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Chapter 6: Conservation of Energy. Introduction l Energy – ability to perform work. l Unit of energy (and of work) – Joules (J) 1 J = 1 N- m = 1 kg- m.

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Presentation on theme: "Chapter 6: Conservation of Energy. Introduction l Energy – ability to perform work. l Unit of energy (and of work) – Joules (J) 1 J = 1 N- m = 1 kg- m."— Presentation transcript:

1 Chapter 6: Conservation of Energy

2 Introduction l Energy – ability to perform work. l Unit of energy (and of work) – Joules (J) 1 J = 1 N- m = 1 kg- m 2 / s 2 Forms of energy:  Light, chemical, nuclear, mechanical, electrical, sound, heat, kinetic, elastic, magnetic etc.  Each type of energy is calculated differently.

3 Law of Conservation of Energy l In any natural process, total energy is always “conserved”, i.e. energy can not be created nor destroyed.  Can be transformed from one form to another.  Can be transferred from one system to another. In science, any law of conservation is a very powerful tool in understanding the physical universe.

4 Work by Constant Force l Work – transfer of energy as a result of force. l Constant force – its magnitude and direction unchanged. l The force acts on the object throughout the process. l Only component of force parallel to direction of motion performs work!

5 Work by Constant Force Work done by force F in moving the object at constant speed through displacement  x : W = (F cos  x F cos  component of F parallel to  x. Work is a scalar quantity. Unit = Joule (J). 1 J = 1 N-m F xx  f N W

6 Work by Constant Force W = (F cos  x 1. If  x = 0, then W = 0  If you held a 50 kg bag on your head without moving for 3 hours, would you have done work?

7 W = (F cos  x 2. If  = 0, cos  ie F and  x are parallel and W = F  x xx F

8 W = (F cos  x 3. If  = 90 o, cos 9    ie F and  x are perpendicular to each other and W = 0 F xx  f N W Work done by the normal force N is zero

9 W = (F cos  x 4. If  = 180 o, cos 18    ie F and  x are antiperpendicular to each other and W = negative. F xx  f N W Work done by the frictional force f is - F  x

10 Example: A 30 N chest was pulled 5 m across the floor at a constant speed by applying a force of 50 N at an angle of 30 o. How much work was done by tension T? 30 o 50 N W = F  x cos   m) cos (30) = 217 Joules T mg N f

11 90 o xx W = mg  x cos   30 x 5 cos(90) =  T mg N f Example: A 30 N chest was pulled 5 m across the floor at a constant speed by applying a force of 50 N at an angle of 30 o. How much work was done by gravity? 30 o 50 N mg f

12 To find the magnitude of f: Consider x-component of forces acting. Since constant velocity: F net = 0 So Tcos30 o – f = 0 Thus f = 43.3 N. W = 43.3 x (cos180 o ) x 5 = -217 J T mg N f Example: A 30 N chest was pulled 5 m across the floor at a constant speed by applying a force of 50 N at an angle of 30 o. How much work was done by friction f? f xx 180 o 50 N mg f

13 Work by Variable Force W = F x  x Work = area under F vs x plot Force x F xx x F = kx Spring F = k x Area = ½ k (x) 2 = W spring

14 Example A dart gun with a spring constant k = 400.0 N/m is compressed 8.0 cm. How much energy will it transfer to a dart when the spring is released? W = ½ k(  x) 2

15 Example As you ride in an elevator going upward at constant speed, work done on you by the normal force from the floor of the elevator (A)is negative. (B) is zero. (C) is positive. (D) is zero, negative or positive depending on the speed of the elevator.

16 Kinetic Energy Suppose a constant force F is applied horizontally on a small rigid object of mass m. Its velocity will change, say from v o to v as it moves through a distance  x (without rotation). Work done on the object W = F x  x [F parallel to  x] = m a x  x [Recall: v 2 = v o 2 + 2a(x-x o )] = ½ m (v 2 – v 0 2 ) = ½ mv 2 - ½ mv 0 2 The quantity ½. mass. (velocity) 2 is called kinetic energy of the object.

17 An object of mass m, moving with velocity v, has kinetic energy K = ½ mv 2  K  m: If m is doubled, K will double.  K  v 2 If v is doubled, K will quadruple.  Can K ever be zero?  Can K ever be negative?  W =  K Can W ever be negative?

