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3). work & energy (finished)
- 2. Work and Energy Principle of Conservation of Energy “Energy can neither be created nor destroyed in any process. It can be converted from one form to another or transferred from one body to another, but the total amount remains constant” Work ;the product of a force times the distance through which it acts 𝑊 = 𝐹𝑠 ***𝐹 and 𝑠 are both vectors, but 𝑊 is scalar Multiplication of vectors ; Dot product (scalar) ; Cross product (vector) Dot product Suppose 𝐴 = 𝑎 𝑥 𝑖 + 𝑎 𝑦 𝑗 + 𝑎 𝑧 𝑘 𝐵 = 𝑏 𝑥 𝑖 + 𝑏 𝑦 𝑗 + 𝑏 𝑧 𝑘 𝐴 ∙ 𝐵 = 𝑎 𝑥 𝑏 𝑥 + 𝑎 𝑦 𝑏 𝑦 + 𝑎 𝑧 𝑏 𝑧 𝐴 ∙ 𝐵 = 𝐴 𝐵 𝑐𝑜𝑠𝜃 𝐴 𝐵 𝜃 𝐴 ∙ 𝐵 = 𝐴 𝐵 𝑐𝑜𝑠𝜃 𝐹 𝜃 𝐹 𝑐𝑜𝑠𝜃 𝐹 𝑠𝑖𝑛𝜃 ∆𝑟 From the diagram, 𝑊 = 𝐹∆𝑟𝑐𝑜𝑠𝜃 𝑊 = 𝐹 ∆𝑟 𝑐𝑜𝑠𝜃 𝑊 = 𝐹 ∙ ∆𝑟 *** 𝑊 = 𝐹 ∙ ∆𝑟 is more accurate ***In reality, 𝐹 is not applied constantly and ∆𝑟 is not always a straight line Concept ***For a curve path, see the curve as the combination of increadibly tiny straight lines, 𝑑𝑟 4 types of Work i. 𝐹; constant magnitude and direction ii. 𝐹; inconstant magnitude and constant direction iii. 𝐹; constant magnitude and inconstant direction iv. 𝐹; inconstant magnitude and direction 𝑊 = 𝐹 ∙ ∆𝑟 𝑑𝑊 = 𝐹 ∙ 𝑑𝑟 𝑊𝑡𝑜𝑡𝑎𝑙 = 𝑑𝑊 = 𝐹 ∙ 𝑑𝑟
- 3. Uniform Force ;both magnitude and direction are constant 𝑊𝐴→𝐵 = 𝑑𝑊 𝐵 𝐴 = 𝐹 ∙ 𝐵 𝐴 𝑑𝑟 Since, 𝐹 is uniform, 𝑊𝐴→𝐵 = 𝐹 ∙ 𝑑𝑟 𝐵 𝐴 𝑊𝐴→𝐵 = 𝐹 ∙ (𝑟𝐵 − 𝑟𝐴) 𝑊𝐴→𝐵 = 𝐹 ∙ ∆𝑟𝐴→𝐵 ***𝑊 is independent on the path, but is depend on the displacement when 𝐹 is