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Chapter 5

Work. Chapter 5. The product of the magnitudes of the component of a force along the direction of displacement and the displacement. Units-Force x Length N x m = Nm = J (Joules) W=Fd (force times displacement). Work. If an object is not moved, NO WORK is done.

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Chapter 5

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  1. Work Chapter 5

  2. The product of the magnitudes of the component of a force along the direction of displacement and the displacement. • Units-Force x Length • N x m = Nm = J (Joules) • W=Fd (force times displacement) Work

  3. If an object is not moved, NO WORK is done. If you hold a heavy object, your body has work being done inside, but no work is done on the object. For W = Fd, the force must be the component in the same direction as the displacement. Work

  4. You push on a van directly horizontal • W = (your force) x (distance moved horizontally) • You push down on a bumber at an angle • Find the force of the horizontal component • Use cos θ ; so W = Fd cosθ Example

  5. You pull a crate by attaching a rope at an angle of 25⁰. You then pull with 100 N, and it moves 10m. Calculate the work done? • Given: • F=100 N • Θ=25⁰ • d=10 m • Work: • W = Fd cosθ • W = (100 N)(10 m) cos(25⁰) • W=900 J • Homework: Practice 5A Page 162 #1-4 Sample Problem

  6. Problem #1 • Given: • F= 5.00 x 103N • d= 3.00Km=3000 m • Work: • W=Fd • W=(5.00 x 103N)(3000m) • W=1.50 x 107J Practice 5A

  7. Problem #2 • Given: • F= 350 N • d= 2.00m • Work: • W=Fd • W=(350N)(2.00 m) • W=7.0 x 102 J Practice 5A

  8. Problem #3 • Given: • F= 35 N • d= 50.0 m • Θ= 25⁰ • Work: • W=Fd cosθ • W=(35N)(50.0 m)cos(25⁰) • W=1.6 x 103J (1600 J) Practice 5A

  9. Not what you might think • Scalar, but can be + or – • Positive-component force in same direction as displacement. • Negative- If opposite (Kinetic Friction Force) • Cos θ is negative for angles greater than 90⁰ but less than 270⁰ • When speed is changed by work: • If sign is positive, increases • If sign is negative, decreases • Positive Work - work done ON the object • Negative Work -work done BY the object Sign of work

  10. Energy associated with an objects motion • Does not depend on direction; scalar • Depends on mass and speed • KE = (1/2) mv2 • m=mass • v=velocity • Units • Kg (m/s)2 • Kg m2/s2 = Joule (J) Kinetic Energy

  11. Stored energy; depends on properties of the object AND it position (reference to environment) • Two Types • Gravitational (PEg) • Elastic (PEelastic) • Total PE= PEg + PE (elastic) POTENTIAL eNERGY

  12. The energy associated with an object due to the object’s position relative to a gravitational source. PEg = mgh m=mass g=9.81m/s2 h=height (In measuring height, choose arbitrary zero level.) Units=Joules (J) Note: Gravity and Free-fall acceleration are not properties of the object Gravitational

  13. Potential energy in a stretched or compressed elastic object. PE (elastic) = (1/2) Kx2 k-spring constant, specific to the elastic object unit N/m x-distance stretched/compressed from resting Unit-N/m Elastic

  14. A 700kg stuntman is attached to a bungee cord with an unstretched length of 15.0m. He jumps off a bridge from a height of 50.0m. When he finally stops, the cord has a stretched length of 44.0m. Disregard weight of cord K = 71.8 N/m. What is total PE relative to the water when the man stops falling? Sample Problem

  15. Given: • m=70.0 Kg • h=50.0 m – 44.0 m = 6.0m • x=44.00 m - 15.0 m = 29.0 m • PE = 0 at River Level • Work: • PEg = mgh = (70.0)(9.81)(6.0) = 4100 J • PE(elastic) = (1/2)kx 2= (1/2)(71.8)(29.0)2 = 30,200J • PE total = PEg + PE (elastic) • PE = 34, 300 J Sample Problem

  16. Mechanical Energy-the sum of kinetic energy and all forms of potential energy • ME = KE + PE • In the absence of friction, ME remains constant; CONSERVATION of MECHANICAL ENERGY • MEi = MEf (no friction) • 1. If only force is gravity: • 1/2mvi2 + mghi = 1/2mvf2 + mghf • 2. What if you had gravity and a spring? Conservation of Energy

  17. Sample ProblemAir hockey puck on incline, no friction. What velocity does it slide with at the bottom? • PE = mgh • hi = 1.1 m • hf = 0 m • m= 2.0Kg • g = 9.81 m/s2 • KE = ½ mv2 • Vf= ? • Vi = 0

  18. PEgi + KEi = PEgf +KEf mghi + (1/2)mv2 = mghf + (1/2)mvf2 (2.0)(9.81)(1.1) + 0 = 0 + (1/2)(2.0)(Vf)2 Vf = √(21.56) Vf = 4.7 m/s Sample Problem (cont.)

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