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Introduction to Work Thursday, Jan 6, 2011! Happy New Year.

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1 Introduction to Work Thursday, Jan 6, 2011! Happy New Year.

2 Energy and Work A body experiences a change in energy when one or more forces do work on it. A body must move under the influence of a force or forces to say work was done. A force does positive work on a body when the force and the displacement are at least partially aligned. Maximum positive work is done when a force and a displacement are in exactly the same direction. If a force causes no displacement, it does zero work. Forces can do negative work if they are pointed opposite the direction of the displacement.

3 Calculating Work – Step 1, constant F If a force on an object is at least partially aligned with the displacement of the object, positive work is done by the force. The amount of work done depends on the magnitude of the force, the magnitude of the displacement, and the degree of alignment. W= F r cos r F F

4 Forces can do positive or negative work. When the load goes up, gravity does negative work and the crane does positive work. When the load goes down, gravity does positive work and the crane does negative work. Ranking Task 11 mgmg F

5 Units of Work SI System: Joule (N m)

6 Problem: A droplet of water of mass 50 mg falls at constant speed under the influence of gravity and air resistance. After the drop has fallen 1.0 km, what is the work done by a) gravity and b) air resistance?

7 Problem: A sled loaded with bricks has a mass of 20.0 kg. It is pulled at constant speed by a rope inclined at 25 o above the horizontal, and it moves a distance of 100 m on a horizontal surface. If the coefficient of kinetic friction between the sled and the ground is 0.40, calculate a)The tension in the rope. b)The work done by the rope on the sled c)The work done by friction on the sled.

8 Work and a Pulley System A pulley system, which has at least one pulley attached to the load, can be used to reduce the force necessary to lift a load. Amount of work done in lifting the load is not changed. The distance the force is applied over is increased, thus the force is reduced, since W = Fd. F m

9 Work as a Dot Product

10 Calculating Work a Different Way Work is a scalar resulting from the multiplication of two vectors. We say work is the dot product of force and displacement. W = F r dot product representation W= F r cos useful if given magnitudes and directions of vectors W = F x r x + F y r y + F z r z useful if given unit vectors

11 The scalar product of two vectors is called the dot product The dot product is one way to multiply two vectors. (The other way is called the cross product.) Applications of the dot product WorkW = F d PowerP = F v Magnetic FluxΦ B = B A The quantities shown above are biggest when the vectors are completely aligned and there is a zero angle between them.

12 Why is work a dot product? s W = F r W = F r cos Only the component of force aligned with displacement does work. F

13 Problem: Vector A has a magnitude of 8.0 and vector B has a magnitude of The two vectors make an angle of 40 o with each other. Find AB.

14 Problem: A force F = (5.0i + 6.0j – 2.0k)N acts on an object that undergoes a displacement of r = (4.0i – 9.0j + 3.0k)m. How much work was done on the object by the force?

15 Problem: A force F = (5.0i – 3.0j) N acts upon a body which undergoes a displacement d = (2.0i – j) m. How much work is performed, and what is the angle between the vectors?

16 Homework 1/6/11 Page : 18, 19, 21, 51, 56

17 Work by Variable Forces Sometimes, forces are not constant. Examples, spring forces, air resistant forces. How do we calculate work in these situations?

18 Work and Variable Forces For constant forces W = F r For variable forces, you cant move far until the force changes. The force is only constant over an infinitesimal displacement. dW = F dr (dr, dW = small, small sample where force could be considered constant) To calculate work for a larger displacement, you have to take an integral W = dW = F dr

19 Work and variable force The area under the curve of a graph of force vs displacement gives the work done by the force. F(x) x xaxa xbxb W = F(x) dx xaxa xbxb

20 Problem: Determine the work done by the force as the particle moves from x = 2 m to x = 8 m. F (N) x (m) We already know this technique manually with a graph; we can now call this manual integration

21 Tutorial on the Integral Mathematically, the integral is used to calculate a sum composed of many, many tiny parts. In physics, the integral is used to find a measurable change resulting from very small incremental changes. Its useful to think of it as a fancy addition process or a multiplication process, depending on the situation. Heres a specific example. Velocity times time gives displacement. If the velocity is changing with time, but a very tiny time change is used to calculate a very tiny displacement, we can nonetheless assume the velocity was constant during that tiny time change. If we calculate tiny displacements this way (recalculating our velocity for each time increment), then add the tiny displacements up to get a larger displacement, we have done integration. 21

