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Lecture 15

Lecture 15. Chapter 10 Define Potential Energy of a Hooke’s Law spring Use Energy diagrams Chapter 11 (Work) Define Work Employ the dot product. Goals:. Heads up: Exam 2 7:15 PM Thursday, Nov. 3 th. m. Energy for a Hooke’s Law spring.

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Lecture 15

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  1. Lecture 15 • Chapter 10 • Define Potential Energy of a Hooke’s Law spring • Use Energy diagrams • Chapter 11 (Work) • Define Work • Employ the dot product Goals: Heads up: Exam 2 7:15 PM Thursday, Nov. 3th

  2. m Energy for a Hooke’s Law spring • Associate ½ ku2 with the “potential energy” of the spring

  3. m Energy for a Hooke’s Law spring • Ideal Hooke’s Law springs are conservative so the mechanical energy is constant

  4. Emech K Energy U y Energy diagrams • In general: Ball falling Spring/Mass system Emech K Energy U 0 0 u = x - xeq

  5. 1 2 h 3 0 mass: m -x Energy (with spring & gravity) • Emech = constant (only conservative forces) • At 1:y1 = h ; v1y = 0At 2:y2 = 0 ; v2y= ?At 3:y3 = -x ; v3 = 0 • Em1 = Ug1 + Us1 + K1 = mgh + 0 + 0 • Em2 = Ug2 + Us2 + K2 = 0 + 0 + ½ mv2 • Em3 = Ug3 + Us3 + K3 = -mgx + ½ kx2 + 0 Given m, g, h & k, how much does the spring compress?

  6. 1 2 h 3 0 mass: m -x Energy (with spring & gravity) • Emech = constant (only conservative forces) • At 1:y1 = h ; v1y = 0At 2:y2 = 0 ; v2y= ?At 3:y3 = -x ; v3 = 0 • Em1 = Ug1 + Us1 + K1 = mgh + 0 + 0 • Em2 = Ug2 + Us2 + K2 = 0 + 0 + ½ mv2 • Em3 = Ug3 + Us3 + K3 = -mgx + ½ kx2 + 0 • Given m, g, h & k, how much does the spring compress? • Em1 = Em3 = mgh = -mgx + ½ kx2 Given m, g, h & k, how much does the spring compress?

  7. 1 2 h 3 0 mass: m -x Energy (with spring & gravity) • Emech = constant (only conservative forces) • At 1:y1 = h ; v1y = 0At 2:y2 = 0 ; v2y= ?At 3:y3 = -x ; v3 = 0 • Em1 = Ug1 + Us1 + K1 = mgh + 0 + 0 • Em2 = Ug2 + Us2 + K2 = 0 + 0 + ½ mv2 • Em3 = Ug3 + Us3 + K3 = -mgx + ½ kx2 + 0 • Given m, g, h & k, how much does the spring compress? • Em1 = Em3 = mgh = -mgx + ½ kx2 Solve ½ kx2 - mgx - mgh = 0 Given m, g, h & k, how much does the spring compress?

  8. Energy (with spring & gravity) 1 mass: m • When is the child’s speed greatest? (A) At y = h (top of jump) (B) Between h & 0 (C) At y = 0 (child first contacts spring) (D) Between 0 & -x (E) At y = -x (maximum spring compression) 2 h 3 0 -x

  9. Energy (with spring & gravity) 1 • When is the child’s speed greatest?(D) Between y2 & y3 • A: Calc. soln. Find v vs. spring displacement then maximize (i.e., take derivative and then set to zero) • B: Physics: As long as Fgravity > Fspring then speed is increasing Find whereFgravity- Fspring= 0 -mg = kxVmaxor xVmax = -mg / k So mgh = Ug23 + Us23 + K23 = mg (-mg/k) + ½ k(-mg/k)2 + ½ mv2  2gh = 2(-mg2/k) + mg2/k+ v2 2gh + mg2/k = vmax2 2 h 3 kx mg 0 -x

  10. Chapter 11: Energy & Work • Impulse (Force over a time) describes momentum transfer • Work (Force over a distance) describes energy transfer • Any single acting force which changes the potential or kinetic energy of a system is said to have done work W on that system DEsys = W W can be positive or negative depending on the direction of energy transfer • Net work reflects changes in the kinetic energy Wnet = DK This is called the “Net” Work-Kinetic Energy Theorem

  11. Work performs energy transfer • If only conservative forces present then work either increases or decreases the mechanical energy. • It is important to define what constitutes the system

  12. v Circular Motion • I swing a sling shot over my head. The tension in the rope keeps the shot moving at constant speed in a circle. • How much work is done after the ball makes one full revolution? (A) W > 0 Fc (B) W = 0 (C) W < 0 (D) need more info

  13. Examples of “Net” Work (Wnet) DK = Wnet • Pushing a box on a smooth floor with a constant force; there is an increase in the kinetic energy Examples of No “Net” Work DK = 0 = Wnet • Pushing a box on a rough floor at constant speed • Driving at constant speed in a horizontal circle • Holding a book at constant height This last statement reflects what we call the “system”

  14. Again a constantF along a line (now x-dir) • Recall eliminating t gives • Multiply by m/2 • And m ax = Fx (which is constant)

  15. F|| Constant force along x, y & z directions Then for each x y z component Fx Fy Fz there is work W = FxDx + FyDy + FzDz Notice that if  is the angle between F and : |F|cos  is the compoent of F parallel to Dr |F|cos  |Dr| = W = FxDx + FyDz + FzDz So here we introduce the “dot” product F Dr≡ |F|cos  |Dr| = FxDx + FyDz + FzDz= DK = Wnet A Parallel Force acting Over a Distance does Work F q Δr

  16. A q Ay Ax î Scalar Product (or Dot Product) • Useful for finding parallel components A î= Ax î  î = 1 î j = 0 • Calculation can be made in terms of components. A B = (Ax )(Bx) + (Ay )(By ) + (Az )(Bz ) Calculation also in terms of magnitudes and relative angles. A B≡ | A | | B | cosq You choose the way that works best for you!

  17. Scalar Product (or Dot Product) Compare: A B = (Ax )(Bx) + (Ay )(By ) + (Az )(Bz ) with A as force F, B as displacement Dr Notice if force is constant: F Dr = (Fx )(Dx) + (Fy )(Dz ) + (Fz )(Dz) FxDx +FyDy + FzDz = DK So here F Dr = DK = Wnet Parallel Force acting Over a Distance does Work

  18. Net Work: 1-D Example (constant force) • A force F= 10 Npushes a box across a frictionless floor for a distance x= 5 m. • Net Work is F x= 10 x 5 N m = 50 J • 1 Nm ≡ 1 Joule and this is a unit of energy • Work reflects energy transfer Finish Start F q = 0° x

  19. Recap Next time: Power (Energy / time) Start Chapter 12 Assignment: • HW7 due Tuesday, Nov. 1st • For Wednesday: Read Chapter 12, Sections 1-3, 5 • do not concern yourself with the integration process

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