomplished only when a force acts on a body and this force is able to move the body.
Work Done is defined is either of two ways:
1. It is the product of the force and the component of the displacement in the direction of the force.
2. The product of the displacement and the component of the force in the direction of the displacement.
Units of
Work
James Prescott Joule FRS (24 December 1818 – 11 October 1889) was an English physicist and brewer, born in Salford, Lancashire. Joule studied the nature of heat, and discovered its relationship to mechanical work

7. What is Work?
In physics and engineering, work is accomplished only when a force acts on a body and this force is able to move the body.
Work Done is defined is either of two ways:
1. It is the product of the force and the component of the displacement in the direction of the force.
2. The product of the displacement and the component of the force in the direction of the displacement.
p.1
p.2
Work is done on the object. m 𝐹
𝑾= 𝑭 ∙ 𝒔
𝑾=𝑭𝒔𝒄𝒐𝒔𝜽 𝑠
𝑾=𝑭𝒔𝒄𝒐𝒔𝟎
𝑾=𝑭𝒔 𝐹 m 𝜃
p.1
p.2
Work is done on the object.
𝑾=𝑭𝒔𝒄𝒐𝒔𝜽
If: 𝜃= 30 𝑜
𝑠=20𝑚
Calculate workdone by the force
𝑊 𝐹 = 100𝑁 20𝑚 𝑐𝑜𝑠 30 𝑜 𝑠
𝑊 𝐹 =1, 𝑁𝑚 or
𝑊 𝐹 =1, 𝐽

8. 𝑾=𝑭𝒔𝒄𝒐𝒔𝜽 𝑠 Calculate Work done on the object If: P=100𝑁 @ 𝜃= 30 𝑜𝑦 𝑥 𝑦 𝑥
Calculate Work done on the object
If: 𝜃= 30 𝑜
𝑠=20𝑚, 𝜇 𝑘 =0.15
𝑾=𝑭𝒔𝒄𝒐𝒔𝜽 𝑁 𝑃
m=40kg 𝜃 𝑓 𝑤 𝑠
𝜇 𝑘 =0.15
Step 5: Calculate Work done on the object by summing up all work done by forces acting on the object.
Σ 𝐹 𝑦 =𝑁+ 𝑃 𝑦 −𝑚𝑔=0
Step 2: Calculate Work done by gravity
𝑁=𝑚𝑔− 𝑃 𝑦 =𝑚𝑔−𝑃𝑠𝑖𝑛𝜃
𝑁= 40𝑘𝑔 𝑚 𝑠 2 −100𝑠𝑖𝑛30 =342.4𝑁
𝑊 𝑔 =𝑚𝑔𝑠𝑐𝑜𝑠𝜃
𝑊 𝑔 = 40𝑘𝑔 𝑚 𝑠 𝑚 𝑐𝑜𝑠 270 𝑜 =0
𝑊 𝑜𝑏𝑗 = 𝑊 𝑃 + 𝑊 𝑔 + 𝑊 𝑁 + 𝑊 𝑓
𝑓= 𝑁
=51.36𝑁
Step 3: Calculate Work done by Normal force N
𝑊 𝑜𝑏𝑗 =1, −1,027.2
𝑊 𝑜𝑏𝑗 = 𝐽
𝑊 𝑜𝑏𝑗 = 𝑊 𝑃 + 𝑊 𝑔 + 𝑊 𝑁 + 𝑊 𝑓
𝑊 𝑁 =𝑁𝑠𝑐𝑜𝑠𝜃
Another way to calculate Work done on the object by 𝑊 𝑜𝑏𝑗 =Σ 𝐹 𝑛𝑒𝑡 ∙𝑠
𝑊 𝑁 =342.4𝑁 20𝑚 𝑐𝑜𝑠 90 𝑜 =0
Step 1: Calculate Work done by pulling force 𝑃
Step 4: Calculate Work done by friction force 𝑓
Σ 𝐹 𝑛𝑒𝑡 =𝑃𝑐𝑜𝑠𝜃−𝑓
𝑊 𝑃 =𝑃𝑠𝑐𝑜𝑠𝜃
Σ 𝐹 𝑛𝑒𝑡 =100𝑐𝑜𝑠30−51.36=35.25N
𝑊 𝑃 = 100𝑁 20𝑚 𝑐𝑜𝑠 30 𝑜
𝑊 𝑓 =𝑓𝑠𝑐𝑜𝑠𝜃
𝑊 𝑜𝑏𝑗 =(35.25N)(20𝑚)
𝑊 𝑃 =1, 𝐽
𝑊 𝑓 =51.36𝑁 20𝑚 𝑐𝑜𝑠 180 𝑜 =−1,027.2𝐽
𝑊 𝑜𝑏𝑗 = 𝐽