22 Velocity as a Function of Time Displacement Velocity can be represented as follows: Rearrangement of this expression yields: What this means is that we can calculate a tiny displacement dx from the velocity v at a given time times a tiny time increment dt. 22

23 Summing the Displacements When we sum up these tiny displacements, we use the following notation: This notation indicates that we are summing up all the little displacements dx starting at position x o at time t o until we reach a final position and time, x f and t f. The velocity v may be a function of time, and may be slightly different for one time increment dt and the next time increment. 23

24 Evaluating Integrals We will evaluate polynomial integrals by reversing the process we used in taking a polynomial derivative. This process is sometimes called anti- differentiation, or doing an anti-derivative. The general method for doing an anti-derivative is: Can you see how this is the reverse of taking a derivative? (Note: This indefinite integral requires us to add a constant C to compensate for constants that may have been lost during differentiation…but we will do definite integrals that do not require C.) 24

25 Evaluating Definite Integrals When a definite integral with limits of integration is to be evaluated, first do the anti-derivative. Then evaluate the resulting functions using the top limits, evaluate them again using the bottom limits, then subtract the two values to get the answer. (Note: do each side of the equation separately.) 25

26 Evaluating Definite Integrals When a definite integral with limits of integration is to be evaluated, first do the anti-derivative. Then evaluate the resulting functions using the top limits, evaluate them again using the bottom limits, then subtract the two values to get the answer. (Note: do each side of the equation separately.) 26

27 Evaluating Definite Integrals When a definite integral with limits of integration is to be evaluated, first do the anti-derivative. Then evaluate the resulting functions using the top limits, evaluate them again using the bottom limits, then subtract the two values to get the answer. (Note: do each side of the equation separately.) 27

28 Evaluating Definite Integrals When a definite integral with limits of integration is to be evaluated, first do the anti-derivative. Then evaluate the resulting functions using the top limits, evaluate them again using the bottom limits, then subtract the two values to get the answer. (Note: do each side of the equation separately.) 28

29 Evaluating Definite Integrals When a definite integral with limits of integration is to be evaluated, first do the anti-derivative. Then evaluate the resulting functions using the top limits, evaluate them again using the bottom limits, then subtract the two values to get the answer. (Note: do each side of the equation separately.) 29

30 Sample problem: Consider a force that is a function of time: F(t) = (3.0 t – 0.5 t 2 )N If this force acts upon a 0.2 kg particle at rest for 3.0 seconds, what is the resulting velocity and position of the particle? 30

31 Problem: A force acting on a particle is F x = (4x – x 2 )N. Find the work done by the force on the particle when the particle moves along the x-axis from x= 0 to x = 2.0 m.

32 Sample problem: Consider a force that is a function of time: F(t) = (16 t 2 – 8 t + 4)N If this force acts upon a 4 kg particle at rest for 1.0 seconds, what is the resulting change in velocity of the particle? 32

33 Try these samples on a clean sheet of paper. 1. Calculate the work done by a force that applies F(x) = 3x + x 3, while it moves an object from 3 to 8 meters. 2. A tractor pulls a trailer with a force that varies according to F(x) = x – 3x. Calculate the work done from rest to 2.0 m. 3.

34 Problem: Derive an expression for the work done by a spring as it is stretched from its equilibrium position

35 Problem: How much work does an applied force do when it stretches a nonlinear spring where the force varies according to the expressions F = (300 N/m) x – (25 N/m 2 ) x 2 from its equilibrium length to 20 cm?

36 Work Energy Theorem

37 Net Work or Total Work An object can be subject to many forces at the same time, and if the object is moving, the work done by each force can be individually determined. At the same time one force does positive work on the object, another force may be doing negative work, and yet another force may be doing no work at all. The net work, or total, work done on the object (W net or W tot ) is the scalar sum of the work done on an object by all forces acting upon the object. W net = ΣW i

38 The Work-Energy Theorem W net = ΔK When net work due to all forces acting upon an object is positive, the kinetic energy of the object will increase. When net work due to all forces acting upon an object is negative, the kinetic energy of the object will decrease. When there is no net work acting upon an object, the kinetic energy of the object will be unchanged. (Note this says nothing about the kinetic energy.)

