# Law of Conservation of Mechanical Energy Law of Conservation of Mechanical Energy Law of Conservation of Mechanical Energy Law of Conservation of Mechanical

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Law of Conservation of Mechanical Energy Law of Conservation of Mechanical Energy Law of Conservation of Mechanical Energy Law of Conservation of Mechanical Energy LLLL aaaa wwww o o o o ffff C C C C oooo nnnn ssss eeee rrrr vvvv aaaa tttt iiii oooo nnnn o o o o ffff MMMM eeee cccc hhhh aaaa nnnn iiii cccc aaaa llll E E E E nnnn eeee rrrr gggg yyyy Slide 2 Principle of Conservation of Mechanical Energy Law of Conservation of Mechanical Energy Conservation of Mechanical Energy Equation Conservation of Mechanical Energy: Mathematical Problem Slide 3 Law of Conservation of Mechanical Energy Principle of Conservation of Mechanical Energy According to the law of conservation of mechanical energy, in an isolated system, that is, in the absence of non-conservative forces like friction, the initial total energy of the system equals to the total energy of the system. Simply stated, the total mechanical energy of a system is always constant (in case of absence of non- conservative forces). For instance, if a ball is rolled down a frictionless roller coaster, the initial and final energies remain constant. Conservative forces are those that don't depend on the path taken by an object. For example, gravity, spring and electrical forces are examples of mechanical energy. Slide 4 Slide 5 Law of Conservation of Mechanical Energy Conservation of Mechanical Energy Equation The quantitative relationship between work and energy is stated by the mechanical energy equation. U T = K i + P i + W ext = K f + P f, where, U T = Total mechanical energy K i = Initial kinetic energy K f = Final kinetic energy P i = Initial potential energy P f = Final potential energy W ext = External work done Slide 6 JoJaRiRheChas Munchkins Caraga Regional Science high School Surigao City Law of Conservation of Mechanical Energy This is a general equation for mechanical energy conservation. In case, there are some external or internal forces acting on the object, that is the forces are non-conservative like friction, air resistance, etc, then only W ext is considered. In absence of such forces, W ext = 0 and so the mechanical energy conservation equation takes the form: U T = K i + P i = K f + P f Slide 7 JoJaRiRheChas Munchkins Caraga Regional Science high School Surigao City Law of Conservation of Mechanical Energy Conservation of Mechanical Energy: Mathematical Problem Let us consider a mathematical problem that involves the use of law of conservation of mechanical energy in finding the values of unknown quantities. Question : A 20 g stone is put in a sling shot with a spring constant of 100 N/m and it is stretched back to 0.7 m. Determine the maximum velocity that the stone will acquire and the speed of stone when it is shot straight up? Slide 8 JoJaRiRheChas Munchkins Caraga Regional Science high School Surigao City Law of Conservation of Mechanical Energy Solution : In this problem, we ignore the air resistance and heat effects that are present while operating the sling shot. This makes external work done zero, that means we can easily apply the law of conservation of mechanical energy formula. Total energy in the beginning of the event E i = K i + Gravitational potential energy (mgh) + spring force ( kx 2 ). Here, K i = (0.5 mv 2 ) = (0.5)m (0) 2 = 0 (Since v = 0 initially) Gravitational potential energy = mg(0) = 0 (since h = 0 initially) Spring force = kx 2 = (0.5)(100)(0.7) 2 = 24.5 J = E i Once out of the sling shot, the stone gains some maximum velocity before it reaches some altitude. Slide 9 JoJaRiRheChas Munchkins Caraga Regional Science high School Surigao City Law of Conservation of Mechanical Energy E f = 0.5 mv 2 + mgh + kx 2 = (0.5)(0.02)(v) 2 + mg(0) + (0.5)k(0) 2 = 0.001v 2 Since E i = E f Therefore, 24.5 J = 0.001v 2 = 24,500 = v 2. Therefore, v = 156.1 m/s (approximate value) At the highest point, the velocity of stone is zero. Therefore, E f = 24.5 J = 0.5 mv 2 + mgh + kx 2 24.5 J = 0.5mv(0) 2 + mgh + 1/2k(0) 2 = 24.5 J = (0.02)(9.8 N/Kg)h = 125 m. Answer: Velocity attained = 156.1 m/s and height attained = 125 m Slide 10 ##### Conservation Area Law Terraces Conservation Area Appraisal 2.0 Conservation Area Context 4 2.1 Current
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