9. What is Energy? Energy (𝐸) is often associated with work.
When work is done on the body, a change is produced in the body.
1. Change in its motion or inertia
2. Change in position
3. Change in temperature (work to overcome friction)
The change produced in the body by the application of work is often times called energy so that energy may be thought of as the ability or capacity to do work.
An agent is said to possess energy if it is also to do work. Energy being the maximum work, then the units of energy is the same as the units of work.
There are many forms of energy:
Mechanical, Chemical, Electrical, Heat or Thermal, Electromagnetic, Nuclear, Solar, etc.
Our daily observations make us realize that transformations occur from one form of energy to another.

10. Forms of mechanical energy:
1. Potential Energy (𝑈) – energy at rest
2. Kinetic Energy (𝐾) - energy in motion
Kinds of P-E:
1. Gravitational ( 𝑈 𝑔 ) - energy possessed due to its position or elevation.
2. Elastic ( 𝑈 𝑒𝑙 ) - due to its state of strain or configuration/
3. Chemical ( 𝑈 𝑐ℎ𝑒𝑚 ) - due to its composition

11. m Gravitational Potential Energy Calculating work done by gravity
𝑊=𝐹𝑠𝑐𝑜𝑠𝜃
Def’n. of Work done
𝑤=𝑚𝑔
p.1
𝑦 1
𝑊 𝑔 =𝑚𝑔ℎ𝑐𝑜𝑠𝜃
ℎ=−( 𝑦 2 − 𝑦 1 )
𝑊 𝑔 =𝑚𝑔 − 𝑦 2 − 𝑦 1 cos0
𝑊 𝑔 =−𝑚𝑔 𝑦 2 − 𝑦 1 (1)
𝑊 𝑔 =− 𝑚𝑔 𝑦 2 −𝑚𝑔 𝑦 1
p.2
𝑦 2
Def’n of Grav’l potential energy
𝑈 𝑔 =𝑚𝑔𝑦
𝑊 𝑔 =− 𝑈 2 − 𝑈 1 𝑦 𝑥 𝑜
𝑊 𝑔 =−∆𝑈
Therefore, workdone by gravity is equal to the negative change in grav’l potential energy.

12. Elastic potential Energy
𝐹=𝑘𝑥
Where:
F = applied force [N, dyne, poundal]
x = stretched/ compressed length [m, cm, ft]
k = spring constant [N/m, dyne/cm, poundal/ft
Definition of elastic potential energy. 𝑈 𝑒𝑙 = 1 2 𝑘 𝑥 2
Work done by elastic
𝑊 𝑒𝑙 =− ∆𝑈 𝑒𝑙 =− 𝑈 𝑒𝑙 2 − 𝑈 𝑒𝑙 1 =− 𝑘 𝑥 2 2 − 𝑘 𝑥 1 2

13. Kinetic Energy (𝐾) 𝑦 𝑥 𝑜 𝑦 𝑥 𝑜
Suppose the net external force 𝐹=500𝑁, displacement 𝑠=20𝑚, & moving mass 𝑚=15𝑘𝑔. What is the final velocity of moving object if it starts from rest.?
𝑣 1
𝑣 2 m 𝐹 𝑠
𝑊=∆𝐾
𝑊=𝐹𝑠𝑐𝑜𝑠𝜃
Def’n. of Work done
Substitute Eq.2 in Eq.1
𝑊=𝑚 𝑣 2 2 − 𝑣
𝑊= 𝐾 2 − 𝐾 1
𝑊=𝐹𝑠𝑐𝑜𝑠0
𝑊=𝐹𝑠
𝑊= 1 2 𝑚 𝑣 2 2 − 1 2 𝑚 𝑣 1 2
𝑊= 1 2 𝑚 𝑣 2 2 − 1 2 𝑚 𝑣 1 2
But: 𝐹=𝑚𝑎
500𝑁 20𝑚 = 𝑘𝑔 𝑣 2 2 −0
𝑊=𝑚𝑎𝑠
Eq.1
Def’n of Kinetic energy:
𝐾= 1 2 𝑚 𝑣 2
From Kinematics: 𝑣 𝑓 2 = 𝑣 𝑖 2 +2𝑎𝑠
𝑣 2 = 2(500𝑁) 20𝑚 15𝑘𝑔 =36.51 𝑚 𝑠
𝑣 2 2 = 𝑣 𝑎𝑠
𝑊= 𝐾 2 − 𝐾 1
This is Work-Energy Theorem
𝑎𝑠= 𝑣 2 2 − 𝑣
𝑊=∆𝐾
Eq.2