39 Kinetic Energy Kinetic energy is one form of mechanical energy, which is energy we can easily see and characterize. Kinetic energy is due to the motion of an object. K = ½ m v 2 K: Kinetic Energy in Joules. m: mass in kg v: speed in m/s In vector form, K = ½ m vv Ranking Tasks

40 Problem: A net force of 320 N acts over 1.3 m on a 0.4 kg particle moving at 2.0 m/s. What is the speed of this particle after this interaction?

41 Problem: Calculate the kinetic energy change of a 3.0 kg object that changes its velocity from (2.0 i j -1.0 k) m/s to (-1.0 i j k) m/s. How much net work done on this object?

42 Problem: A force of F 1 = (4.0 i + j) N and another of F 2 = -4.0 j N act upon a 1 kg object at rest at the origin. What is the speed of the object after it has moved a distance of 3.0 m?

43 Power Power is the rate of which work is done. No matter how fast we get up the stairs, our work is the same. When we run upstairs, power demands on our body are high. When we walk upstairs, power demands on our body are lower. P ave = W / t P inst = dW/dt P = F v

44 Units of Power Watt = J/s ft lb / s horsepower 550 ft lb / s 746 Watts

45 Problem: A 1000-kg space probe lifts straight upward off the planet Zombie, which is without an atmosphere, at a constant speed of 3.0 m/s. What is the power expended by the probes engines? The acceleration due to gravity of Zombie is ½ that of earths.

46 Problem: Develop an expression for the power output of an airplane cruising at constant speed v in level flight. Assume that the aerodynamic drag force is given by F D = bv 2. By what factor must the power be increased to increase airspeed by 25%?

47 How We Buy Energy… The kilowatt-hour is a commonly used unit by the electrical power company. Power companies charge you by the kilowatt- hour (kWh), but this not power, it is really energy consumed.

48 Problem: Using what you know about units, calculate how many Joules is in a kilowatt-hour.

49 Conservative and Non- Conservative Forces

50 More about force types Conservative forces: Work in moving an object is path independent. Work in moving an object along a closed path is zero. Work is directly related to a negative change in potential energy Ex: gravity, electrostatic, magnetostatic, springs Non-conservative forces: Work is path dependent. Work along a closed path is NOT zero. Work may be related to a change in mechanical energy, or thermal energy Ex: friction, drag, magnetodynamic

51 Potential Energy, U A type of mechanical energy possessed by an object by virtue of its position or configuration. Represented by the letter U. Examples: Gravitational potential energy, U g. Electrical potential energy, U e. Spring potential energy, U s. The work done by conservative forces is the negative of the potential energy change. W = -ΔU

52 Gravitational Potential Energy (U g ) The change in gravitational potential energy is the negative of the work done by gravitational force on an object when it is moved. For objects near the earths surface, the gravitational pull of the earth is roughly constant, so the force necessary to lift an object at constant velocity is equal to the weight, so we can say ΔU g = -W g = mgh Note that this means we have defined the point at which U g = 0, which we can do arbitrarily in any given problem close to the earths surface. mgmg F app h

53 Spring Potential Energy, U s Springs obey Hookes Law. F s (x) = -kx F s is restoring force exerted BY the spring. W s = F s (x)dx = -k xdx W s is the work done BY the spring. U s = ½ k x 2 Unlike gravitational potential energy, we know where the zero potential energy point is for a spring.

54 Problem: Three identical springs (X, Y, and Z) are hung as shown. When a 5.0-kg mass is hung on X, the mass descends 4.0 cm from its initial point. When a 7.0-kg mass is hung on Z, how far does the mass descend? XY Z

55 Sample problem: Gravitational potential energy for a body a large distance r from the center of the earth is defined as shown below. Derive this equation from the Universal Law of Gravity. Hint 1: dW = F(r)dr Hint 2: ΔU = -W c (and gravity is conservative!) Hint 3: U g is zero at infinite separation of the masses.