uniform. A uniform force is considered “Conservative Force” 𝑊 > 0; work done by the system 𝑊 < 0; work done on the system 𝑟𝐴 𝑥 𝑦 𝑜 𝑟𝐵 ∆𝑟 𝐵 𝐴 𝑊𝐴→𝐵 = 𝐹 ∙ ∆𝑟𝐴→𝐵 Non-uniform Force ;inconstant magnitude and constant direction Ex; Spring At (a), x=0, equilibrium point At (b), an external force (𝐹𝑒𝑥𝑡) pulls an object to the right, restoring force (𝐹𝑠) will be experienced by the object also ∴ 𝐹𝑒𝑥𝑡 = 𝐹𝑠 At (b),𝐹𝑒𝑥𝑡 = 𝑘𝑥1 At (c),𝐹𝑒𝑥𝑡 = 𝑘𝑥2 𝑑𝑊 = 𝐹 ∙ 𝑑𝑟 𝑊 𝑏 →(𝑐) = 𝑑𝑊 (𝑐) (𝑏) = 𝐹𝑒𝑥𝑡 ∙ (𝑐) 𝑏 𝑑𝑟 = 𝐹𝑒𝑥𝑡 (𝑐) 𝑏 𝑑𝑟 𝑐𝑜𝑠𝜃 = 𝑘𝑥 𝑐 𝑏 𝑑𝑥 (𝑐𝑜𝑠0°) m m m (𝑐) (𝑏) (𝑎) 𝑥1 𝑥 = 0 𝐹𝑒𝑥𝑡 𝐹𝑠 𝐹𝑠 = 𝑟𝑒𝑠𝑡𝑜𝑟𝑖𝑛𝑔 𝑓𝑜𝑟𝑐𝑒 𝑥2
- 4. 𝑊 𝑏 →(𝑐) = 𝑘𝑥 𝑐 𝑏 𝑑𝑥 (𝑐𝑜𝑠0°) 𝑊 𝑏 →(𝑐) = 1 2 𝑘𝑥2 2 − 1 2 𝑘𝑥1 2 Since, 1 2 𝑘𝑥2 is Elastic Potential Energy *** 𝑊𝐴→𝐵 = 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑏𝑦 𝐹𝑒𝑥𝑡 ***𝑊 𝐹𝑠 = 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑏𝑦𝐹𝑠 Non-uniform Force ;constant magnitude and inconstant direction Ex; Circular motion 𝑑𝑊 = 𝐹 ∙ 𝑑𝑟 = 𝐹 𝑑𝑟 𝑐𝑜𝑠𝜃 Since at every point, 𝜃 = 90° 𝑊 = 0 𝑊𝐴→𝐵 = 𝐸𝑃𝐸 𝐵 − 𝐸𝑃𝐸𝐴 𝑊 𝐹𝑒𝑥𝑡 = −𝑊 𝐹𝑠 𝑣0 𝑣0 𝐹𝐴 𝐹𝐵 𝐴 𝐵 Non-uniform Force ;inconstant magnitude and direction ***non-conservative force; work done by force depends on the paths taken by a particle Ex; a particle moves from A to B in AEB and ADCB paths. The equation of force in the field is 𝐹 = −𝑦𝑖 + 𝑥𝑗 Newton. Find the work done in the 2 paths. The external force acting on the particle need to be equal to the force of the field and has opposite direction to cancel out. ∴ 𝐹𝑒𝑥𝑡 = −𝐹 𝑊𝐴→𝐸→𝐵 = 𝐹𝑒𝑥𝑡 ∙ 𝑑𝑟 𝐵 𝐴 = −𝐹 ∙ 𝑑𝑟 𝐵 𝐴 = − −𝑦 𝑥 ∙ 𝑑𝑥 𝑑𝑦 𝐵 𝐴 = 𝑦𝑑𝑥 𝐵 𝐴 − 𝑥𝑑𝑦 𝐵 𝐴 = 𝑦𝑑𝑥 𝑥=1 𝑥=1 − 1 𝑑𝑦 𝑦=−1 𝑦=1 = 0 − −1 − 1 = 2 𝐽𝑜𝑢𝑙𝑒𝑠 𝑥 𝑦 𝐴(1,1) 𝐷(2,1) 𝐵(1, −1) 𝐶(2, −1) 𝐸 0
