Study of work done by a variable force. Overview of energy. Study of work done by a constant force. Understanding of energy conservation.

Transcription

1 Chap. 7: Work and Energy Overview of energy. Study of work done by a constant force as defined in physics. Relation between work and kinetic energy. Study of work done by a variable force. Study of potential energy. Understanding of energy conservation. Including time and study of the relationship of work to power. 1 Old Exam2 We studied motion (F = ma) in previous chapters We introduce energy as the next step. Work-Energy Theorem Energy Conservation 2

2 Old Exam2 FBD Work Energy Theorem Conservation of Energy Internal Energy 3 Work done by Force F May the Force be with you You will be trained to be Jedi Master: What is force? How to use force? 4

3 Work (W) Done by a Constant Force Energy Transfer from Initial State to Final State F.B.D. on sled [phys201 Q] Is the force effectively pulling a sled of firewood? F T F f = 3500 N F N d = 20 m F G = 14,700 N [A] Both magnitude (F) and directions ( ) must be taken into account: F cos = 5,000 x cos(36.9 o ) W = F T cos x (Distance) 5 Work (W) Done by a Constant Force Energy Transfer from Initial State to Final State F.B.D. on sled F N F T F f = 3500 N d = 20 m F G = 14,700 N 6

4 F f = 3500 N F N d = 20 m F G = 14,700 N [Q] What is this? 7 Work Energy Theorem Energy Transfer from Initial State to Final State [A] That is a sum of W tension and W friction (= W net ). Work-Energy Theorem W net = K f K i You measure Force(s) and Distance. You measure Velocities. 8

5 An Overview of Energy Energy is conserved. Energy can be transferred. Kinetic Energy describes motion and relates to the mass of the object and it s velocity squared. Energy on earth originates from the sun. Energy on earth is stored thermally and chemically. Chemical energy is released by metabolism. Energy is stored as potential energy in object (through elastic deformation or in height and mass). Energy can be dissipated as heat and noise. 9 Intentionally left with blank page 10

6 Looking back at Old Exam2 Work Energy Theorem Known: m, f, d, v 0 FBD Also see Conceptual Analysis 7.6 P.7-47, 7-54, 11 Find Velocity Using W-E Theorem W net = K f K i W net = W i = W grav = (mg)*(d) = (2 kg)(9.80 m/s 2 )(60.0 m) = 1176 J K f = (1/2)(2 kg)(v f2 ) K i = (1/2)(2 kg)(10.0 m/s) 2 = 100 J 1176 = (1/2)(2 kg) (v f2 ) 100 J v f = 35.7 m/s 12

7 Potential Energy from W-E Theorem W net = K f K i W net = W i = W grav = (mg)*(d) = (mg)*(y i y f ) = (mg)*(y i ) (mg)*(y f ) = U i U f 13 Conservation of K + U This means W g can also be calculated using Gravitational Potential Energy Function U(y) at y = y 1 relative to at y = y 2 Gravitational Potential Function The work done by the gravitational force depends only on the initial and final positions. Conservative Force W g < 0 if y 2 > y 1 W g > 0 if y 2 < y 1 14

8 Looking back at Old Exam2 15 Study of Energy Transformation This transformation begins as elastic potential energy in the elastomer. It then becomes kinetic energy as the projectile flies upward. During the upward flight, kinetic energy becomes potential until at the top of the flight, all the energy is potential. Finally, the stored potential energy changes back to kinetic energy as the projectile falls. 16

9 Intentionally left with blank page 17 Energy can be lost as heat. Energy can be dissipated by heat (motion transferred at the molecular level. This is referred to as dissipation. 18

10 Looking back at Old Exam2 19 Non-conservative Forces 20

11 (Non-)Conservative Force Gravitational al force as Conservative Force Vs. Friction force as Non-Conservative Force Start from rest, sliding down by the same height of 30.0 m. [Q1] No friction Which is the fastest? [Q2] k = 0.05 Which is the fastest? Goal A B C v A v B v C 0CAcQ_AUoAWoVChMIna3upq6oyAIVB4-ACh2uHwFn&biw=1093&bih=453#imgrc=lbA11AwEJceIpM%3A 21 Intentionally left with blank page 22

12 What was Work? Potential Energy Kinetics Energy Heat Work = Energy Transformation 23 Glossary 1. K: Energy associated with the motion of an object. 2. U: Energy stored in a system of objects Can either do work or be converted to K. 3. Q: Thermal Energy (Internal Energy) The energy of atoms and molecules that make up a body. 24

13 Summary In previous chapters we studied motion Sometimes force and motion are not enough to solve a problem. Energy was introduced. Work-Energy Theorem: W = K f - K i Energy Conservation: K f + U f = K i + U i Power = W/ t 25 Appendix 26

14 Example 1 A 50.0-kg crate is pulled 40.0 m by a constant force exerted (F P = 100 N and = 37.0 o ) by a person. Assume a coefficient of friction force k = Determine the work done by each force acting on the crate and its net work. Find the final velocity of the crate if d = 40 m and v i = 0 m/s. W net = W i = 1302 [J] (> 0) W net = K f K i = (1/2) m v f Example 1 Workbook 28

15 F.B.D. Example 1 Solution 180 o 50.0 kg d 90 o 29 Example 1 Solution F.B.D. 180 o Find F N and F f Use Newton s 2 nd Law: F P sin37 o + F N mg = 0 Thus F N = 430 N ; F f = 47.3 N d 90 o 30

