Chapter 6. Work and Energy

Transcription

1 Chapter 6 Work and Energy

2 ENERGY IS THE ABILITY TO DO WORK = TO APPLY A FORCE OVER A DISTANCE= Example: push over a distance, pull over a distance. Mechanical energy comes into 2 forms: Kinetic energy KE or energy of motion. The kinetic energy changes if the speed changes. KE = 0.5 mv 2 Potential energy PE like the gravitational potential energy: PEg=mgh h is the height above the ground. PE is Stored energy. It depends on the position of The object. (height for example). You need to define a reference level for which the energy =PE=0. So a book on a table has potential energy. If it drops it can hurt. You also have elastic potential energy stored in a spring (stretched or Compressed). Ready to push or pull over a distance. The energy depends On the position of the spring when compressed or stretched relative to its equilibrium position. Note: fat is chemical potential energy. Ready to be burnt and transformed.

3 We will come back to energy in 6.3. But the point it to define the word WORK in Physics. It has a special meaning. WORK is done on a system by a force only if the effect of the force Is to speed up pr slow done the system. If the system is not moving When acted upon by the force, no work is done. If, because of the force, the speed of the system increases the WORK is positive. If the speed decreases, the WORK is negative. Note that more than 1 force can act on an object. So 1 force can speed up The object while another force slows it down (like friction/air resistance). In that case you need to find the net force to find the change in energy Of motion. Examples: 1.You push on a wall. Puffing, swearing, sweating. Are you doing work on the wall. (is your push is?) Ask is the energy of motion the wall increases? (did it move?) 2.You are lifting your computer from the table to the shelf. What are the 2 forces acting on the computer. Which one is doing a positive work? And a negative work?

4 You are pushing a cart. The cart moves as a consequence. Are you doing work on the car? Is it positive? The frictional force slows down cars. Is it doing work? Is it positive or negative? What about air resistance on a parachute? A drive engine on a plane? When work is done (positive or negative), A force F is applied over a Displacement x. Say you are carrying a suitcase around. Moving at a constant speed. The suitcase stays at the same level. Is the lifting force doing work? (perpendicular to the displacement). So always ask your self if the force under consideration (or one of the component of the force) is increasing the velocity of an object and in which direction.

5 Let's take the simple case: The force is along the x-axis and the displacement is also along the x-axis. How to compute the work W done on an object by the force F If the object moves by a distance s. What do you think? think Do you expect the work done to increase with the distance The object moves and with the force applied?

6 6.1 Work Done by a Constant Force W = 1 N m = Fs 1 joule ( J) THIS quantity W measured the amount of energy transferred from You to the car. It measured how many calories you burnt to push the car. Actually we are not a very efficient machine so most of the calories burnt Are lost to heat. 1 food calorie is 1 Cal is about 4,200 joules.

7 To find the work done by a force: I- identify the displacement direction II- Find the component of the force along the displacement. If the force is perpendicular to the displacement work=0 If there is no displacement work=0 Multiply the component by the displacement : Long the x-axis Fx(X) or along the y-axis Fy(Y) II- If the force slows down the motion the work is negative. If the force speeds up the object, the work is positive.

8 6.1 Work Done by a Constant Force

9 W Consider this example above. The force has 2 components. 1 along the X-axis and 1 along the y-axis. Which component is doinf The work? Which component is making the suitcase move? Fy FN Fx Along the Xaxis : Fx increases the velocity of the suitcase Fx-fk = mass x acceleration Along the Y-axis Fy and Fn are balanced by W, no position change, no Increase in speed.

10 SO in that case: WORK done by F on the suitcase is: W by F = horizontal component of F x W by F = Fx (X) displacement X Note that the work done by Fx is positive The work done by the normal FN is (none) The work done by fk is (slows down the object) The work done by the weight is (none) The work done by Fy is (none) SO ONLY THE COMPONENT // DISPLACEMENT is doing work. so: - no displacement MEANS WORK=0 no transfer of energy - component perpendicular to displacement means WORK = 0

11 NET WORK DONE on an object = WORK done by the net force. For the suit case: NET FORCE = Fx fk NET WORK = (Fx fk) x displacement. Example: Find the net work done on the suitcase if the horizontal component of F is 10N@right And the frictional force is 3N@left and the displacement is 5m. This quantity is the increase of energy Of the suitcase. Amount of energy Transferred to the suitcase. We can compute its increase in speed. Its change in kinetic energy.

12 Easy examples First draw the the free-body diagram and identify the components That do the work.. Source: Physics of every day phenomena. Griffith. McGraw Hill 1. A woman does 160 J of work to move a table 4 m across the floor. What is the magnitude of the force that the woman applied to the table if this Force is applied in the horizontal direction? 2. A force of 60N used to push a chair across a room does 300J of work. How far does the chair move in this process? 3. A rope applies a horizontal force of 180N to pull a crate a distance of 2m Across the floor. A frictional force of 120N opposes this motion. A) What is the work applied by the rope. B)What is the work done by the frictional force. C)What is the total work done on the crate?

13 A force of 50N is used to drag a crate 4m across a floor. The force is directed at an angle upward from the crate so the vertical Component is 30N and the horizontal component is 40N. 50N 30N 40N 4m What is the work done by the horizontal component of the force? What is the work done by the vertical component of the force? What is the total work done?

