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Work, Energy, and Energy Conservation

Work, Energy, and Energy Conservation. Chapter 5, Sections 1 - 3 Pg. 168-186. Σ F. d. Work. Any force that causes a displacement on an object does work (W) on that object. W= ∑ Fd. d. Work. Work is done only when components of a force are parallel to a displacement. F. θ. Σ F.

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Work, Energy, and Energy Conservation

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  1. Work, Energy, and Energy Conservation Chapter 5, Sections 1 - 3 Pg. 168-186

  2. ΣF d Work Any force that causes a displacement on an object doeswork (W)on that object. W=∑Fd

  3. d Work Work is done only when components of a force are parallel to a displacement. F θ ΣF W=∑Fd(cos θ) Work is expressed in Newton • meters (N•m) = Joules (J)

  4. Sample Problem d 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 W=∑Fd(cos θ) = (50.0 N x 3.0 m)(cos 30.0°) W = 130 J

  5. 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.

  6. Kinetic Energy Kinetic energy depends on the speed and the mass of the object. KE = ½ mv²

  7. 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

  8. 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 = ½mvf² - ½mvi²

  9. 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 Vf = 4.0 m/s Wnet = ½mvf² - ½mvi² ∑F≈ 27 N ∑Fd = ½mvf² ∑F = (½mvf²)/d = [(½)(50kg)(4.0m/s)²]/ 15 m

  10. 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

  11. 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

  12. 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 simplychanges form.

  13. h Total mechanical energy remains constant in the absence of friction. 100 % PE 0 % KE 50 % PE 50 % KE 0 % PE 100 % KE

  14. 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

  15. 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 = 7.67 m/s Vf ² = 58.9 m²/s²

  16. Mechanical Energy in the presence of friction KE KE f Fapp KE KE In the presence of friction, measured energy values at start and end points will differ. Total energy, however, will remain conserved.

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