18 Kinetic Energy If more than one force acts on the object, the change in kinetic energy is the total work done by all the forces acting on the object. Total work W total =  K This is called the Work- Energy Principle F vovo v xx

19 Total Work Total work W total =  K = change in K.E. W total = First find work done by each force, then add them. = first add all the forces and then then calculate work done by the net force Q: What is the total work done on an object moving with constant velocity? (A) Positive (B) Negative (C) Zero

20 Example 1: A 30 N chest was pulled 5 m across the floor at a constant speed by applying a force of 50 N at an angle of 30 o. How much work was done by all the forces acting on it? 30 o T = 50 N W T = F  x cos  m) cos (30) = 217 J W N = 0, W mg = 0 W f = -217J W total = 217 + 0 +0 +(-217) = 0 T mg N f

21 Example 2: A 30 N chest was pulled 5 m across the floor by applying a force of 50 N at an angle of 30 o. How much work was done by all the forces acting on it if its speed changed from 2 m/s to 5 m/s in the process?

22 Example You are towing a car up a hill with constant velocity. The work done on the car by the normal force is: 1. positive 2. negative 3. zero T W FNFN V correct Work done by gravity? Work done by tension? Since the direction of the force is positive the value of work will be positive. it's negative because it's trying to slow down the car. The normal force does no work because it acts in a direction perpendicular to the displacement of the car.

23 Example: Block w/ friction l A block is sliding on a surface with an initial speed of 5 m/s. If the coefficient of kinetic friction between the block and table is 0.4, how far does the block travel before stopping? 5 m/s mg N f x y Y direction: F=ma N-mg = 0 N = mg Work W N = 0 W mg = 0 W f = f  x cos(180) = -  mg  x W =  K -  mg  x = ½ m (v f 2 – v 0 2 ) -  g  x = ½ (0 – v 0 2 )  g  x = ½ v 0 2  x = ½ v 0 2 /  g = 3.1 meters

24 POTENTIAL ENERGY Potential Energy (U) = stored energy. It is the energy an object possess due to its position or its configuration. There are different types of potential energy: Gravitational potential energy. Elastic potential energy. Electrical potential energy.

25 Gravitational Potential Energy To move mass m from initial point to final point at constant velocity, an external force F EXT, equal but opposite to the force of gravity must be applied. W EXT = F cos   y = mgh = (mg)(cos0)h = mgh W g = mgcos(180 o )(h) = -mgh mg Y o = 0 Y f = h F EXT

26 Gravitational Potential Energy l The quantity mgh is called gravitational potential energy near the earth’s surface: U = mgh where U = 0 wherever we have chosen h = 0 More general way of writing gravitational potential energy is mg U = 0 U = mg h F EXT

27 The gravitational potential energy of an object of mass m at a height h is U g = mgh 1. U g = mgh and W g = -mgh ie U g = - W g. 2. U g  m 3. U g  h 4. A reference level where h = 0 can be chosen to be at any convenient location. 5. Only change in U g is important.  U g = mg  h = mg(h f – h i )

28 Gravitational Potential Energy Example 3: A block slid down a plane If a block is slid down a plane. Distance moved down the plane =  s Vertical height moved = h How much work will the force of gravity do? W = F cos   x W g = mg. cos .  S But  S =  y /cos  W g = (mg)(cos  )(h/cos  ) = mgh h  mg S

29 Elastic Potential Energy Elastic (Spring) force F = -kx Work done by spring force W = ½ kx 2 Energy stored in the spring when compressed or stretched by distance x Elastic Potential Energy U = ½ kx 2

30 Example A dart gun with a spring constant k = 400.0 N/m is compressed 8.0 cm. (a) How much energy will it transfer to a dart when the spring is released? (b) With what speed will the dart (mass 200 g) leave the gun?

31 Example Emil throws an orange straight up and catches it at the same point it was thrown. (a) How much work is done during the orange’s free fall? (b) If the orange was thrown upward from a 1.2 m height above the ground and falls to the ground, how much work is done by gravity? [Mass of orange =.25 kg]

32 Conservative and Non Conservative Forces Thrown h  mg Dropped Sliding down a plane Work done by force of gravity, W g = - mgh Independent of path

33 Gravitational Potential Energy To move mass m from initial point to final point at constant velocity, an external force F EXT, equal but opposite to the force of gravity must be applied. W EXT = F cos   y = mgh = (mg)(cos0)h = mgh W g = mgcos(180 o )(h) = -mgh mg Y o = 0 Y f = h F EXT

