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Energy and Energy Conservation Energy Energy is the ability to do work. Two types of Energy: 1. Kinetic Energy (KE) - energy of an object due to its motion 2. Potential Energy (PE) - energy associated with an object due to the position of the object. Kinetic Energy Kinetic energy depends on the speed and the mass of the object. KE = ½ mv² Sample Problem What is the kinetic energy of a 0.15 kg baseball moving at a speed of 38.8 m/s? KE = ½ mv² KE = (½)(0.15 kg)(38.8m/s)² KE = 113 J Work-Kinetic Energy Theorem The net work done on an object is equal to the change in kinetic energy of an object. Wnet = ΔKE Wnet = ½mvf ² - ½mvi² Sample Problem A 50 kg sled is being pulled horizontally across an icy surface. After being pulled 15 m starting from rest, it’s speed is 4.0 m/s. What is the net force acting on the sled? Vi = 0 m/s Wnet = ½mvf ² - ½mvi² ∑Fd = ½mvf ² Vf = 4.0 m/s ∑F ≈ 27 N ∑F = (½mvf ²)/d = [(½)(50kg)(4.0m/s)²]/ 15 m Potential Energy Potential energy (PE) is often referred to as stored energy. Gravitational potential energy (PEg) depends on the height (h) of the object relative to the ground. PEg= mgh Sample Problem What is the gravitational potential energy of a 0.25 kg water balloon at a height of 12.0 m? PEg= mgh PEg= (0.25 kg)(9.81 m/s²)(12.0 m) PEg= 29.4 J Potential Energy Elastic potential energy (PEelastic) is the potential energy in a stretched or compressed elastic object. PEelastic = ½ kx² “X” is referred to as the distance of the spring compressed or stretched. “K” is the spring constant and is expressed in N/m. Sample Problem Calculate the elastic potential energy of a block spring, with a spring constant of 2.3 N/m, that has a compressed length of 0.15 m and a maximum stretch length of 0.55 m? x = 0.55 m – 0.15 m = 0.40 m PEelastic = ½ kx² = ½ (2.3 N/m)(0.40 m)² PEelastic = 0.18 J Conservation of Mechanical Energy Mechanical energy (ME) is the sum of kinetic and all forms of potential energy. ME = KE +∑PE Law of conservation of energy: Energy is neither created or destroyed. It simply changes form. Total mechanical energy remains constant in the absence of friction. 100 % PE 0 % KE 50 % PE h 50 % KE 0 % PE 100 % KE Sample Problem Starting from rest, a child zooms down a frictionless slide from an initial height of 3.00 m. What is the child’s speed at the bottom of the slide? The child’s mass is 25.0 kg. m = 25.0 kg hi = 3.00 m vi = 0 m/s hf = 0 m vf = ? m/s m = 25.0 kg hi = 3.00 m vi = 0 m/s hf = 0 m vf = ? m/s ½ mvi² + mghi = ½ mvf² +mghf (25.0 kg) (9.81 m/s²) (3.00 m) = (½)(25.0 kg) (Vf)² 736 J / (12.5 kg) = Vf ² Vf ² = 58.9 m²/s² Vf = 7.67 m/s Mechanical Energy in the presence of friction In the presence of friction, measured energy values at start and end points will differ. KE f KE Fapp KE Total energy, however, will remain conserved. KE Work Work Any force that causes a displacement on an object does work (W) on that object. ΣF d W=∑Fd(cos θ) Work Work is done only when components of a force are parallel to a displacement. F θ ΣF d W=∑Fd(cos θ) Work is expressed in Newton • meters (N•m) = Joules (J) Sample Problem How much work is done on a box pulled 3.0 m by a force of 50.0 N at an angle of 30.0° above the horizontal? 50.0 N 30.0° ΣF d W=∑Fd(cos θ)= (50.0 N x 3.0 m)(cos 30.0°) W = 130 J Efficiency Efficiency is a measure of how much of the work put into a machine is changed into useful work by the machine. Efficiency = (Wout/Win) x 100 % Sample Problem A man expends 200 J of work to move a box up an inclined plane. The amount of work produced is 40 J. What is the efficiency of the inclined plane? Efficiency = (Wout/Win) x 100 % = (40 J/ 200 J) x 100 = 20 % Momentum and Impulse Momentum and Impulse Momentum is a measure on how difficult it is to stop a moving object. Momentum is a vector quantity. p = mν Measured in kg • m/s Objects with a high momentum can have a greater mass, velocity, or both! 1 2 ν1 = ν 2 m1 > m2 Falling Object 1 Falling Object 2 ν1 > ν 2 m1 = m2 A change in momentum takes force and time. FΔt = Δp This product of force and the time over which it acts on an object is known as an impulse (FΔt). FΔt = mvf – mvi Impulse-Momentum Theorem p1 p2 Wall exerts an impulse on the moving ball, thereby causing a change in momentum Sample Problem A 1400 kg car moving westward with a velocity of 15 m/s collides with a pole and is brought to rest in 0.30 s. What is the magnitude of the force exerted on the car during the collision? (Pg. 211) FΔt = mνf – mνi F (0.30 s) = (1400 kg)(0 m/s) – (1400 kg)(- 15 m/s) F (0.30 s) = 21,000 kg • m/s F = 7.0 x 104 N to the East Δt = 0.30 s m = 1400 kg νf = 0 m/s νi = -15 m/s