1. Physics 207: Lecture 15, Pg 1 Lecture 15 Goals: Chapter 10 Chapter 10  Employ conservation of energy principle Chapter 11 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) Assignment: l HW7 due Tuesday, Nov. 2 nd l 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. Physics 207: Lecture 15, Pg 2 Energy scaling question l I have a block a distance y above the floor (U g ≡ 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., U g = U s = 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. Physics 207: Lecture 15, Pg 3 Mechanical Energy l Potential Energy (U) l Kinetic Energy (K) l If “conservative” forces (e.g, gravity, spring) then E mech = constant = K + U During  U spring +K 1 +K 2 = constant = E mech l Mechanical Energy conserved Before During After 1 2

4. Physics 207: Lecture 15, Pg 4 Energy (with spring & gravity) l E mech = constant (only conservative forces) l At 1: y 1 = h ; v 1y = 0 At 2: y 2 = 0 ; v 2y = ? At 3: y 3 = -x ; v 3 = 0 l E m1 = U g1 + U s1 + K 1 = mgh + 0 + 0 l E m2 = U g2 + U s2 + K 2 = 0 + 0 + ½ mv 2 l E m3 = U g3 + U s3 + K 3 = -mgx + ½ kx 2 + 0 1 3 2 h 0 -x mass: m Given m, g, h & k, how much does the spring compress?

5. Physics 207: Lecture 15, Pg 5 Energy (with spring & gravity) l E mech = constant (only conservative forces) l At 1: y 1 = h ; v 1y = 0 At 2: y 2 = 0 ; v 2y = ? At 3: y 3 = -x ; v 3 = 0 l E m1 = U g1 + U s1 + K 1 = mgh + 0 + 0 l E m2 = U g2 + U s2 + K 2 = 0 + 0 + ½ mv 2 l E m3 = U g3 + U s3 + K 3 = -mgx + ½ kx 2 + 0 l Given m, g, h & k, how much does the spring compress? l E m1 = E m3 = mgh = -mgx + ½ kx 2  Solve ½ kx 2 – mgx +mgh = 0 1 3 2 h 0 -x mass: m Given m, g, h & k, how much does the spring compress?

6. Physics 207: Lecture 15, Pg 6 Energy (with spring & gravity) l E mech = constant (only conservative forces) l At 1: y 1 = h ; v 1y = 0 At 2: y 2 = 0 ; v 2y = ? At 3: y 3 = -x ; v 3 = 0 l E m1 = U g1 + U s1 + K 1 = mgh + 0 + 0 l E m2 = U g2 + U s2 + K 2 = 0 + 0 + ½ mv 2 l E m3 = U g3 + U s3 + K 3 = -mgx + ½ kx 2 + 0 l Given m, g, h & k, how much does the spring compress? l E m1 = E m3 = mgh = -mgx + ½ kx 2  Solve ½ kx 2 – mgx +mgh = 0 1 3 2 h 0 -x mass: m Given m, g, h & k, how much does the spring compress?

7. Physics 207: Lecture 15, Pg 7 Energy (with spring & gravity) l When is the child’s speed greatest? (A) At y 1 (top of jump) (B) Between y 1 & y 2 (C) At y 2 (child first contacts spring) (D) Between y 2 & y 3 (E) At y 3 (maximum spring compression) 1 3 2 h 0 -x mass: m

8. Physics 207: Lecture 15, Pg 8 Inelastic Processes l If non-conservative” forces (e.g, deformation, friction) then E mech is NOT constant l After  K 1 + K 2 < E mech (before) l Accounting for this “loss” we introduce l Thermal Energy (E th, new) where E sys = E mech + E th = K + U + E th Before During After 1 2

9. Physics 207: Lecture 15, Pg 9 Chapter 11: Energy & Work l Impulse (Force over a time) describes momentum transfer l Work (Force over a distance) describes energy transfer l Any single acting force which changes the potential or kinetic energy of a system is said to have done work W on that system  E sys = W W can be positive or negative depending on the direction of energy transfer l Net work reflects changes in the kinetic energy W net =  K This is called the “Net” Work-Kinetic Energy Theorem

10. Physics 207: Lecture 15, Pg 10 Work performs energy transfer l If only conservative force present then work either increases or decreases the mechanical energy. l It is important to define what constitutes the system

11. Physics 207: Lecture 15, Pg 11 Circular Motion l I swing a sling shot over my head. The tension in the rope keeps the shot moving at constant speed in a circle. l How much work is done after the ball makes one full revolution? v (A) W > 0 (B) W = 0 (C) W < 0 (D) need more info FcFc

12. Physics 207: Lecture 15, Pg 12 Examples of “Net” Work(W net ) Examples of No “Net” Work  K = W net l Pushing a box on a rough floor at constant speed l Driving at constant speed in a horizontal circle l 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. )   K = W net l Pushing a box on a smooth floor with a constant force; there is an increase in the kinetic energy

13. Physics 207: Lecture 15, Pg 13 Changes in K with a constant F along a linear path l If F is constant

14. Physics 207: Lecture 15, Pg 14 Net Work: 1-D Example (constant force) = 10 x 5 N m = 50 J Net Work is F  x = 10 x 5 N m = 50 J l 1 Nm ≡ 1 Joule and this is a unit of energy l Work reflects energy transfer xx A force F = 10 N pushes a box across a frictionless floor for a distance  x = 5 m. F  = 0° Start Finish