14. For Gravitational Forces only For the effect of other Forces
𝑊 𝑡𝑜𝑡 =𝑊 𝑔 + 𝑊 𝑜𝑡ℎ𝑒𝑟
𝑊 𝑔 =−∆𝑈
Eq.1
Work Done by gravity
𝑊=∆𝐾
Work-Energy Theorem
𝑊 𝑔 + 𝑊 𝑜𝑡ℎ𝑒𝑟 = 𝑊 𝑡𝑜𝑡
For gravitational forces only
−∆𝑈+ 𝑊 𝑜𝑡ℎ𝑒𝑟 =∆𝐾
𝑊 𝑔 =∆𝐾
Eq.2
− 𝑈 2 − 𝑈 1 + 𝑊 𝑜𝑡ℎ𝑒𝑟 = 𝐾 2 − 𝐾 1 m
Substitute Eq.2 in Eq.1
− 𝑈 2 + 𝑈 1 + 𝑊 𝑜𝑡ℎ𝑒𝑟 = 𝐾 2 − 𝐾 1
∆𝐾=−∆𝑈
𝐾 1 + 𝑈 1 + 𝑊 𝑜𝑡ℎ𝑒𝑟 = 𝐾 2 + 𝑈 2
𝐾 2 − 𝐾 1 =− 𝑈 2 − 𝑈 1
1 2 𝑚 𝑣 𝑚𝑔 𝑦 1 + 𝑊 𝑜𝑡ℎ𝑒𝑟 = 1 2 𝑚 𝑣 𝑚𝑔 𝑦 2
𝐾 2 − 𝐾 1 =− 𝑈 2 + 𝑈 1
𝐾 1 + 𝑈 1 = 𝐾 2 + 𝑈 2
The law of conservation of mechanical energy
But: 𝐸=𝐾+𝑈
𝐸 1 = 1 2 𝑚 𝑣 𝑚𝑔 𝑦 1 + 𝑊 𝑜𝑡ℎ𝑒𝑟
𝑬 𝟏 = 𝑬 𝟐
𝐸 2 = 1 2 𝑚 𝑣 𝑚𝑔 𝑦 2
This equation means that total energy 𝐸 is constant through out the travel.
Therefore the total initial energy is equal to the total final energy, hence total energy is constant.
𝑬 𝟏 = 𝑬 𝟐

15. Calculate the final velocity of the box.
1 2 𝑚 𝑣 𝑚𝑔 𝑦 1 + 𝑊 𝑜𝑡ℎ𝑒𝑟 = 1 2 𝑚 𝑣 𝑚𝑔 𝑦 2
Application problem
𝑣 1 =0
30kg
Assume the bottom of the plane be our ref. point, hence 𝑦=0 at this point.
𝜃= 40 𝑜
𝐿=20𝑚
𝜇 𝑘 =0.20
0+𝑚𝑔 𝑦 1 + 𝑊 𝑜𝑡ℎ𝑒𝑟 = 1 2 𝑚 𝑣
Work Done by the other 𝑊 𝑜𝑡ℎ𝑒𝑟 is only friction
0+𝑚𝑔 𝑦 1 +(𝑓𝐿𝑐𝑜𝑠 180 𝑜 )= 1 2 𝑚 𝑣
𝑣 2 =?
0+𝑚𝑔 𝑦 1 −𝑓𝐿= 1 2 𝑚 𝑣
0+𝑚𝑔 𝑦 1 − 𝜇 𝑘 𝑚𝑔𝑐𝑜𝑠𝜃∙𝐿= 1 2 𝑚 𝑣
Calculate the final velocity of the box.
0+𝑚𝑔 𝐿𝑠𝑖𝑛𝜃 − 𝜇 𝑘 𝑚𝑔𝑐𝑜𝑠𝜃∙𝐿= 1 2 𝑚 𝑣
𝑬 𝟏 = 𝑬 𝟐
𝑔 𝐿𝑠𝑖𝑛𝜃 − 𝜇 𝑘 𝑔𝑐𝑜𝑠𝜃∙𝐿= 𝑣 2 2
𝐾 1 + 𝑈 1 + 𝑊 𝑜𝑡ℎ𝑒𝑟 = 𝐾 2 + 𝑈 2
𝑣 2 = 2𝑔𝐿 𝑠𝑖𝑛𝜃− 𝜇 𝑘 𝑐𝑜𝑠𝜃
𝑣 2 =13.86 𝑚 𝑠