- 5. 𝑊𝐴→𝐷→𝐶→𝐵 = 𝑊𝐴→𝐷 + 𝑊𝐷→𝐶 + 𝑊𝐶→𝐵 = 𝑦𝑑𝑥 𝐷 𝐴 − 𝑥𝑑𝑦 𝐷 𝐴 + 𝑦𝑑𝑥 𝐶 𝐷 − 𝑥𝑑𝑦 𝐶 𝐷 + 𝑦𝑑𝑥 𝐵 𝐶 − 𝑥𝑑𝑦 𝐵 𝐶 = (1)𝑑𝑥 𝑥=2 𝑥=1 − 𝑥𝑑𝑦 𝑦=1 𝑦=1 + 𝑦𝑑𝑥 𝑥=2 𝑥=2 − (2)𝑑𝑦 𝑦=−1 𝑦=1 + (−1)𝑑𝑥 𝑥=1 𝑥=2 − 𝑥𝑑𝑦 𝑦=−1 𝑦=−1 = (1)𝑑𝑥 𝑥=2 𝑥=1 − 0 + 0 − (2)𝑑𝑦 𝑦=−1 𝑦=1 + (−1)𝑑𝑥 𝑥=1 𝑥=2 − 0 = (1)𝑑𝑥 𝑥=2 𝑥=1 − 2 𝑑𝑦 𝑦=−1 𝑦=1 + (−1)𝑑𝑥 𝑥=1 𝑥=2 = 2 − 1 − −2 − 2 + −1 − −2 = 1 + 4 + 1 = 6 Joules 𝑥 𝑦 𝐴(1,1) 𝐷(2,1) 𝐵(1, −1) 𝐶(2, −1) 𝐸 0 Prove that 𝐹 = 2𝑥𝑖 + 3𝑦2 𝑗 is a conservative force. Let the initial and final point be 𝐴 and 𝐵 respectively 𝑊𝐴→𝐵 = 𝐹 ∙ 𝑑𝑟 𝐵 𝐴 = 2𝑥 3𝑦2 ∙ 𝑑𝑥 𝑑𝑦 𝐵 𝐴 = 2𝑥𝑑𝑥 𝐵 𝐴 + 3𝑦2 𝑑𝑦 𝐵 𝐴 = 𝑥2 + 𝑦3 𝐴 𝐵 Since, the value of 𝑊𝐴→𝐵 depends on only the values of 𝑥 and 𝑦 at 𝐴 and 𝐵. It does not depend on the other points on the path. Thus, 𝐹 is a conservative force. ***How to check whether 𝑭 is a conservative force? For 𝐹 = 𝐹𝑥 𝑖 + 𝐹𝑦 𝑗 + 𝐹𝑧 𝑘 𝐹 is a conservative force, when 𝜕𝐹𝑥 𝜕𝑦 = 𝜕𝐹𝑦 𝜕𝑥 And 𝜕𝐹𝑦 𝜕𝑧 = 𝜕𝐹𝑧 𝜕𝑦 And 𝜕𝐹𝑧 𝜕𝑥 = 𝜕𝐹𝑥 𝜕𝑧 Power ; average and instantaneous power 𝑃 = 𝑊 ∆𝑡 𝑃 = 𝑑𝑊 𝑑𝑡
- 6. 𝑃 = 𝑑𝑊 𝑑𝑡 𝑃 = 𝑑𝑊 𝑑𝑡 Since, 𝑊 = 𝐹 ∙ 𝑑𝑟 𝑃 = 𝑑( 𝐹 ∙ 𝑑𝑟) 𝑑𝑡 Assume that 𝐹 is uniform, 𝑃 = 𝐹 ∙ 𝑑𝑟 𝑑𝑡 = 𝐹 ∙ 𝑣 Energy ;the capacity to do work Mechanical Energy 2 types; Potential Energy & Kinetic Energy Potential Energy (i) Elastic Potential Energy, 𝑈𝑠 For spring, 𝑊 1 →(2) = 1 2 𝑘𝑥2 2 − 1 2 𝑘𝑥1 2 (ii) Gravitational Potential Energy, 𝑈𝑔 An object initially at 𝑦𝑎 is lifted up by 𝐹 𝑎𝑝𝑝 to 𝑦 𝑏. 