1 / 16

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.

africa
Télécharger la présentation

Work, Energy, and Energy Conservation

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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.

More Related