16. Summary
Work (𝑊) can be done if their is force applied to the object and able to move the object.
The product of the displacement and the component of the force in the direction of the displacement. 𝑊=𝐹𝑠𝑐𝑜𝑠𝜃
The unit of Work: Joule (𝐽), erg, foot-poundal
Work is scalar physical quantity. 𝑊= 𝐹 ∙ 𝑠
Energy (𝐸) is the ability to do work, hence Energy have the same unit with Work.
Two kinds of mechanical energy: The Potential (𝑈) and Kinetic (𝐾) Energy.
Definition of gravitational potential energy. 𝑈=𝑚𝑔y
Definition of elastic potential energy. 𝑈 𝑒𝑙 = 1 2 𝑘 𝑥 2
Definition of kinetic energy. 𝐾= 1 2 𝑚 𝑣 2
Work done by gravity 𝑊 𝑔 =−∆𝑈=− 𝑈 2 − 𝑈 1 =− 𝑚𝑔 𝑦 2 −𝑚𝑔 𝑦 1
Work done by elastic 𝑊 𝑒𝑙 =− ∆𝑈 𝑒𝑙 =− 𝑈 𝑒𝑙 2 − 𝑈 𝑒𝑙 1 =− 𝑘 𝑥 2 2 − 𝑘 𝑥 1 2
Work-Energy theorem 𝑊=∆𝐾
The total energy at any point of travel is constant 𝐸 1 = 𝐸 2 = 𝐸 3 = 𝐸 𝑛 …
The law of conservation of mechanical energy 𝑚 𝑣 𝑚𝑔 𝑦 1 + 𝑊 𝑜𝑡ℎ𝑒𝑟 = 1 2 𝑚 𝑣 𝑚𝑔 𝑦 2

17. PROBLEM

18. ASSIGNMENT #1

19. ASSIGNMENT #2

20. ASSIGNMENT #3
A 2.00kg package is released on a 53.1 incline, 4.00 m from a long spring with force constant 1.20x102 N/m that is attached at the bottom of the incline (Fig). The coefficients of friction between the package and incline are µs = and µk = The mass of the spring is negligible. (a) What is the maximum compression of the spring? (b) The package rebounds up the incline. How close does it get to its original position? (c) What is the change in the internal energy of the package and incline from the point at which the package is released until it rebounds to its maximum height?

21. Power What is Power? The rate of doing work.
It is the amount of energy consumed per unit time.
Having no direction, it is a scalar quantity.
In SI system, the unit of power is the joule per second (J/s), known as the watt in honor of James Watt, the eighteenth-century developer of the steam engine.
𝑃 𝑎𝑣 = 𝑊 ∆𝑡
Where:
𝑃 𝑎𝑣 = average power, in (watt) or in 𝐽 𝑠
𝑊 = quantity of work or work done, in (𝐽)
∆𝑡 = time interval, in (𝑠)
𝑃 𝑎𝑣 = 𝐹∆𝑠 ∆𝑡
𝑃 𝑎𝑣 =𝐹∙ ∆𝑠 ∆𝑡
1 𝐻𝑝=746 𝑤𝑎𝑡𝑡𝑠
1 𝐻𝑝=550 𝑓𝑡−𝑙𝑏 𝑠 = 2545 Btu/h
𝑃 𝑎𝑣 =𝐹∙𝑣
Note: Horse power (𝐻𝑝) is another unit of power

22. Power
Problem:
A 50kg marathon runner runs up the stairs to the top of of Chicago’s 443-m-tall Willis Tower, the tallest building in the United States (Figure, right). To lift herself to the top in 15 minutes, what must be her average power output? Express your answer in watts, in kilowatts, and in horse power.
𝑊=𝑚𝑔ℎ=50𝑘𝑔 𝑚 𝑠 𝑚
= 217, 𝐽
Solutions:
Another method: Solve the velocity, then multiply by the required force.
Time to reach the top is 15 min
ℎ=443𝑚
∆𝑡=15𝑚𝑖𝑛 60 𝑠 1 𝑚𝑖𝑛 =900 𝑠
𝑣= ∆𝑠 ∆𝑡 = 443 𝑚 900 𝑠 = 𝑚/𝑠
𝑃 𝑎𝑣 = 217, 𝐽 900 𝑠 = 𝑊
𝑃 𝑎𝑣 = 𝑊 1𝑘𝑊 1000 𝑤𝑎𝑡𝑡𝑠 =0.241 𝑘𝑊
𝑃 𝑎𝑣 =𝐹∙𝑣=𝑚𝑔∙𝑣
𝑃 𝑎𝑣 =50𝑘𝑔(9.81 𝑚 𝑠 2 )( 𝑚 𝑠 )
= 𝑊
𝑤=𝑚𝑔
𝑃 𝑎𝑣 = 𝑊 1 𝐻𝑝 746𝑊 =0.324 𝐻𝑝

