# ENERGY.pptx

Jun. 1, 2023
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### ENERGY.pptx

• 1. ENERGY  LAW OF CONSERVATION OF ENERGY  CONSERVATION OF MECHANICAL ENERGY (BY GRAVITATIONAL FORCE)
• 2. LAW OF CONSERVATION OF ENERGY The law of conservation of energy states that energy can neither be created nor be destroyed but can be transformed from one form to the other. The total energy before and after the transformation remains the same. Our bodies convert the chemical energy in our food into mechanical energy for us to do work. The chemical energy in battery gets converted into light energy in torches. The electrical energy gets converted into heat energy for bread toasts.
• 3. • Mechanical energy = Potential energy + Kinetic energy • Suppose a ball of mass ‘m’ is dropped from a height(H). • Let the velocity of ball at point A = v(A)=0 m/s. • The velocity of the ball at point B=v(B). • The velocity of the ball at ground level (C)=v(C). m v(A)=0 A C H CONSERVATION OF MECHANICAL ENERGY B
• 4. MECHANICAL ENERGY AT POINT ‘A’ • At present the ball is at point ‘A’ at a height H from the ground level. The ball is at rest in point ‘A’. • Since the ball is at rest the v=0. Thus, the KE is also equal to 0. • Therefore, the ME possessed by the ball at point ‘A’ is equal to the PE possessed by the ball at ‘A’ PEA=mgH KEA=0 (v=0) MEA=PEA+KEA MEA=mgH+0=mgH m v(A)=0 A C H
• 5. MECHANICAL ENERGY AT POINT ‘B’ • At present the ball is at point ‘B’ fallen from height H to have fallen through (H-h). • The ball at present has gained velocity[v(B)]. • Therefore, at point ‘B’ the ball possesses partial kinetic energy and partial potential energy. PEB=mgh KEB=1/2m [v(B)]2 v2-u2=2gh [v(B)]2-0(because it has fallen from rest)=2g(H-h) [v(B)]2=2g(H-h) KEB=1/2m[2g(H-h)] KEB=mgH-mgh MEB=PEB+KEB =mgh+mgH-mgh =mgH  MEB=mgH v(A)=0 A C H B [v(B)] H-h h m PN:The red colour denotes cancellation.
• 6. MECHANICAL ENERGY AT POINT ‘C’ • The ball is now at point ‘C’ and has the highest velocity i.e, v(C). • The ball is now at ground level which implies that there is no height due to which there is absence of PE. • Therefore, the ME possessed by the ball at ‘C’ is equal to the KE energy possessed by the ball at ‘C’. PEC=0 KEC=1/2m[v(C)]2 v2-u2=2gh [v(C)]2-0(fallen from rest)=2gH [v(C)]2=2gH KEC=1/2m(2gH) KEC=mgH MEC=PEC+KEC =0+mgH  MEC=mgH v(A)=0 A C H m v(C)
• 7. CONCLUSION • The sum of potential energy and kinetic energy is same at all points, which implies: PE+KE=constant mgh+1/2mv2=constant • The mechanical energy(ME) of the ball at all points [A,B&C]=mgH which implies that the mechanical energy is conserved.
• 8. SAMPLE PROBLEM Use the law of conservation of energy (assume no friction) to fill in the blanks at the various marked positions for a 1000-kg roller coaster car.Find the value of A,B,C,D,E,F,G and H only.
• 9. SOLUTIONS • A: h = 45.9 m • B: v = 0 m/s • C: KE = 250 000 J • D: h = 20.4 m • E: v = 22.4 m/s • F: KE = 450 000 J • G: PE = 0 J • H: v = 30 m/s