1 / 19

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.

dasan
Download Presentation

Lecture 15

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. 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

More Related