56 Conservation of Mechanical Energy

57 System Boundary

58 Law of Conservation of Energy E = U + K + E int = Constant No mass can enter or leave! No energy can enter or leave! Energy is constant, or conserved! The system is isolated and boundary allows no exchange with the environment.

59 Law of Conservation of Mechanical Energy E = U + K = Constant We only allow U and K to interchange. We ignore E int (thermal energy)

60 Law of Conservation of Mechanical Energy E = U + K = C or E = U + K = 0 for gravity U g = mgh f - mgh i K = ½ mv f 2 - ½ mv i 2 (What assumptions are we making here?) for springs U s = ½ kx f 2 - ½ kx i 2 K = ½ mv f 2 - ½ mv i 2 (What assumptions are we making here?) Ranking Tasks 11

61 h Pendulum Energy ½mv mgh 1 = ½mv mgh 2 For any points two points in the pendulums swing

62 Spring Energy mm -x m x 0 ½ kx ½ mv 1 2 = ½ kx ½ mv 2 2 For any two points in a springs oscillation

63 Problem: A single conservative force of F = (3i + 5j) N acts on a 4.0 kg particle. Calculate the work done if the particle if the moves from the origin to r = (2i - 3j) m. Does the result depend on path? What is the speed of the particle at r if the speed at the origin was 4.0 m/s? What is the change in potential energy of the system?

64 Sample Problem: A bead slides on the loop-the-loop shown. If it is released from height h = 3.5 R, what is the speed at point A? How great is the normal force at A if the mass is 5.0 g?

65 Non-conservative Forces and Conservation of Energy

66 Non-conservative forces Non-conservative forces change the mechanical energy of a system. Examples: friction and drag W tot = W nc + W c = K W nc = K – W c W nc = K + U

67 Sample Problem: A 2,000 kg car starts from rest and coasts down from the top of a 5.00 m long driveway that is sloped at an angel of 20 o with the horizontal. If an average friction force of 4,000 N impedes the motion of the car, find the speed of the car at the bottom of the driveway.

68 Problem: A parachutist of mass 50 kg jumps out of a hot air balloon 1,000 meters above the ground and lands on the ground with a speed of 5.00 m/s. How much energy was lost to friction during the descent?

69 Force and Potential Energy In order to discuss the relationships between potential energy and force, we need to review a couple of relationships. W c = F x (if force is constant) W c = Fdx = - dU = - U (if force varies) Fdx = - dU F = -dU/dx

70 Stable Equilibrium – 1 st and 2 nd Derivatives U x

71 Unstable Equilibrium – 1 st and 2 nd Derivatives U x

72 Neutral Equilibrium – 1 st and 2 nd Derivatives U x

73 More on Potential Energy and Force In multiple dimensions, you can take derivatives of each dimension separately. F(r) = -dU(r)/dr F x = - U/ x F y = - U/ y F z = - U/ z F = F x i + F y j + F z k

74 Molecular potential energy diagrams R U

75 Problem: The potential energy of a two-particle system separated by a distance r is given by U(r) = A/r, where A is a constant. Find the radial force F that each particle exerts on the other.

76 Problem: A potential energy function for a two-dimensional force is of the form U = 3x 3 y – 7x. Find the force acting at a point (x,y).

77 Linear Momentum October 31, 2008

78 Announcements Turn in homework due today: Chapter 8, problems 28,29,31 Next week, W-F, Rocket Project

79 Linear Momentum Momentum is a measure of how hard it is to stop or turn a moving object. p = mv (single particle) P = Σp i (system of particles)

80 Problem: How fast must an electron move to have the same momentum as a proton moving at 300 m/s?

81 Problem: A 90-kg tackle runs north at 5.0 m/s and a 75-kg quarterback runs east at 8.0 m/s. What is the momentum of the system of football players?

82 Impulse (J) Impulse is the integral of force over a period of time. J = Fdt The change in momentum of a particle is equal to the impulse acting on it. p = J

83 Problem: Restate Newtons 2 nd Law in terms of impulse.

84 time Force Impulsive forces are generally of high magnitude and short duration.

85 Problem: A 150-g baseball moving at 40 m/s 15 o below the horizontal is struck by a bat. It leaves the bat at 55 m/s 35 o above the horizontal. What is the impulse exerted by the bat on the ball? If the collision took 2.3 ms, what was the average force?

