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Types of Energy How many types/forms of energy can you list? Gravitational Potential Energy (PE) The work done to lift (or lower) something is equal to its change in gravitational potential energy PE = Fd = (mg)Δh = mg (h2 – h1) m h PE=mgh PE: in Joules h=0 (reference point) m: in kg h: in m Example: Find the potential energy of a 60.0 kg student sitting on a 80.0 cm high stool… a) relative to the floor in the room b) relative to the floor in the room 1 floor below (3.00 m lower) c) relative to the ceiling 2.50 m above the student Example: Find the potential energy of a 60.0 kg student sitting on a 80.0 cm high stool… a) relative to the floor in the room. b) relative to the floor in the room 1 floor below (3.0 m lower) c) relative to the ceiling 2.50 m above the student a) PE = mgh = 60.0 x 9.8 x 0.800 = 470 Joules b) PE = mgh = 60.0 x 9.8 x 3.8 = 2200 Joules c) PE = mgh = 60 x 9.8 x (-2.5) = -1470 Joules Kinetic Energy (KE) The energy of motion. Moving things have kinetic energy. Proportional to mass of the object and the square of its velocity. KE = ½ KE: Kinetic Energy (J) m: mass (kg) v: velocity (m/s) 2 mv Example 1 Which has more kinetic energy: a 1200 kg car moving at 20 km/hr or a transport truck (m=60 t) moving at 3 km/hr? Example 1. 2. How much energy does a 4500 kg car have if it is traveling at 35 km/hr? If that same car had 1 MJ of energy, how fast is it moving? 1. 2. How much energy does a 4500 kg car have if it is traveling at 35 km/hr? KE = ½ mv2 = 0.5 * 4500 * (35/3.6)2 = 212673 J = 2.1 x 105 J If that same car had 1 MJ of energy, how fast is it moving? KE = ½ mv2 1,000,000 = ½ * 4500 * v2 v2 = 444 m2/s2 v = 21.08 m/s = 76 km/hr To Do… Read pgs. 217-224 Do Ch. 11 P.P. (1-8) Skateboard Push How do you get a person to go from zero velocity to some non-zero velocity? You do some work! Energy & Work Consider an object that is accelerated from initial velocity, vo, to velocity, v, by a force, F (net force), over a distance, d. What is the work done? F = ma F·d = ma·d F·d = m(a·d) 2ad = v2 – vo2 F·d = m(v2 - vo2) 2 W = ½mv2 - ½mvo2 Kinetic Energy (energy of motion) KE = ½mv2 W = KE - KEo W = ΔKE (Work – Energy Theorem) Conservative (Internal) vs. NonConservative (External) Forces Conservative Forces Total Energy will be conserved Non-Conservative Forces Total Energy will not be conserved (amount of total energy will change) FGravity FSpring Force Fapplied Fair Ffriction Work – Energy Theorem W = ΔKE The work done on an object by the net force equals the change in kinetic energy of that object. Example: A 1000 kg car going 20 m/s brakes and comes to a stop. How much work is done? What happens to the energy? KE = ½mvo2 = ½(1000)(20)2 = 200000 J W = ΔKE = 0 – 200000 J = –200000 J (lost energy) Energy is dissipated as heat. Conservation of Mechanical Energy Total Mechanical Energy (TE): the sum of an object’s Potential Energy (PE) and Kinetic Energy (KE). TE = PE + KE TE = mgh + ½mv2 Law of Conservation of Energy: Within a closed (no external forces), isolated system, the total energy of an object is conserved. E1 = E2 Einitial = Efinal Example: 1. A pink soccer ball of mass 0.50 kg is kicked with an initial v velocity of 15 m/s. Find its… a) velocity when it is 4.0 m high. h b) height when it is going 5.0 m/s. 15m/s a) Einital = Efinal mgh1 + ½mv12 = mgh2 + ½mv22 0 + ½(0.5)(15)2 = (0.5)(9.8)(4.0) + ½(0.5)v2 56.3 = 19.6 + 0.25v2 v = 12 m/s b) 56.3 = (0.5)(9.8)h + ½(0.5)(5.0)2 h = 10 m 2. Tarzan, a 75 kg Ape-man, swings from a branch 3.0 m above the ground with an initial speed of 5.0 m/s. a) Find his velocity when he swings past ground level. b) Find the maximum height he swings to. c) In a separate jump, what should his initial speed be if he is to just reach a branch 4.0 m high? b) & c) 3.0 m a) h=0 Note: in the physics jungle there is no air resistance and the vines have no mass and are unstretchable! Note: The vine does NO WORK because it is always applying its force perpendicular to the direction of motion a) mgho + ½mvo2 = mgh + ½mv2 gho + ½vo2 = 0 + ½v2 m cancels out; h = 0 v = √(vo2 + 2gho) v = √((5.0)2 + 2(9.8)(3.0)) v = 9.2 m/s horizontal b) gho + ½vo2 = 0 + gh h = vo2 + ho 2g v=0 h = 4.3 m c) gho + ½vo2 = 0 + gh vo = √(2gh - 2gho) vo = 4.4 m/s v=0 2. Tarzan’s less famous brother, George, swings from the same 3.0 m-high branch with an initial speed of 3.5 m/s. Unfortunately, he hits a tree 1.5 m above the ground. Find his speed just before impact and the work done by the tree. His mass is 90.0 kg. 3.0 m 1.5 m mgho + ½mvo2 = mgh + ½mv2 m cancels out gho + ½vo2 = gh + ½v2 v = √(vo2 + 2gho – 2gh) v = √((3.5)2 + 2(9.8)(3.0 – 1.5)) v = 6.5 m/s W W W W = ΔKE = KE - KEo = ½mv2 – ½mvo2 = 0 – ½(90.0)(6.5)2 = -1900 J Work – Energy Theorem W = ΔTotal Energy The work done on an object by a net force equals the change in total energy of that object. W = ΔKE + ΔPE W = (KEf – KEo) + (PEf – PEo) Question: How much heat energy is produced when a “speed” skier loses 300.0 m in elevation if he starts from rest and finishes up going 40.0 m/s? His mass (including gear) is 90.0 kg. Heat = Work done by friction 0m/s Work = ΔPE + ΔKE W = (PEf – PEo) + (KEf – KEo) W = (mgh – mgho) + (½mv2 – 0) = (0 – (90)(9.8)(300)) + (½(90)(40)2 – 0) 300m W = -1.93 x 105 J 40m/s h=0 193000 J of heat are produced EXAMPLE: A 3.0 kg block slides down a frictionless 58o incline with a slant length of 4.0 m. a) Assuming no friction, what is the velocity at the bottom of the plane? b) Suppose the co-efficient of friction is 0.25. What is the velocity of the block at the bottom? EXAMPLE: A 3.0 kg block slides down a frictionless 58o incline with a slant length of 4.0 m. a) Assuming no friction, what is the velocity at the bottom of the plane? b) Suppose the co-efficient of friction is 0.25. What is the velocity of the block at the bottom? h = 4.0*sin58o h = 3.39 m PEtop = KEbottom mgh = ½ mv2 (3.0)(9.8)(3.39) = 0.5(3.0)v2 v = 8.1 m/s b) Work done by friction=FFd FF = μFN = μ mgcos58o =0.25(3.0)(9.8)cos58o =4.07 N W = (4.07)(-4.0) = -16.3 J (opposite direction of motion W = (KEf – KEo) + (PEf – PEo) -16.3 = 0.5(3.0)v2 – (3.0)(9.8)(3.39) v = 7.4 m/s Homework