Chapter 4 work, energy and power
- 1. CHAPTER 4 WORK, ENERGY AND POWER DBS 1012 ENGINEERING SCIENCE 1
- 2. SUMMARY • This topic explains about the relationship between work, energy and power. Students will exposed to energy changes and mechanical efficiency. 2
- 3. LEARNING OUTCOME: 4.1 Understand the concept of work, energy and power 4.1.1Define work, energy and power 4.1.2 calculate form of energy by using formulas: a) Kinetic Energy, 𝐸 𝑘 = 1 2 m𝑣2 b) Potential Energy, 𝐸 𝑝=𝑚𝑔ℎ 4.1.3 State Principle of Conservation of energy 4.1.4 Describe conversion of energy from one form of energy to another. 4.1.5 Apply the concept and formula of work, energy and power in solving the related problems’ 4.1.6 Calculate the efficiency of mechanical system. Efficiency = 𝑃𝑜 𝑃 𝑖𝑛 × 100% 𝑎𝑛𝑑 Efficiency = 𝐸 𝑜 𝐸 𝑖𝑛 × 100% 3
- 4. 4.1.1 DEFINE WORK, ENERGY AND POWER • Work done is the product of an applied force and the displacement of an object in the direction of the applied force. 𝑾 = 𝑭 × 𝒔 F = force ( Newton) s = displacement (m) Unit work = J @ Nm s 𝐹 𝐹 4 Figure 1 : force and the displacement in the same direction
- 5. 5 4.1.1 DEFINE WORK, ENERGY AND POWER s 𝐹 𝚹 𝚹 𝐹 𝑭𝒄𝒐𝒔 𝚹 𝑭𝒄𝒐𝒔 𝚹 Figure 2 : force and the displacement in different direction In this situation, we use: 𝑾 = 𝑭𝒔𝒄𝒐𝒔 𝜭 F = force ( Newton) s = displacement (m) Unit work = J @ Nm
- 6. No work is done Object is stationary No force is applied Direction of motion is perpendicular to applied force 6 4.1.1 DEFINE WORK, ENERGY AND POWER
- 7. NO WORK IS DONE 7 Figure 3 : No work is done Figure 3(a) Pushing a wall Figure 3(b) a satellite orbiting in space
- 8. DEFINITION ENERGY, E • Energy is capacity to do work • Unit : Joule (J) 8 Form of energy Solar energy Electric energy Chemical energy Nuclear energy Gravitational potential energy Kinetic energy
- 9. GRAVITATIONAL POTENTIAL ENERGY, 𝐸 𝑝 • Definition : energy of an object due to its higher position in the gravitational field. 9 m = mass (kg) g = gravity acceleration =9.81m/𝑠2 h= height 𝑬 𝒑 = 𝒎𝒈𝒉 m Figure 4: Gravitational Potential Energy
- 10. KINETIC ENERGY, 𝐸 𝑘 • Definition : energy of an object due to its motion • Unit : Joule (J) 10 𝑣 𝑚 𝑬 𝒌 = 𝟏 𝟐 𝒎𝒗 𝟐 𝑚= mass (kg) 𝑣 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑚 𝑠 Figure 5: kinetic energy
- 11. DEFINITION POWER, P • Definition : the rate at which work is done or the amount of work done per second • Unit : J/s @ Watt 11 𝑷 = 𝑾 𝒕 = 𝒎𝒈𝒉 𝒕 = 𝑭 × 𝒗 𝑚= mass (kg) 𝑣 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑚 𝑠 𝑊 = 𝑤𝑜𝑟𝑘 𝐽 𝑡 = 𝑡𝑖𝑚𝑒 𝑠 𝐹 = 𝑓𝑜𝑟𝑐𝑒 𝑁 𝑔 = 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 9.81 𝑚/𝑠2
- 12. 4.1.2 CALCULATE FORM OF ENERGY USING FORMULA 𝐸 𝑝 AND 𝐸 𝑘 A car is moving with the velocity of 10 m/s and is having mass of 250 Kg. Calculate its Kinetic energy? 12 EXAMPLE 250 Kg 𝑣 = 10 𝑚/𝑠 Solution : The car posses kinetic energy so use formula 𝑬 𝒌 = 𝟏 𝟐 𝒎𝒗 𝟐 Then, 𝑬 𝒌 = 𝟏 𝟐 𝒎𝒗 𝟐 = 𝟏 𝟐 𝟐𝟓𝟎 𝟏𝟎 𝟐 = 𝟏𝟐𝟓𝟎𝟎 𝑱