14 6.1 Work Done by a Constant Force SO WORK done a force F on an object = Component of force along the displacement x displacement. Let's go back to the example of the suit case again. The object is moved only along the x-axis. So only the horizontal component of F is doing work. W = Fx (X) or

15 6.1 Work Done by a Constant Force W = ( F cosθ ) X cos0 = 1 cos90 = 0 cos180 = 1 Work is positive and max No work is done Work is negative and minimum

16 NET WORK = Sum of the work of the forces. NO net work = no increase in speed, no change in height, no stretching Of the spring In which one of the following situations is zero net work done? a) a bunch of bananas are placed on a spring scale in the supermarket Bananas are placed on a scale. The spring stretches. b) a sky diver falls from an airplane before opening her parachute Weight is doing work on the diver because the speed increases. (at a slower rate) c) a horse pulls a wagon at a constant velocity The horse is doing a positive work, but friction is doing a negative work such as they cancel each other. d) a snowball rolls down a hill Height is changing because 1 component of the weight is pulling along slope. e) a skateboarder steps on a skateboard and begins to roll When you start moving, speed increases

17 Work may be expressed using all of the following units except: a) watt b) joule c) N m d) erg e) ft lb

18 A 5.0-kg ball on the end of a chain is whirled at a constant speed of 1.0 m/s in a horizontal circle of radius 3.0 m. What is the work done by the centripetal force during one revolution? a) 2.5 J b) 1.7 J c) 1.2 J d) 0.56 J e) zero J

19 Which one of the following combinations of units is equal to the joule? a) kg m 2 /s b) kg m c) kg m/s d) kg m 2 /s 2 e) kg s

20 In which one of the following circumstances does the force do positive work on the object? a) The direction of the force is perpendicular to the object s displacement. b) The direction of the force is in the opposite direction to the object s displacement. c) No matter the direction of the force, positive work will be done if there is a displacement of the object. d) The direction of the force is in the same direction as the object s displacement. e) The object s displacement is zero meters as the force is applied.

21 REMEMBER: If no displacement = no work If force (or component of a force) is perpendicular to displacement = no work Work measures the transfer of energy from 1 system to another.

22 6.1 Work Done by a Constant Force Example 1 Pulling a Suitcase-on-Wheels Find the work done if the force is 45.0-N, the angle is 50.0 degrees, and the displacement is 75.0 m.

23 W = ( ) [ F cosθ s = ( 45.0 N) cos 50.0 ]( 75.0 m) = 2170 J

24 6.1 Work Done by a Constant Force The weight lifter is bench-pressing a barbell. The weight of the barbell is 710N. He raises the barbell a distance 0f 0.65m Then he lowers the barbell in the same direction Work is done on if the energy of the barbell increases. Find the work done in each case. It can be 0, positive or negative. Convert to Kcal (1 kcal = 1 food cal = 4,200 joules.

25 6.1 Work Done by a Constant Force W = 0 W = 460J W = 460J 460J= 0.1 Kcal You have to lift and lower 2000 times to lose 1 soda can (about 220Kcal)!!!!

26 6.1 Work Done by a Constant Force Example 3 Accelerating a Crate The truck is accelerating at a rate of m/s 2. The mass of the crate is 120-kg and it does not slip. The magnitude of the displacement is 65 m. Newton's first law says, a mass stays at rest unless acted upon By a force. How come the crate does not slip? It should according to inertia's principle? First do the free-body diagram for the crate.

27 6.1 Work Done by a Constant Force Example 3 Accelerating a Crate The truck is accelerating at a rate of m/s 2. The mass of the crate is 120-kg and it does not slip. The magnitude of the displacement is 65 m. Find the net force acting on the crate. What is the total work done on the crate by all of the forces acting on it?

28 6.1 Work Done by a Constant Force The angle between the displacement and the friction force is 0 degrees. f s = ma = ( )( kg 1.5m s ) = 180N W = [( ) ]( ) 4 180N cos0 65 m = J

29 Assignments with work: 1. A. motorist runs out of gas on a level road 200m from a gas station. The driver Pushed the 1,200kg car to the station. If a 150N force is required to keep the car Moving, how much work does the driver do? 2. p.188 number 1. You need to focus on the skier and draw forces doing work In a free-body diagram. Find the work done by the tension in the rope.use a top view. (source: Johnson and Cutnell, Wiley Physics 8e) Hint: first you need to find The displacement covered in 12s. Speed is constant = 9.30 m/s X Then you need to do find the component Of thetension (135N) that is doing the Work. (responsible for keeping The skier moving). Note: negative work is done by the Friction between the water and the ski. Net work =0 since there is no increase In speed.

30 3.number 3 p.188. Work is negative when the force oppose the increase in energy, That is decrease the energy. 4. number 4 p.188 same idea as in number 5 p.188 let's go back to the suit case. The angle is unknown. The magnitude of F is 30N, the distance covered is 50m The work done by the horizontal component of F is 1,100 J. Find the angle.

31 6. number 6 p.188. Like for the suit case example. In (b) F has no component along the vertical/. 7. number 8 p N F 29 (a) remember since the cart is moving At a constant speed that means: Left = right or friction = Fx 48=Fcos(29) solve for F (b) work done by Fx (d=22m) work done by fk (d) work done by the weight

32 8.Number7 p.188. VERY GOOD EXAMPLE TO DEVELOP STRATEGY (a) what make more sense to you? (b)follow steps: Only the x-component of The thrust and the x-component of the weight are doing Work. I.First for each case find the x-components of F and W (see below) II.Find the work done by each component for each case. (work <0 when opposes motion) III. Find the net work for each case. IV. Find the difference. 25 case1 Wx= - Tx= Work done by Wx= Work done by T = Net work = Case2!!angle between W and y-axis is 25 Wx= - Tx= Work done by Wx= Work done by T = Net work = Find the difference. You don't need to know T, since you are computing the difference, the work done by T will cancel.