1 / 28

Lecture 15

Lecture 15. Chapter 10 Employ conservation of energy principle Chapter 11 Employ the dot product Employ conservative and non-conservative forces Relate force to potential energy Use the concept of power (i.e., energy per time). Goals:. Assignment: HW7 due Tuesday, Nov. 2 nd

swansond
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 • Employ conservation of energy principle • Chapter 11 • Employ the dot product • Employ conservative and non-conservative forces • Relate force to potential energy • Use the concept of power (i.e., energy per time) Goals: Assignment: • HW7 due Tuesday, Nov. 2nd • For Wednesday: Read Chapter 12, Sections 1-3, 5 & 6 do not concern yourself with the integration process in regards to “center of mass” or “moment of inertia”

  2. Energy scaling question • I have a block a distance y above the floor (Ug≡ 0 when the block is on the floor). There is a spring attached to the block and, at y, the potential energy of the spring is equal to the gravitational energy (i.e., Ug = Us = mgy). I now raise the block to a distance 2y above the floor. The total potential energy of the system (spring and block) increases by a factor of A 2 B 3 C 4 D 5 E Can’t be determined with the information given

  3. 2 1 Mechanical Energy • Potential Energy (U) • Kinetic Energy (K) • If “conservative” forces (e.g, gravity, spring) then Emech = constant = K + U During  Uspring+K1+K2 = constant = Emech • Mechanical Energy conserved Before During After

  4. 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?

  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? • Em1 = Em3 = mgh = -mgx + ½ kx2 Solve ½ kx2 – mgx +mgh = 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 Solve ½ kx2 – mgx +mgh = 0 Given m, g, h & k, how much does the spring compress?

  7. Energy (with spring & gravity) 1 mass: m • When is the child’s speed greatest? (A) At y1 (top of jump) (B) Between y1 & y2 (C) At y2 (child first contacts spring) (D) Between y2 & y3 (E) At y3 (maximum spring compression) 2 h 3 0 -x

  8. Before During 2 1 After Inelastic Processes • If non-conservative” forces (e.g, deformation, friction) then Emech is NOT constant • After  K1 + K2 < Emech (before) • Accounting for this “loss” we introduce • Thermal Energy (Eth , new) where Esys = Emech + Eth = K + U + Eth

  9. 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

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

  11. 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

  12. 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 = 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” ( Dropping a book is more complicated because it involves changes in U and K, U is transferred to K. The answer depends on what we call the system. )

  13. Changes in K with a constant F along a linear path • If F is constant

  14. Finish Start q = 0° F 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 x

  15. Units: Force x Distance = Work Newton x [M][L] / [T]2 Meter = Joule [L][M][L]2 / [T]2 mks cgs Other BTU = 1054 J calorie = 4.184 J foot-lb = 1.356 J eV = 1.6x10-19 J Dyne-cm or erg = 10-7 J N-m or Joule

  16. Net Work: 1-D 2nd Example (constant force) • Net Work is F x= -10 x 5 N m = -50 J • Work again reflects energy transfer • A forceF= 10 Nis opposite the motion of a box across a frictionless floor for a distance x = 5 m. Finish Start q = 180° F x

  17. Work in 3D…. • x, y and z with constant F:

  18. Work: “2-D” Example (constant force) • An angled force, F= 10 N,pushes a box across a frictionless floor for a distance x= 5 m and y= 0 m • (Net) Work is Fxx= F cos(-45°) x = 50 x 0.71 Nm = 35 J • Work reflects energy transfer Finish Start F q = -45° Fx x

  19. 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!

  20. 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 Again: A Parallel Force acting Over a Distance does Work

  21. “Scalar or Dot Product” Definition of Work, The basics Ingredients:Force (F ), displacement ( r) Work, W, of a constant force F acts through a displacement  r: W = F· r(Work is a scalar) F  r  displacement If we know the angle the force makes with the path, the dot product gives usF cos qandDr If the path is curved at each point and

  22. v ExerciseWork in the presence of friction and non-contact forces • A box is pulled up a rough (m > 0) incline by a rope-pulley-weight arrangement as shown below. • How many forces (including non-contact ones) are doing work on the box ? • Of these which are positive and which are negative? • State the system (here, just the box) • Use a Free Body Diagram • Compare force and path • 2 • 3 • 4 • 5

  23. Work and Varying Forces (1D) Area = Fx Dx F is increasing Here W = F·r becomes dW = Fdx • Consider a varying force F(x) Fx x Dx Finish Start F F q = 0° Dx Work has units of energy and is a scalar!

  24. to vo m spring at an equilibrium position x F V=0 t m spring compressed Example: Work Kinetic-Energy Theorem with variable force • How much will the spring compress (i.e. x) to bring the box to a stop (i.e., v = 0 ) if the object is moving initially at a constant velocity (vo) on frictionless surface as shown below ? Notice that the spring force is opposite the displacement For the mass m, work is negative For the spring, work is positive

  25. Uf rf rf ò U = Uf - Ui = - W = -F•dr ri Ui ri Conservative Forces & Potential Energy • For any conservative force F we can define a potential energy function U in the following way: The work done by a conservative force is equal and opposite to the change in the potential energy function. • This can be written as: ò W=F·dr= - U

  26. Exercise: Work & Friction • Two blocks having mass m1 and m2 where m1 > m2. They are sliding on a frictionless floor and have the same kinetic energy when they encounter a long rough stretch (i.e. m > 0) which slows them down to a stop. • Which one will go farther before stopping? • Hint: How much work does friction do on each block ? (A)m1(B)m2(C)They will go the same distance m1 v1 v2 m2

  27. Conservative Forces and Potential Energy • So we can also describe work and changes in potential energy (for conservative forces) DU = - W • Recalling (if 1D) W = Fx Dx • Combining these two, DU = - Fx Dx • Letting small quantities go to infinitesimals, dU = - Fx dx • Or, Fx = -dU / dx

  28. Recap Next time: Ch. 11 Power Start Chapter 12 Assignment: • HW7 due Wednesday, March 10 • For Tuesday: Read Chapter 12, Sections 1-3, 5 & 6 do not concern yourself with the integration process

More Related