𝑊 = 𝐹 𝑎𝑝𝑝 ∙ ∆𝑟 = 𝑚𝑔𝑗 ∙ [(𝑦 𝑏 − 𝑦𝑎)𝑗] 𝑊 = 𝑚𝑔𝑦 𝑏 − 𝑚𝑔𝑦𝑎 𝑊 = ∆𝑈𝑔 𝑈𝑔 = 𝐺𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 ***𝑈𝑔 depends on only vertical force(s). Kinetic Energy 𝑤 = 𝐹 ∙ 𝑑𝑥 𝑥 𝑓 𝑥 𝑖 𝑃 = 𝐹 ∙ 𝑣 ∆𝑟 𝑦𝑎 𝑦 𝑏 𝐹 𝑎𝑝𝑝 𝑚𝑔 0 𝑦 𝑥 𝑈𝑠 = 1 2 𝑘𝑥2 𝑈𝑔 = 𝑚𝑔𝑦 𝐹 ∆𝑥 𝑣𝑖 𝑣 𝑓
- 7. 𝑤 = 𝐹 ∙ 𝑑𝑥 𝑥 𝑓 𝑥 𝑖 𝑤 = 𝐹 𝑑𝑥 𝑐𝑜𝑠𝜃 𝑥 𝑓 𝑥 𝑖 Since, 𝜃 = 0° 𝑤 = 𝐹 𝑑𝑥 𝑥 𝑓 𝑥 𝑖 𝐹 𝑑𝑥 𝑥 𝑓 𝑥 𝑖 = 𝑚𝑔 𝑑𝑥 𝑥 𝑓 𝑥 𝑖 𝑚𝑔 𝑑𝑥 𝑥 𝑓 𝑥 𝑖 = 𝑚 𝑑𝑣 𝑑𝑡 𝑑𝑥 𝑥 𝑓 𝑥 𝑖 Using chain rule, 𝑚 𝑑𝑣 𝑑𝑡 𝑑𝑥 𝑥 𝑓 𝑥 𝑖 = 𝑚 𝑑𝑣 𝑑𝑥 𝑑𝑥 𝑑𝑡 𝑑𝑥 𝑥 𝑓 𝑥 𝑖 𝑚 𝑑𝑣 𝑑𝑥 𝑑𝑥 𝑑𝑡 𝑑𝑥 𝑥 𝑓 𝑥 𝑖 = 𝑚 𝑑𝑣 𝑑𝑥 𝑣 𝑑𝑥 𝑥 𝑓 𝑥 𝑖 ∴ 𝑊 = 𝑚𝑣 𝑑𝑣 𝑥 𝑓 𝑥 𝑖 = 1 2 𝑚𝑣𝑓 2 − 1 2 𝑚𝑣𝑖 2 ∴ 𝑊 = 𝐾𝐸𝑓 − 𝐾𝐸𝑖 = ∆𝐾𝐸 Work Energy Theorem 𝐾𝐸 = 1 2 𝑚𝑣2 𝑊 = 𝐾𝐸𝑓 − 𝐾𝐸𝑖 𝑊𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 𝑓𝑜𝑟𝑐𝑒 + 𝑊𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 𝑓𝑜𝑟𝑐𝑒 = 𝐾𝐸𝑓 − 𝐾𝐸𝑖 𝑊𝑐𝑜𝑛 + 𝑊𝑛𝑜𝑛−𝑐𝑜𝑛 = 𝐾𝐸𝑓 − 𝐾𝐸𝑖 𝑊𝑐𝑜𝑛, work due to conservative forces consist of the change in Elastic and Gravitational Potential Energy, ∆𝑈𝑠 + ∆𝑈𝑔 ***For Elastic PE, Work done by 𝐹𝑒𝑥𝑡 is equal to negative work done by 𝐹𝑠 (𝑘𝑥) ***For Gravitational PE, Work done by 𝐹𝑎𝑝𝑝 is equal to negative work done by 𝑚𝑔 Thus, − 𝑈𝑠 𝑓 − 𝑈𝑠 𝑖 + − 𝑈𝑔 𝑓 − 𝑈𝑔 𝑖 + 𝑊𝑛𝑜𝑛−𝑐𝑜𝑛 = 𝐾𝐸𝑓 − 𝐾𝐸𝑖 𝑊𝑛𝑜𝑛−𝑐𝑜𝑛 + 𝑈𝑠 𝑖 + 𝑈𝑔 𝑖 + 𝐾𝐸𝑖 = 𝑈𝑠 𝑓 + 𝑈𝑔 𝑓 + 𝐾𝐸𝑓 Since, 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 = 𝑃𝐸 + 𝐾𝐸 𝑊𝑛𝑜𝑛−𝑐𝑜𝑛 + 𝐸𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝐸𝑓𝑖𝑛𝑎𝑙 Law Of Conservation Of Energy 𝑊𝑛𝑜𝑛−𝑐𝑜𝑛 + 𝐸𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝐸𝑓𝑖𝑛𝑎𝑙
