Work and Energy. Chapter 7

Transcription

1 Work and Energy Chapter 7

2 Scalar Product of Two Vectors Definition of the scalar, or dot, product: A B A Alternatively, we can write: x B x A y B y A z B z

3 Work

4 Work Done by a Constant Force The work done by a constant force is defined as the distance moved times the component of the force in the direction of the displacement vector: In the SI system, the units of work are joules:

5 Work Done by a Constant Force Example: As long as this person does not lift or lower the bag of groceries, he is doing no work on it. The force he exerts has no component in the direction of motion. Remember, forces need to act over some distance to do work!

6 ConcepTest 7.1 To Work or Not to Work Is it possible to do work on an object that remains at rest? 1) yes ) no

7 ConcepTest 7.1 To Work or Not to Work Is it possible to do work on an object that remains at rest? 1) yes ) no Work requires that a force acts over a distance. If an object does not move at all, there is no displacement, and therefore no work done.

8 ConcepTest 7.a Friction and Work I A box is being pulled across a rough floor at a constant speed. What can you say about the work done by friction? 1) friction does no work at all ) friction does negative work 3) friction does positive work N Displacement f Pull mg

9 ConcepTest 7.a Friction and Work I A box is being pulled across a rough floor at a constant speed. What can you say about the work done by friction? 1) friction does no work at all ) friction does negative work 3) friction does positive work Friction acts in the opposite direction to the displacement, so the work is negative. Or using the f N Displacement Pull definition of work (W = F d cos q ), because q = 180 º, then W < 0. mg

10 ConcepTest 7.c Play Ball! In a baseball game, the catcher stops a 90-mph pitch. What can you say about the work done by the catcher on the ball? 1) catcher has done positive work ) catcher has done negative work 3) catcher has done zero work

11 ConcepTest 7.c Play Ball! In a baseball game, the catcher stops a 90-mph pitch. What can you say about the work done by 1) catcher has done positive work ) catcher has done negative work 3) catcher has done zero work the catcher on the ball? The force exerted by the catcher is opposite in direction to the displacement of the ball, so the work is negative. Or using the definition of work (W = F d cos q ), because q = 180 º, then W < 0. Note that because the work done on the ball is negative, its speed decreases. Follow-up: What about the work done by the ball on the catcher?

12 ConcepTest 7.3 Force and Work A box is being pulled up a rough incline by a rope connected to a pulley. How many forces are doing work on the box? 1) one force ) two forces 3) three forces 4) four forces 5) no forces are doing work

13 ConcepTest 7.3 Force and Work A box is being pulled up a rough incline by a rope connected to a pulley. How many forces are doing work on the box? 1) one force ) two forces 3) three forces 4) four forces 5) no forces are doing work Any force not perpendicular to the motion will do work: N does no work T does positive work f does negative work mg does negative work T N f mg

14 Work Done by a Varying Force Example: Below is the path of a particle acted on by a varying force. Clearly, F d is not constant!

15 Work Done by a Varying Force The work done by a varying force can be approximated by 1.Dividing the distance up into small pieces,. Finding the work done during each, 3.Adding them up.

16 Work Done by a Varying Force In the limit that the pieces become infinitesimally narrow, the work is the area under the curve: Or:

17 Ideal Springs and Hook s Law

18 Work Done by a Spring The force exerted by a spring is given by:. This is Hook s Law

19 Work Done by a Spring Plot of F vs. x. Work done is equal to the shaded area.

20 Kinetic Energy and the Work-Energy Principle

21 Energy Energy was traditionally defined as the ability to do work. We now know that not all forces are able to do work; however, we are dealing in these chapters with mechanical energy, which does follow this definition. Alternate definition: Energy is a measure of an object s presence in the universe

22 Kinetic Energy If we write the acceleration in terms of the velocity and the distance, we find that the work done here is We define the kinetic energy as:

23 Work-Energy Principle This means that the work done is equal to the change in the kinetic energy: If the net work is positive, the kinetic energy increases. If the net work is negative, the kinetic energy decreases. Since work and kinetic energy can be equated, they must have the same units (Joules).

24 ConcepTest 7.5a Kinetic Energy I By what factor does the kinetic energy of a car change when its speed is tripled? 1) no change at all ) factor of 3 3) factor of 6 4) factor of 9 5) factor of 1

25 ConcepTest 7.5a Kinetic Energy I By what factor does the kinetic energy of a car change when its speed is tripled? 1) no change at all ) factor of 3 3) factor of 6 4) factor of 9 5) factor of 1 Because the kinetic energy is mv, if the speed increases by a factor of 3, then the KE will increase by a factor of 9. 1 Follow-up: How would you achieve a KE increase of a factor of?

26 ConcepTest 7.5b Kinetic Energy II Car #1 has twice the mass of car #, but they both have the same kinetic energy. How do their speeds compare? 1) v 1 = v ) v = v 1 3) 4v 1 = v 4) v 1 = v 5) 8v 1 = v

27 ConcepTest 7.5b Kinetic Energy II Car #1 has twice the mass of car #, but they both have the same kinetic energy. How do their speeds compare? 1) v 1 = v ) v = v 1 3) 4v 1 = v 4) v 1 = v 5) 8v 1 = v Because the kinetic energy is 1 mv, and the mass of car #1 is greater, then car # must be moving faster. If the ratio of m 1 /m is, then the ratio of v values must also be. This means that the ratio of v /v 1 must be the square root of.