23. Power
Assignment 1: A 20 Hp engine is used to lift gravel from the ground to the top of a building 60 ft high. Neglecting loss of energy due to friction how many tons of gravel can be lifted in 50 minutes.
Assignment 2: What weights can a 6 Hp engine pulls along a level road at 15 mi/hr if the coefficient of friction between the weight and the road is 0.2?
Assignment 3: A 20.0-kg rock is sliding on a rough, horizontal surface at 8.00 m/s and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is What average power is produced by friction as the rock stops?
Assignment 4: When its 75-kW (100-hp) engine is generating full power, a small single-engine airplane with mass 700 kg gains altitude at a rate of 2.5 m/s (150 m/min, or 500 ft/min). What fraction of the engine power is being used to make the airplane climb? (The remainder is used to overcome the effects of air resistance and of inefficiencies in the propeller and engine.)

24. Problems
1. A force of 8 𝑙𝑏𝑠. pulls a body along a horizontal surface to a distance of 10 𝑓𝑡. a) How much work is done, b) If the force acts at an angle of 30o above the horizontal, how much work is done?
2. A 100𝑔 object is dragged with a uniform velocity along a plane inclined 30o with the horizontal by a force parallel to the inclined. If the coefficient friction between the object and the plane is 0.20, how much work is done when the object is moved a distance of 40 𝑐𝑚 along the plane?
3. A man weighing 120 𝑙𝑏𝑠, climb up a stairway inclined 45𝑜 consisting of 20 𝑠𝑡𝑒𝑝𝑠 each step 6" high. What is his potential energy at the top.
4. A body a mass 10 𝑠𝑙𝑢𝑔𝑠 is thrown with a velocity of 6 𝑓𝑡/𝑠. along a horizontal floor. The coefficient of friction between the body and the floor 𝑖𝑠 Find a) Kinetics Energy and velocity of the body after travelling a distance of 2 𝑓𝑡. b) How far will the body travel before it comes to rest.

25. Problems
5. A body weighing 64 𝑙𝑏 slides down from rest at the top of a plane 18 𝑓𝑡 long and inclined 30 𝑜 above the horizontal. The coefficient of friction is Find the velocity of the body as it reaches the bottom of the plane.
6. What weights can a 6 𝐻𝑝 engine pulls along a level road at 15 𝑚𝑖/ℎ𝑟 if the coefficient of friction between the weight and the road is 0.20?

26. Summary
Work (𝑊) can be done if their is force applied to the object and able to move the object.
The product of the displacement and the component of the force in the direction of the displacement. 𝑊=𝐹𝑠𝑐𝑜𝑠𝜃
The unit of Work: Joule (𝐽), erg, foot-poundal
Work is scalar physical quantity. 𝑊= 𝐹 ∙ 𝑠
Energy (𝐸) is the ability to do work, hence Energy have the same unit with Work.
Two kinds of mechanical energy: The Potential (𝑈) and Kinetic (𝐾) Energy.
Definition of gravitational potential energy. 𝑈 𝑔 =𝑚𝑔y
Definition of kinetic energy. 𝐾= 1 2 𝑚 𝑣 2
Work done by gravity 𝑊 𝑔 =−∆𝑈
Work-Energy theorem 𝑊=∆𝐾
The total energy at any point of travel is constant 𝐸 1 = 𝐸 2 = 𝐸 3 = 𝐸 𝑛 …
The law of conservation of mechanical energy 𝑚 𝑣 𝑚𝑔 𝑦 1 + 𝑊 𝑜𝑡ℎ𝑒𝑟 = 1 2 𝑚 𝑣 𝑚𝑔 𝑦 2
Average power: The rate of doing work. It is the amount of energy consumed per unit time. 𝑃 𝑎𝑣 = ∆𝑊 ∆𝑡

27. eNd