- 13. A man is carrying a trolley of mass 6 kg and having Kinetic energy of 40 J. Calculate its velocity with which he is running? 13 4.1.2 CALCULATE FORM OF ENERGY USING FORMULA 𝐸 𝑝 AND 𝐸 𝑘 EXAMPLE 6 kg 𝑣 =? Solution : The man posses kinetic energy so use formula 𝑬 𝒌 = 𝟏 𝟐 𝒎𝒗 𝟐 Then, 𝑬 𝒌 = 𝟏 𝟐 𝒎𝒗 𝟐 𝟒𝟎 𝑱 = 𝟏 𝟐 𝟔 𝒗 𝟐 𝒗 𝟐 = 𝟒𝟎 ×𝟐 𝟔 = 𝟏𝟑. 𝟑𝟑 𝑣 = 13.33 = 3.65 𝑚 𝑠
- 14. 4.1.3 STATE PRINCIPLE OF CONSERVATION OF ENERGY PRINCIPLE OF CONSERVATION OF ENERGY Energy cannot be created or destroyed Energy can be transformed from one form to another Total energy in a system is constant 14
- 15. 4.1.4 DESCRIBE CONVERSION FROM ONE FORM OF ENERGY TO ANOTHER 15 ℎ 50 % 𝐸 𝑝 50 % 𝐸 𝑘 100 % 𝐸 𝑘 ½ of ℎ ℎ ℎ = 0 100 % 𝐸 𝑝 𝐸 𝑘 = 0 Figure 6(a): Conversion of Energy
- 16. 16 4.1.4 DESCRIBE CONVERSION FROM ONE FORM OF ENERGY TO ANOTHER 𝑬 𝒑 maximum 𝑬 𝒌 = 0 𝑬 𝒑 is transferred 𝒕𝒐 𝑬 𝒌 𝑬 𝒑 𝒊𝒔 increasing 𝑬 𝒌 maximum 𝑬 𝒑 = 0 𝑬 𝒌 is decreasing And 𝑬 𝒑 is gaining 𝑬 𝒑 minimum 𝑬 𝒌 maximum Figure 6(b): Conversion of Energy
- 17. 17 4.1.4 DESCRIBE CONVERSION FROM ONE FORM OF ENERGY TO ANOTHER Figure below shows a simplified route of a roller coaster. Initially, the cart runs down from the starting point . Neglecting the effects of friction, find its speed at (a) the "valley" V (b) at the top of second hill L (c) If friction is taken in consideration, will the answers be smaller or larger than that found in part (a) and (b)? EXAMPLE
- 18. 18 4.1.4 DESCRIBE CONVERSION FROM ONE FORM OF ENERGY TO ANOTHER Solution : a) 𝐸 𝑝 𝑎𝑡 𝐻 = 𝐸 𝑘 ( 𝑎𝑡 𝑉) 𝑚𝑔ℎ = 1 2 𝑚𝑣2 𝑣2 = 2𝑔ℎ 𝑣 = 2𝑔ℎ = 2 9.81𝑚 𝑠 20 = 19.81 𝑚 𝑠 b) 𝐸 𝑝 𝑎𝑡 𝐻 + 𝐸 𝐾 𝑎𝑡 𝐻 = 𝐸 𝑝 𝑎𝑡 𝐿 + 𝐸 𝐾 𝑎𝑡 𝐿 𝑚 9.81 20 + 0 = 𝑚(9.81)(10) + 1 2 𝑚𝑣2 196.2 𝑚 = 98.1 𝑚 + 1 2 𝑚𝑣2 1 2 𝑚𝑣2 = 196.2m − 98.1m = 98.1m 𝑚𝑣2 = 196.2𝑚 𝑣 = 196.2 = 14.0 𝑚 𝑠
- 19. 19 c) The results would be smaller than found in part (a) and (b), because some energy is lost by friction. 4.1.4 DESCRIBE CONVERSION FROM ONE FORM OF ENERGY TO ANOTHER
- 20. 4.1.5 APPLY THE CONCEPT AND FORMULA OF WORK, ENERGY AND POWER IN SOLVING THE RELATED PROBLEMS 20 A box is dragged across a floor by a 100N force directed 60o above the horizontal. How much work does the force do in pulling the object 8m? EXAMPLE Solution : 𝑊 = 𝐹𝑠 cos 𝜃 = 100 × 8 cos 60° = 400𝑁𝑚 = 400 𝐽
- 21. 21 EXAMPLE 4.1.5 APPLY THE CONCEPT AND FORMULA OF WORK, ENERGY AND POWER IN SOLVING THE RELATED PROBLEMS Calculate the work done by the weight lifter in lifting the weights? Solution: 𝐸 𝑝 = F × 𝑠 = mg = 25kg + 25 kg 9.81m 𝑠2 2m = 981 J
- 22. 22 A tired squirrel (mass of approximately 1 kg) does push-ups by applying a force to elevate its center-of-mass by 5 cm in order to do a mere 0.50 Joule of work. If the tired squirrel does all this work in 2 seconds, then determine its power. EXAMPLE Solution : 𝑃 = 𝑊 𝑡 = 0.50𝐽 2𝑠 = 0.25 𝑊 4.1.5 APPLY THE CONCEPT AND FORMULA OF WORK, ENERGY AND POWER IN SOLVING THE RELATED PROBLEMS
