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Ramp Velocity - Inquiry in motion Velocity measured as cart is at the end of the track height to bottom of track at front of cart 5o Height height X10 10o .09 15o .19 .27 .9 1.9 2.7 Velocity Velocity2 1.3 1.9 2.4 1.69 3.61 5.76 V2/2 .85 1.8 2.9 5o 10o .075 .75 1.1 1.21 .6 15o .17 1.7 1.8 3.24 1.6 .26 2.6 2.2 4.84 2.4 The relationship between height and velocity is 10 times height equals velocity squared divided by 2 10 X height = Velocity2 2 10 X mass X Height = Mass X Velocity2 2 Gravitational Potential Kinetic Energy Energy More accurately gravitational potential energy is 9.8 m/s2 X mass (kg) X height (m) Mechanical Energy Energy of moving or potentially moving macroscopic objects Kinetic energy energy of a moving object Kinetic energy = mass X 2 velocity2 K.E. = (1/2)mv2 Gravitational potential Energy (GPE) Potential energy due to a gravitational field GPE = Mass X acceleration of gravity X height GPE = mgh g = 9.8m/s2 Spring Potential Energy or Elastic Potential Energy Potential energy due to a compressed or extended spring or elastic material Problems from inquiry on 11/7 Angle Height(m) 5o 0.075 10o 0.17 o 15 0.26 Velocity(m/s) 1.1 1.8 2.2 What is the kinetic energy of the cart at 5o? m= 0.5kg v = 1.1m/s KE = (1/2)m v2 = (1/2) (0.5kg)(1.1m/s)2 = 0.30 kg m2/s2 Units for energy kg m2/s2 known as a joule (J) What was the potential energy of the cart at the top of the track at 15o? m = 0.5kg height = .26m g = 9.8m/s2 GPE = mgh = (0.5kg)(9.8m/s2)(.26m) = 1.3 kg m2/s2 Units for energy kg m2/s2 known as a joule (J) Units for both kinetic and gravitational potential energy are the same. Conservation of Energy A second important concept from the original data is concept of conservation of mechanical energy in an isolated system Isolated system - defined so that neither matter nor energy enter or leave the system 10 X mass X Height =(1/2) Mass X Velocity2 Gravitational Potential Kinetic Energy Energy This shows that for any given height, the potential energy at the top equals the kinetic energy at the bottom. Another Example: A 0.2-kg ball thrown in the air at 5m/s, how high does the ball go? Maximum height so maximum potential energy Maximum velocity at bottom so maximum kinetic energy KE =(1/2) mv2 =(1/2) (.2kg)(5m/s)2 = 2.5J Potential energy at top = Kinetic energy at bottom GPE = mgh 2.5J =(.2kg)(10m/s2)h 1.25m = h If it is caught at the same height, what would its maximum speed be? 5m/s if no energy is lost then kinetic energy is the same Example 3: A 5kg rock is on a cliff 40 m in the air. The rock would have (5kg) (10m/s2)(40m) = 2000J of potential energy. Conservation of energy means it would have 2000J of kinetic energy at the instant it hit the ground. "At the instant" is just before the ground can slow the object down You could then determine the velocity of the rock. KE = (1/2)mv2 2000J =(1/2) (5kg)v2 4000 = 5v2 800 = v2 take the square root of both sides 28.3m/s = v Example 4: A 0.05 kg toy rocket is launched using a spring with 2.5 J of elastic potential energy. What is the Kinetic Energy of the rocket at the instant it leaves the spring? EPE = KE = 2.5J What is the Gravitational potential energy at the highest point in the rockets flight? KE = GPE = 2.5J What is the maximum height of the rocket? GPE = mgh 2.5J = (0.05kg)(10m/s2)h 5m=h What was the velocity of the rocket at the instant it leaves the spring? KE = (1/2) mv2 2.5J = (1/2) (0.05kg)v2 100m2/s2 = v2 10m/s = v Energy transfers Cart springing up ramp Against block - spring potential energy After release - kinetic energy On the way up - loses K.E. and gains gravitational potential energy At top - gravitational potential energy Ball thrown into the air As it is launched - Kinetic Energy Traveling up - K.E. and G.P.E. Top - G.P.E. Falling down - G.P.E. and K.E. Before caught - K.E. Pendulum A At A - maximum GPE A - B - both KE and GPE B - maximum KE B - C both kE and GPE C - Maximum GPE C B Spring C A A B A is equilibrium point No energy to start B - spring potential energy A - Kinetic energy C - gravitational potential energy