86 Problem: An 85-kg lumberjack stands at one end of a floating 400-kg log that is at rest relative to the shore of a lake. If the lumberjack jogs to the other end of the log at 2.5 m/s relative to the shore, what happens to the log while he is moving?

87 Problem: Two blocks of mass 0.5 kg and 1.5 kg are placed on a horizontal, frictionless surface. A light spring is compressed between them. A cord initially holding the blocks together is burned; after this, the block of mass 1.5 kg moves to the right with a speed of 2.0 m/s. A) What is the speed and direction of the other block? B) What was the original elastic energy in the spring?

88 Conservation of Linear Momentum The linear momentum of a system is conserved unless the system experiences an external force.

89 Collision Review Collisions in 1 and 2 Dimension

90 Collision review In all collisions, momentum is conserved. Elastic Collisions: No deformation occurs Kinetic energy is also conserved. Inelastic Collisions: Deformation occurs Kinetic energy is lost. Perfectly Inelastic Collisions Objects stick together, kinetic energy is lost. Explosions Reverse of perfectly inelastic collision, kinetic energy is gained.

91 1 Dimensional Problem: A 1.5 kg cart traveling at 1.5 m/s collides with a stationary 0.5 kg cart and sticks to it. At what speed are the carts moving after the collision?

92 1 Dimensional Problem: A 1.5 kg cart traveling at 1.5 m/s collides elastically with a stationary 0.5 kg cart. At what speed are each of the carts moving after the collision?

93 1 Dimensional Problem: What is the recoil velocity of a 120- kg cannon that fires a 30-kg cannonball at 320 m/s?

94 For 2-dimensional collisions Use conservation of momentum independently for x and y dimensions. You must resolve your momentum vectors into x and y components when working the problem

95 2 Dimensional Problem: A pool player hits a cue ball in the x-direction at 0.80 m/s. The cue ball knocks into the 8-ball, which moves at a speed of 0.30 m/s at an angle of 35 o angle above the x-axis. Determine the angle of deflection of the cue ball.

96 Collision Problems Practice Day

97 Problem 1 A proton, moving with a velocity of v i i, collides elastically with another proton initially at rest. If the two protons have equal speeds after the collision, find a) the speed of each in terms of v i and b) the direction of the velocity of each.

98 Center of Mass

99 The center of mass of a system is the point at which all the mass can be assumed to reside. Sometimes the system is an assortment of particles and sometimes it is a solid object. Mathematically, you can think of the center of mass as a weighted average.

100 Center of Mass – system of points Analogous equations exist for velocity and acceleration, for example

101 Sample Problem Determine the Center of Mass x y 2 kg 3 kg 1 kg

102 Center of Mass -- solid objects If the object is of uniform density, you pick the geometric center of the object. x x x For combination objects, pick the center of each part and then treat each center as a point. x x

103 Center of Mass Problem Determine the center of mass of this shape. x y 2R

104 Center of Mass for more complicated situations If the shapes are points or simple geometric shapes of constant density, then Anything else is more complicated, and

105 Problem: Find the x-coordinate of the center of mass of a rod of length L whose mass per unit length varies according to the expression = x. L x y

106 Problem: A thin strip of material of mass M is bent into a semicircle of radius R. Find its center of mass. x y R R -R

107 Motion of a System of Particles

108 If you have a problem involving a system of many particles, you can often simplify your problem greatly by just considering the motion of the center of mass. If all forces in the sytem are internal, the motion of the center of mass will not change. The effect of an external force that acts on all particles can be simplified by considering that it acts on the center of mass; that is: F ext = Ma cm

109 Homework problem – collaborative work (15 minutes) This problem is a toughie! However, it gets easier if you use the idea that the center of mass of the system of Romeo, Juliet, and canoe does not change because all forces are internal. The question is paraphrased below: Romeo (77 kg) sits in the back of a canoe 2.70 m away from Juliet (55 kg) who sits in the front. The canoe has a mass of 80 kg. Juliet moves to the rear of the canoe to kiss Romeo. How far does the canoe move forward when she does this? (Assume a canoe that is symmetrical).


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