Chapter 7 Work and Kinetic Energy

Transcription

1 Chapter 7 Work and Kinetic Energy Which one costs energy? Question: (try it) How to throw a baseball to give it large speed? Answer: Apply large force across a large distance! Force exerted through a distance performs Force exerted through a distance performs mechanical work. 1

2 Units of Chapter 7 Work Done by a Constant Force Kinetic Energy The Work-Energy Theorem Work Done by a Variable Force (optional) Power Read Chapter 8, Potential energy before the next lecture. We will finish Chapter 8 in the next lecture. 2

3 7-1 Work Done by a Constant Force When the force is parallel to the displacement: Constant force in direction of motion does work W. (7-1) SI unit: newton-meter (N m) = Joule, J 1 J = 1 N.m If F= 15 N, distance = 2 m, W=30 J 3

4 7-1 Work Done by a Constant Force 1J Joule 1 J How much is that? 4

5 If the force is at an angle to the motion, it does the following work: (7-3) θ is the angle between force and motion direction. Pulling at θ= 20, F=15N, d=2 m W= Fd cos θ =15*2* cos 20 = 28 J 5

6 7-1 Work Done by a Constant Force The work can also be written as the dot product of the force and the displacement: θ is the angle between force and motion direction. 6

7 The work done may be positive, zero, or negative, depending on the angle between the force and the motion: Here for F and d we only use their sizes (absolute value). The sign of the work is determined only by the angle between that force and motion. 7

8 Special cases: When force is perpendicular p to motion direction, it does no work. Examples: Normal force is always perpendicular to surface. Tension of pendulum When force is Opposite to motion direction, cos(180)=-1. Examples: Kinetic friction force does negative work. W fk = f k d 8 Always!

9 If there is more than one force acting on an object, we can find the work done by each force and add them together to find the total work. (7-5) Total work: W total = W 1 + W 2 +W the sum of the work done by each forces. 9

10 Q: Is Work a scalar or a vector? Scalar! It only says how much energy is added d or used, (positive or negative), but doesn t indicate motion directions or force directions. When we add work to get total work, we add as positive and negative simple numbers. We don t add work as vector arrows. f k F pull If F Pull = f k ; a=0, v=constant W Pull = F Pull d ; W fk = f k d ; W total = 0 x If F >f a >0 If >0 ill increase If F Pull > f k ; a >0, If v >0, v will increase. W total = F Pull d f k d = (F Pull f k ) d = F net d 10

11 7-2 Kinetic Energy and the Work-Energy Theorem When positive work is done on an object, its speed increases; when negative work is done, its speed decreases. 11

12 7-2 Kinetic Energy and the Work-Energy Theorem After algebraic manipulations of the equations of motion, we find: The total work done to one object is always equal to the change of its ½mv 2 Therefore, we define the kinetic energy: (7-6) Kinetic energy has SI unit: J (same dimension as Work) 1 kg m 2 /s 2 = 1 (kg m/s 2 )m =1Nm =1 Joule 12

13 7-2 Kinetic Energy and the Work-Energy Theorem Work-Energy Theorem: The total work done on an object is equal to its change in kinetic energy. (7-7) It s true for ALL MOTIONS, no only for constant a motion!!! 13

14 m=1000kg, v 0 =0, µ k =0.2 φ=30 ο, d=0.5m,find d v f 1.Compute work for individual forces first. mg= 9800 (N) N=mgcosφ f k =µ k N=µ k mgcosφ f k = cos30=1697(ν) Problem solving strategy : (work problems) 1C t kf i di id lf fi t 2. Add all work together as scalar numbers. 3. Set equation If you know total work, you can solve v, if you know v, you can solve Work. W mg =mg d cos 60 = 9800*0.5* cos60=2450 J W fk = f k d = 1697*0.5= 849 J ; W N =0 ; W total = = 1601 J 14 K f = K i +W total ½mv f2 =W total =1601J v f =1.79m/s

15 W mg =mg d cos 60 =2450 J W fk = f k d = 849 J ; W total = ( 849) = 1601 J Work is a scalar. To get total t work done by all forces, add work kin Joules directly as simple numbers! You only need to worry about the angles between each force and actual motion when you calculate work done by each force. After the work is calculated. Add them as simple numbers, no worry about direction any more. 15

16 Spring Force: Hooke s Law F spring = - k x k :Hooke s constant (how strong a spring is) x : distance of stretch/compression Force direction: Always in the opposite direction of x Spring force always tries to recover its natural Length Attention: ti This k is not K, Spring constant is not Kinetic Energy. 16

17 7-3 Work Done by a Variable Force If the force is constant, we can interpret the work done graphically: 17

18 7-3 Work Done by a Variable Force If the force takes on several successive constant values: 18

19 7-3 Work Done by a Variable Force We can then approximate a continuously varying force by a succession of constant values. 19

20 7-3 Work Done by a Variable Force The force needed to stretch a spring an amount x is F = kx. Therefore, the work done in stretching the spring is (7-8) 20

21 7-4 Power Power is a measure of the rate at which work is done: (7-10) SI unit: J/s = watt, W 1 horsepower = 1 hp = 746 W 21

22 7-4 Power 22

23 7-4 Power If an object is moving at a constant speed in the face of friction, gravity, air resistance, and so forth, the power exerted by the driving force can be written: (7-13) 23

24 Summary of Chapter 7 If the force is constant and parallel to the displacement, work is force times distance If the force is not parallel to the displacement, The total t work is the work done by the net force: 24

25 Summary of Chapter 7 SI unit of work: the joule, J Total work is equal to the change in kinetic energy: where 25 Kinetic energy is either positive or 0, never negative.

26 Summary of Chapter 7 Work done by a spring force: Power is the rate at which work is done: SI unit of power: the watt, W 26

27 27

28 28