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RK Patrol. Mechanical Energy and its types: DISCOVERED What is meant by Conservation of Mechanical Energy: ANSWERED THINGS TO REMEMBER when solving conservation.

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Presentation on theme: "RK Patrol. Mechanical Energy and its types: DISCOVERED What is meant by Conservation of Mechanical Energy: ANSWERED THINGS TO REMEMBER when solving conservation."— Presentation transcript:

1 RK Patrol

2 Mechanical Energy and its types: DISCOVERED What is meant by Conservation of Mechanical Energy: ANSWERED THINGS TO REMEMBER when solving conservation of Energy problems Formulas to be IMPLANTED on mind and EXAMPLES of Energy Transfers Problem SOLVED The HOT, BURNING News TODAY:

3 Mechanical Energy is the sum of KINETIC and POTENTIAL ENERGY. 3 TYPES: 1.Kinetic Energy a.k.a. KE 2.Gravitational Potential Energy a.k.a PEg 3.Elastic Potential Energy a.k.a. PEs Mechanical Energy and its types: DISCOVERED

4 What is meant by Conservation of Mechanical Energy: ANSWERED

5 THINGS TO REMEMBER when solving conservation of Energy problems

6 KEi + PEi = KEf + PEf PE = mgh KE = ½ mv2 Formulas to be IMPLANTED on mind and EXAMPLES of Energy Transfers

7 KEi + PEi = KEf + PEf PE = mgh KE = ½ mv2 Formulas to be IMPLANTED on mind and EXAMPLES of Energy Transfers

8 KEi + PEi = KEf + PEf PE = mgh KE = ½ mv2 Formulas to be IMPLANTED on mind and EXAMPLES of Energy Transfers

9 KEi + PEi = KEf + PEf PE = mgh KE = ½ mv2 Formulas to be IMPLANTED on mind and EXAMPLES of Energy Transfers

10 Problem: A large chunk of ice with mass 15.0 kg falls from a roof 8.00 m above e the ground a. Find the kinetic energy of the ice when it reaches the ground b. What is the speed of the ice when it reaches the ground? c. Is the answer the same as you would determine by solving as a constant acceleration problem? Problem SOLVED

11 Given: m = 15.0 kg g = 9.80 m/s2 h = 8.00 m KE i = 0 PE f = 0 Problem SOLVED Unknowns: a. Ke f b.V f Basic Equations: KE i + PE i = KE f + PE f Solution: a.KE i + PE i = KE f + PE f 0 + mgh = KE f + 0 KE f = mgh = (15.0 kg) (9.80 m/s2) (8.00 m) = 1.18 x 10^3 J

12 b. KE f = ½ mvf2 v f 2 = 2KE f = 2(1.18 x 10^3 J) =157 m2/s2 m15.0 kg v f = 12.5 m/s c. yes Problem SOLVED

13 THE END

14 Questions to Answer

15


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