- 8. Collisions Law Of Conservation Of Momentum 𝑃𝑖 = 𝑃𝑓 ***The law is applicable for all collisions. Why? ; the Law is based on the 3rd law of Newton, reaction and action forces. According to the 3rd law of Newton, 𝐹12 = −𝐹21 𝐹12 + 𝐹21 = 0 According to the 2rd law of Newton, 𝑚1 𝑎1 + 𝑚2 𝑎2 = 0 𝑚1 𝑑𝑣1 𝑑𝑡 + 𝑚2 𝑑𝑣2 𝑑𝑡 = 0 𝑑𝑚1 𝑣1 𝑑𝑡 + 𝑑𝑚2 𝑣2 𝑑𝑡 = 0 𝑑(𝑚1 𝑣1 + 𝑚2 𝑣2) 𝑑𝑡 = 0 𝐹12 𝑚1 𝑚2 𝐹21 𝑣1 𝑖 𝑣2 𝑖 𝑣1 𝑓 𝑣2 𝑓 𝑑(𝑃1 + 𝑃2) 𝑑𝑡 = 0 ∴ 𝑃1 + 𝑃2 𝑖𝑠 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑃𝑖 = 𝑃𝑓 1 dimension elastic collision From, 𝑃𝑖 = 𝑃𝑓 and 𝐸𝑖 = 𝐸𝑓, we can derive another formula. or 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑎𝑝𝑝𝑟𝑜𝑎𝑐ℎ = 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛 ***The question will state whether the collision is elastic or inelastic. However, if the 2 masses stick together after the collision, it is inelastic collision. Elastic Inelastic 𝑃𝑖 = 𝑃𝑓 𝐸𝑖 = 𝐸𝑓 (Work due to non- conservative force = 0) 𝐸𝑖 ≠ 𝐸𝑓 (Work due to non- conservative force ≠ 0) 𝑣1𝑖 + 𝑣1 𝑓 = 𝑣2 𝑖 + 𝑣2 𝑓 𝑣1𝑖 − 𝑣2 𝑖 = 𝑣2 𝑓 − 𝑣1 𝑓
- 9. 2 dimension elastic collision 𝑃𝑖 = 𝑃𝑓 𝑚1 𝑣1 𝑖 + 𝑚2 𝑣2 𝑖 = 𝑚1 𝑣1 𝑓 + 𝑚2 𝑣2 𝑓 Where, 𝑣 = 𝑣 𝑥 𝑖 + 𝑣 𝑦 𝑗 For the other way, let’s consider 2 axis separately. x-axis; 𝑚1 𝑣1 𝑖𝑥 + 𝑚2 𝑣2 𝑖𝑥 = 𝑚1 𝑣1 𝑓𝑥 + 𝑚2 𝑣2 𝑓𝑥 y-axis; 𝑚1 𝑣1 𝑖𝑦 + 𝑚2 𝑣2 𝑖𝑦 = 𝑚1 𝑣1 𝑓𝑦 + 𝑚2 𝑣2 𝑓𝑦 𝑚1 𝑚2 𝑣1 𝑖 𝑣2 𝑖 𝑣1 𝑓 𝑣2 𝑓 𝜃 𝜙 𝑣1 𝑓 𝑐𝑜𝑠 𝜃 𝑣1 𝑓 𝑠𝑖𝑛 𝜃 𝑣2 𝑓 𝑐𝑜𝑠 𝜙 𝑣2 𝑓 𝑠𝑖𝑛 𝜙