15. Physics 207: Lecture 15, Pg 15 Units: N-m or Joule Dyne-cm or erg = 10 -7 J BTU= 1054 J calorie= 4.184 J foot-lb= 1.356 J eV= 1.6x10 -19 J cgsOthermks Force x Distance = Work Newton x [M][L] / [T] 2 Meter = Joule [L] [M][L] 2 / [T] 2

16. Physics 207: Lecture 15, Pg 16 Net Work: 1-D 2 nd Example (constant force) = -10 x 5 N m = -50 J Net Work is F  x = -10 x 5 N m = -50 J l Work again reflects energy transfer xx F A force F = 10 N is opposite the motion of a box across a frictionless floor for a distance  x = 5 m.  = 180° Start Finish

17. Physics 207: Lecture 15, Pg 17 Work in 3D…. l x, y and z with constant F:

18. Physics 207: Lecture 15, Pg 18 Work: “2-D” Example (constant force) = F cos( = 50 x 0.71 Nm = 35 J ( Net) Work is F x  x = F cos( -45°)  x = 50 x 0.71 Nm = 35 J l Work reflects energy transfer xx F An angled force, F = 10 N, pushes a box across a frictionless floor for a distance  x = 5 m and  y = 0 m  = -45° Start Finish FxFx

19. Physics 207: Lecture 15, Pg 19 l Useful for finding parallel components A  î = A x î  î = 1 î  j = 0 î A AxAx AyAy  A  B = (A x )(B x ) + (A y )(B y ) + (A z )(B z ) l Calculation can be made in terms of components. Calculation also in terms of magnitudes and relative angles. Scalar Product (or Dot Product) A  B ≡ | A | | B | cos  You choose the way that works best for you!

20. Physics 207: Lecture 15, Pg 20 Scalar Product (or Dot Product) Compare: A  B = (A x )(B x ) + (A y )(B y ) + (A z )(B z ) with A as force F, B as displacement  r Notice if force is constant: F   r = (F x )(  x) + (F y )(  z ) + (F z )(  z) F x  x +F y  y + F z  z =  K So here F   r =  K = W net Again: A Parallel Force acting Over a Distance does Work

21. Physics 207: Lecture 15, Pg 21 Definition of Work, The basics Ingredients: Force ( F ), displacement (  r ) “Scalar or Dot Product”   r displacement F Work, W, of a constant force F acts through a displacement  r : (Work is a scalar) W = F ·  r (Work is a scalar) If we know the angle the force makes with the path, the dot product gives us F cos  and  r If the path is curvedat each point and

22. Physics 207: Lecture 15, Pg 22 Exercise Work in the presence of friction and non-contact forces A. 2 B. 3 C. 4 D. 5 A box is pulled up a rough (  > 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 v

23. Physics 207: Lecture 15, Pg 23 Work and Varying Forces (1D) l Consider a varying force F(x) FxFx x xx Area = F x  x F is increasing Here W = F ·  r becomes dW = F dx F  = 0° Start Finish Work has units of energy and is a scalar! F xx

24. Physics 207: Lecture 15, Pg 24 Work Kinetic-Energy Theorem: Net Work done = change in kinetic energy (final – initial)

25. Physics 207: Lecture 15, Pg 25 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 (v o ) on frictionless surface as shown below ? xx vovo m toto F spring compressed spring at an equilibrium position V=0 t m Notice that the spring force is opposite the displacement For the mass m, work is negative For the spring, work is positive

26. Physics 207: Lecture 15, Pg 26 Conservative Forces & Potential Energy l 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. l This can be written as: W = F ·dr = -  U   U = U f - U i = - W = - F dr  riri rfrf riri rfrf UfUf UiUi

27. Physics 207: Lecture 15, Pg 27 Exercise: Work & Friction Two blocks having mass m 1 and m 2 where m 1 > m 2. They are sliding on a frictionless floor and have the same kinetic energy when they encounter a long rough stretch (i.e.  > 0) which slows them down to a stop. l Which one will go farther before stopping? l Hint: How much work does friction do on each block ? (A) (B) (C) (A) m 1 (B) m 2 (C) They will go the same distance m1m1 m2m2 v2v2 v1v1

28. Physics 207: Lecture 15, Pg 28 Home Exercise Work Done by Gravity l An frictionless track is at an angle of 30° with respect to the horizontal. A cart (mass 1 kg) is released from rest. It slides 1 meter downwards along the track bounces and then slides upwards to its original position. l How much total work is done by gravity on the cart when it reaches its original position? (g = 10 m/s 2 ) 1 meter 30° (A) 5 J(B) 10 J(C) 20 J(D) 0 J h = 1 m sin 30° = 0.5 m

29. Physics 207: Lecture 15, Pg 29 Conservative Forces and Potential Energy l So we can also describe work and changes in potential energy (for conservative forces)  U = - W l Recalling (if 1D) W = F x  x l Combining these two,  U = - F x  x l Letting small quantities go to infinitesimals, dU = - F x dx l Or, F x = -dU / dx

30. Physics 207: Lecture 15, Pg 30 Recap Next time: Ch. 11 Ch. 11Power Start Chapter 12 Assignment: l HW7 due Wednesday, March 10 l For Tuesday: Read Chapter 12, Sections 1-3, 5 & 6 do not concern yourself with the integration process