16 F.B.D. Example 1 Solution 180 o W F P d cos ( 37 o P = 3195 [J] ) W F f d cos ( 180 o f = 1892 [J] (< 0) ) W m g d cos ( 90 o g = 0 [J] ) W F N d cos ( 90 o N = 0 [J] ) d 90 o NOTE: You have noticed W FN and W FG are zero. No Work -- What was a special condition to see this? 31 Example 1 Solution (cont d) W net = W i = 1302 [J] (> 0) W net = K f K i = (1/2) m v f 2 0 The body s speed increases. d 32

17 Example: Sliding on a Ramp This is an example of problem that can be solved by W- E theorem. DaD FBD 33 Special Case: No Work [Example] Work done on the bag by the person.. Special case: W = 0 J a) W P = F P d cos ( 90 o ) b) W g = m g d cos ( 90 o ) Nothing to do with the motion 34

18 Work done by Spring Force Depending on the course, work by a changing force is sometimes considered. In the case of spring force W S = ½ k x f 2 + ½ k x i 2 35 Example 2 A person pulls on the spring, stretching it 3.0 cm, which requires a maximum force of 75 N. How much work does the spring do? If, instead, the person compresses the spring 3.0 cm, how much work does the spring do? 36

19 Example 2 Workbook 37 Example 2 Solution (a) Find the spring constant k k = F max / x max = (b) Then, the work done by the spring is W S = ½ k x f 2 + ½ k x i 2 = (c) x f = m W S = 38

20 Example 2 Solution (a) Find the spring constant k k = F max / x max = (75 N) / (0.030 m) = 2.5 x 10 3 N/m (b) Then, the work done by the spring is W S = ½ k x f 2 + ½ k x i 2 = ½ k (+0.03) 2 + ½ k (0.00) 2 = 1.1 J (c) x f = m W S = 1.1 J 39 Example 3 You are weighing 600 N on a bathroom scale containing a stiff spring. In equilibrium the spring is 1.0 cm under your weight. Find the spring constant and the work done by the spring. 40

21 Example 3 Workbook Example 2 (a) Find the spring constant k k = F max / x max = (b) Then, find the work done by the spring (W S ) W S = ½ k x f 2 + ½ k x i 2 41 Example 3 Workbook 42

22 Example 3 Solution (a) Find the spring constant k k = F max / x max = (600 N) / (0.010 m) = 6.0 x 10 4 N/m (b) Then, the work done by the spring is W S = ½ k x f 2 + ½ k x i 2 = J x f = m ; x i = 0 43 Example 4 (1) F.B.D. (2) W by each force (4) W-E Theorem to find v 2. (3) W net 44

23 Example 4 Workbook 45 Example 4 Solution (1) F.B.D. (2) W by each force F N F P F f F G 46

24 (1) F.B.D. Example 4 Solution (2) W by each force F N W P = F P d cos( ) W N = F N d cos( ) F f F G F P F N =? W f = F f d cos( ) W G = F G d cos( ) 47 (1) F.B.D. Example 4 Solution (2) W by each force W P = F P d cos(30 o ) W N = F N d cos(90 o ) F N F f F G F P F N = F G cos(30 o ) + F P sin(30 o ) W f = F f d cos(180 o ) W G = F G d cos(120 o ) 48

25 (1) F.B.D. Example 4 Solution (2) W by each force F N W P = 650 J W N = 0 J F P F f F G F N = F G cos(30 o ) + F P sin(30 o ) W f = 122 J W G = 490 J 49 Example 4 Solution (cont d) (1) F.B.D. (2) W by each force (3) W net ( ) J (4) W-E Theorem to find v 2 : W net = K 2 K 1 ( ) m/s 50

26 Example 4 Solution (cont d) (1) F.B.D. (2) W by each force (3) W net 38 J (4) W-E Theorem to find v 2 : W net = K 2 K m/s 51 Example 4 Wrong Solution (1) F.B.D. (2) W by each force F N W P = F P d /cos(30 o ) W N = F N d F P or F P /cos(30 o ) F f F N = F G cos(30 o ) or F G W f = F f d cos(30 o ) F G W G = F G d cos(90 o ) 52

27 Example 4 Wrong Solution (4) W-E Theorem to find v 2 : 53 A Example 5 A roller coaster sliding without friction along a circular vertical loop (radius R) is to remain on the track at all times. Find the minimum release height h. C B Energy Conservation 54

28 Example 5: Analysis 0 A roller coaster sliding without friction along a circular vertical loop (radius R) is to remain on the track at all times. Find A the minimum release height h. v C (2) (1) (3) B Energy Conservation 55 A Example 5: Analysis 1 A roller coaster sliding without friction along a circular vertical loop (radius R) is to remain on the track at all times. Find the minimum release height h. v C (2) Rotation (1) Energy Conservation (3) F.B.D. B Energy Conservation 56

29 Example 5: Analysis 2 A roller coaster sliding without friction along a circular vertical loop (radius R) is to remain on the track at all times. Find A (1) U A = U C + K C the minimum release height h. v C (2) F = m (v 2 /R) mg F N = 0 B Energy Conservation 57 A Example 5: Analysis 3 A roller coaster sliding without friction along a circular vertical loop (radius R) is to remain on the track at all times. Find (1) mgh = mg(2r) + ½ m v 2 the minimum release height h. v C (2) g = v 2 /R mg F N = 0 Plug in B Energy Conservation 58