- 23. 23 When doing a chin-up, a physics student lifts her 42 kg body a distance of 0.25 meters in 2 seconds. What is the power delivered by the student's biceps ? EXAMPLE Solution : 𝑃 = 𝑊 𝑡 = 𝐹 × 𝑠 𝑡 = 𝑚𝑔 𝑠 𝑡 = (42)( 9.81𝑚 𝑠2 )(0.25𝑚) 2 = 51.5 𝑊 4.1.5 APPLY THE CONCEPT AND FORMULA OF WORK, ENERGY AND POWER IN SOLVING THE RELATED PROBLEMS
- 24. 24 A 300 kg car has a kinetic energy of 500 J. Find its speed. EXAMPLE Solution : 𝐸 𝑘 = 1 2 𝑚𝑣2 500 𝐽 = 1 2 300𝑘𝑔 𝑣2 𝑣2 = 500 ×2 300 = 3.33 𝑣 = 3.33 = 1.83 𝑚/𝑠 4.1.5 APPLY THE CONCEPT AND FORMULA OF WORK, ENERGY AND POWER IN SOLVING THE RELATED PROBLEMS
- 25. 25 A 800g ball is pulled up a slope as shown in the diagram. Calculate the potential energy it gains EXAMPLE 4.1.5 APPLY THE CONCEPT AND FORMULA OF WORK, ENERGY AND POWER IN SOLVING THE RELATED PROBLEMS Solution : 𝐸 𝑝 = 𝑚𝑔ℎ = 800 1000 9.81𝑚 𝑠2 20 100 = 1.5696 J Converts gram to kg Converts cm to meter
- 26. 4.1.6 CALCULATE THE EFFICIENCY OF MECHANICAL SYSTEM • The efficiency of an engine or machine is defined as: 26 Efficiency ,𝞰 = 𝑼𝒔𝒆𝒇𝒖𝒍 𝒆𝒏𝒆𝒓𝒈𝒚 𝒐𝒖𝒕𝒑𝒖𝒕 𝒘𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 𝒆𝒏𝒆𝒓𝒈𝒚 𝒊𝒏𝒑𝒖𝒕 × 𝟏𝟎𝟎% = 𝑬 𝟎 𝑬 𝒊𝒏 × 𝟏𝟎𝟎% @ Efficiency,𝞰 = 𝑼𝒔𝒆𝒇𝒖𝒍 𝒑𝒐𝒘𝒆𝒓 𝒑𝒐𝒘𝒆𝒓 𝒊𝒏𝒑𝒖𝒕 × 𝟏𝟎𝟎% = 𝑷 𝟎 𝑷 𝒊𝒏 × 𝟏𝟎𝟎%
- 27. 27 Find the efficiency of a rope and pulley system if a painter uses 1.93kJ of mechanical energy to pull on the rope and lift a 20kg paint barrel at constant speed to a height of 7.5m above the ground? 4.1.6 CALCULATE THE EFFICIENCY OF MECHANICAL SYSTEM Solution : 𝐸𝑖𝑛 = 1.93 𝑘𝐽 = 1.93 × 1000 = 1930 𝐽 𝐸 𝑜 = 𝑚𝑔ℎ = 20 × 9.81 × 7.5 = 1471.5 𝐽 Efficiency = 𝐸 𝑜 𝐸 𝑖𝑛 × 100 = 1471.5 1930 × 100 = 76.24% EXAMPLE
- 28. 28 EXAMPLE 4.1.6 CALCULATE THE EFFICIENCY OF MECHANICAL SYSTEM A heat engine gives out 400 J of heat energy as the useful work. Calculate the energy given to it as input if its efficiency is 40%? Solution : Efficiency= 𝑬 𝟎 𝑬 𝒊𝒏 × 𝟏𝟎𝟎% 𝟒𝟎% = 𝟒𝟎𝟎𝑱 𝑬 𝒊𝒏 × 𝟏𝟎𝟎% 𝑬𝒊𝒏 = 𝟒𝟎𝟎𝑱 𝟒𝟎% × 𝟏𝟎𝟎% = 𝟏𝟎𝟎𝟎 𝑱
- 29. EXERCISE 29 1) How much work is done by the boy pulling his sister 30.0 m in a wagon as shown in the above figure? Assume no friction acts on the wagon. Answer = 1299.038 J
- 30. 30 2) A cart is loaded with a brick and pulled at constant speed along an inclined plane to the height of a seat-top. If the mass of the loaded cart is 3.0 kg and the height of the seat top is 0.45 meters, then what is the potential energy of the loaded cart at the height of the seat-top? Answer : 𝐸 𝑝 = 13.2 J EXERCISE
- 31. 31 3) Determine the kinetic energy of a 625-kg roller coaster car that is moving with a speed of 18.3 m/s. EXERCISE Answer : 𝐸 𝑘 = 1.05 x105 Joules
- 32. REFERENCE • Keat,Y.E .(1996). Sukses Lengkap SPM fizik. Petaling Jaya: Pustaka Delta Pelajaran Sdn Bhd. • S.L Chang, K.K.(2011). Focus SPM Physics. Bangi: Penerbitan Pelangi Sdn Bhd. 32