28 ConcepTest 7.8a Slowing Down If a car traveling 60 km/hr can brake to a stop within 0 m, what is its stopping distance if it is traveling 10 km/hr? Assume that the braking force is the same in both cases. 1) 0 m ) 30 m 3) 40 m 4) 60 m 5) 80 m

29 ConcepTest 7.8a Slowing Down If a car traveling 60 km/hr can brake to a stop within 0 m, what is its stopping distance if it is traveling 10 km/hr? Assume that the braking force is the same in both cases. 1) 0 m ) 30 m 3) 40 m 4) 60 m 5) 80 m F d = W net = DKE = 0 mv, and thus, F d = mv. Therefore, if the speed doubles, the stopping distance gets four times larger. 1 1

30 ConcepTest 7.8b Speeding Up I A car starts from rest and accelerates to 30 mph. Later, it gets on a highway and accelerates to 60 mph. Which takes more energy, the 0 30 mph, or the mph? 1) 0 30 mph ) mph 3) both the same

31 ConcepTest 7.8b Speeding Up I A car starts from rest and accelerates to 30 mph. Later, it gets on a highway and accelerates to 60 mph. Which takes more energy, the 0 30 mph, or the mph? 1) 0 30 mph ) mph 3) both the same 1 The change in KE ( mv ) involves the velocity squared. So in the first case, we have: m (30 0 ) = m (900) In the second case, we have: m (60 30 ) = m (700) Thus, the bigger energy change occurs in the second case

32 Textbook Problems

33 Problem 7.6 Lever Arm

34 Problem 7.6 Lever Arm We are told that the work input equals the work output, so right away we can write: O l l O l I l I I h O h q q F I F O O I I O O O I I O O I I O O I I O I h h F F h F F h h F h F W W ˆ) ( ˆ) ( ˆ) ( ˆ) ( j j j j h F h F y

35 Problem 7.6 Lever Arm Now, notice that the triangle on the left side of the diagram is geometrically similar to the triangle on the right. This means that the ratios of their corresponding sides must be equal. In other words, hi l I h l O O hi Applying this result to our previous equation, we get: F F O I h h I O h O l l I O

36 Problem 7.9 Here we are talking about the work done by a constant force, which means we can use the formula: W Fx where F is the force moving the box, and x is the displacement (see diagram below). F x

37 Problem 7.9 We are told that the force accelerates the box across the floor, this tells us that the work done by the force is positive (since the force and the displacement are in the same direction). We are not given any angles between the force and the displacement, so we can safely assume that the angle is zero, meaning that they are parallel. Since cos 0º = 1, then: W Fx Fx cos 0 Fx

38 Problem 7.9 We are given mass and acceleration, so we can find the force using Newton s nd law, but we are not given the displacement, so we need to calculate that next. Since the acceleration is constant, we can use the kinematic equations derived in chapter. Specifically, we are given initial velocity, acceleration and time, and we need to find distance, thereby we can use: x x 1 0 v0t at We can assume that the box starts from x 0 = 0, and we are told that it starts from rest, so we get: x 1 at

39 Problem 7.9 Finally, we can substitute our equation for x, and the values for F, to find the work done by the force: W Fx ( ma)( 1 1 ma (6.0 kg)(.0 m/s 588 J 1 t at ) ) (7.0 s) W J

40 Problem 7.10 Helicopter For part (a), we first need to express our vectors into components, and then use Newton s nd Law to find the magnitude of the needed force. Using our diagram, our vectors are: F w ˆj mgˆj H F H F Ma ˆ net j where F H is what we are looking for, and a = 0.10g, the net upwards acceleration of the helicopter. y M F H w

41 Problem 7.10 Helicopter Applying Newton s nd Law to the vertical components, we get: F H F Mg F y H Ma Ma Mg Ma F H 1. 1Mg For part (b), notice that the displacement, and the force applied by the helicopter are parallel. Applying the definition of work, we get: W F H hcos Mgh

42 Problem 7.11 Sliding Piano

43 Problem 7.11 Sliding Piano We can start by using our diagrams to express all our vectors as a sum of their components: w mgsinq ˆi F ˆ P F P i d dˆi mg cosq ˆj In part (a), we are asked to find F P. We are told that the piano is not accelerating, and F P is directed only in the x-direction, so we can apply Newton s nd law in the x-direction to find this force. y x d q q w F N F P m w w y w x

44 Problem 7.11 Sliding Piano F P F mgsinq F x P 0 0 mgsinq (380 kg)(9.80 m/s N )sin 7 Therefore: F P ( N) ˆi

45 Problem 7.11 Sliding Piano For part (b), notice that the displacement, and the force applied by the person are anti-parallel. In other words, the angle between them is 180º. Applying the definition of work, we get: W F P d cos180 J J Therefore: W For part (c), we simply apply the definition of work: W G wd

46 Problem 7.11 Sliding Piano W G ( mgsinq ˆi mgd sinq (380 kg)(9.80 m/s J mg cosq ˆ) j ( dˆ) i )(3.9 m)sin 7 Therefore: W J