34 34 1. As you ride an elevator going upward at constant speed, the work done on you by the normal force from the floor of the elevator is A.negative B.zero C.positive D.Negative, zero or positive depending on the speed of the elevator

35 35 2. What is the total work done on an object moving with constant velocity? A.Positive B.Negative C.Zero

36 36 3. A passenger whose mass is 95 kg is seated in a jet airliner flying 1,200 m above the ground at a speed of 253 m/s. What is the kinetic energy of the passenger? A.931 J B.6.08 x 10 6 J C.3.04 x 10 6 J D.1.12 x 10 6 J

37 37 4. A passenger whose mass is 95 kg is seated in a jet airliner flying 1,200 m above the ground at a speed of 253 m/s. What is the potential energy of the passenger? A.931 J B.6.08 x 10 6 J C.3.04 x 10 6 J D.1.12 x 10 6 J

38 38 5. A spring has an elastic constant of 920 N/m. By how much should it be stretched in order to store 120 J of energy? A.0.36 m B.0.51 m C.0.26 m D.7.7 m E.1.96 m

39 Conservative Forces Forces which perform same amount of work independent of the path taken are called conservative forces. Eg: gravity, elastic force, electric force. 1.Work done by a conservative force depends only on the initial and final position and not on the path taken to get there. 2.Potential energy only goes with conservative forces. There is potential energy associated with gravity, elastic force, electric force. 3.W C = -  U [ U g = mgh and U spring = ½ kx 2 ]

40 Non Conservative Forces 1.Amount of work done by them depend on the path taken. Eg: friction, tension, push/pull. 2. There is no potential energy associated with non conservative forces.

41 Work - Energy w/ Conservative Forces Total work = Work done by all types of forces. = Work by conservative forces + work by non-conservative forces W Total = W C + W NC But W Total =  K and W C = -  U So,  K = -  U + W NC ie, W NC =  K +  U W NC =  K +  U

42 Work - Energy w/ Conservative Forces W NC =  K +  U When only conservative forces are acting; W NC = 0 0 =  K +  U OR 0 = (K f – K i ) + (U f – U i ) OR K f + U f = K i + U i The quantity K + U is called mechanical energy E Thus E f = E i (Conservation of mechanical energy)

43 Skiing Example (no friction) A skier goes down a 78 meter high hill with a variety of slopes. What is the maximum speed she can obtain if she starts from rest at the top? [Assume friction is negligible] Conservation of energy: K i + U i = K f + U f ½ m v i 2 + m g y i = ½ m v f 2 + m g y f 0 + g y i = ½ v f 2 + g y f v f 2 = 2 g (y f -y i ) v f = sqrt( 2 g (y f -y i )) v f = sqrt( 2 x 9.8 x 78) = 39 m/s

44 Two similar objects are released from rest at the same time to slide down two frictionless slopes A and B of different inclines as shown in the figure below. Which of these statements is true about the motion of the two balls? They will reach the bottom with the same (A)Acceleration (B) Speed (C) time duration (D) Same acceleration, speed, time duration A B h 30 o h

45 Example Suppose the initial kinetic and potential energies of a system are 75J and 250J respectively, and that the final kinetic and potential energies of the same system are 300J and -25J respectively. How much work was done on the system by non-conservative forces? 1. 0J 2. 50J 3. -50J 4. 225J 5. -225J Initial = 75 J + 250 J = 325 J. Final = 300 J - 25 J = 275 J. Final -Initial = 275 - 325 = -50 J. No energy goes into or comes out of the system overall, so some non- conservative force has to be accounted for.

46 Example A dart gun with a spring constant k = 400.0 N/m is compressed 8.0 cm. (a) How much energy will it transfer to a dart of mass 200 g when the spring is released? (b) What would be the escape speed of the dart? (c) If the dart was aimed upward, how high would it travel? U spring = ½ k(  x) 2 K = ½ mv 2, U g = mgh

47 Power Average Power P = rate at which work is being done. = rate at which energy is being transformed. P =  W /  t Units: J/s = Watt  W/  t = [F  x cos(  )]/  t = F (v  t) cos(  ) i.e., P = F v cos(  )

48 Example: How much electrical energy will a 75-watt light bulb use if left lit for 5 hours? [1 hour = 3,600 s]

49 Example: A hot plate used 225,000 J of electrical energy in 5 minutes. What is the power (wattage) of the hot plate?


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