Announcements Test Wednesday Closed book 3 page sheet sheet (on web) Calculator Chap 12.6-10, 13.1-6 Principle of Work and Energy - Sections 14.1-3 Today s Objectives: Students will be able to: a) Calculate the work of a force. b) Apply the principle of work and energy to a particle or system of particles. In-Class Activities: Reading quiz Applications Work of a force Principle of work and energy Concept quiz Group problem solving Attention quiz Engr222 Spring 2004 Chapter 14 1
Reading Quiz 1. What is the work done by the force F? A) F s B) F s F s 1 s 2 C) Zero D) None of the above s 2. If a particle is moved from 1 to 2, the work done on the particle by the force, F R will be s A) 2 Σ Ft ds B) s1 s C) 2 Σ Fn ds D) s1 s 2 s1 Σ Ft ds s 2 s1 Σ Fnds Applications A roller coaster makes use of gravitational forces to assist the cars in reaching high speeds in the valleys of the track. How can we design the track (e.g., the height, h, and the radius of curvature, ρ) to control the forces experienced by the passengers? Engr222 Spring 2004 Chapter 14 2
Applications - continued Crash barrels are often used along roadways for crash protection. The barrels absorb the car s kinetic energy by deforming. If we know the typical velocity of an oncoming car and the amount of energy that can be absorbed by each barrel, how can we design a crash cushion? Work and Energy Another equation for working kinetics problems involving particles can be derived by integrating the equation of motion (F = ma) with respect to displacement. By substituting a t = v (dv/ds) into F t = ma t, the result is integrated to yield an equation known as the principle of work and energy. This principle is useful for solving problems that involve force, velocity, and displacement. It can also be used to explore the concept of power. To use this principle, we must first understand how to calculate the work of a force. Engr222 Spring 2004 Chapter 14 3
Work of a Force A force does work on a particle when the particle undergoes a displacement along the line of action of the force. Work is defined as the product of force and displacement components acting in the same direction. So, if the angle between the force and displacement vector is θ, the increment of work du done by the force is du = F ds cos θ By using the definition of the dot product and integrating, the total work can be U 1-2 = written as r 2 r 1 F dr Work of a Force - continued If F is a function of position (a common case) this becomes U 1-2 s 2 = F cos θ ds s 1 If both F and θ are constant (F = F c ), this equation further simplifies to U 1-2 = F c cos θ (s 2 - s 1 ) Work is positive if the force and the movement are in the same direction. If they are opposing, then the work is negative. If the force and the displacement directions are perpendicular, the work is zero. Engr222 Spring 2004 Chapter 14 4
Work of a Weight The work done by the gravitational force acting on a particle (or weight of an object) can be calculated by using y 2 U 1-2 = y 1 - W dy = - W (y 2 - y 1 ) = - W y The work of a weight is the product of the magnitude of the particle s weight and its vertical displacement. If y is upward, the work is negative since the weight force always acts downward. Work of a Spring Force When stretched, a linear elastic spring develops a force of magnitude F s = ks, where k is the spring stiffness and s is the displacement from the unstretched position. The work of the spring force moving from position s 1 to position s 2 is s 2 s 2 U F k s ds = 0.5k(s 2 ) 2 1-2 = s ds - 0.5k(s 1 ) 2 s 1 = s 1 If a particle is attached to the spring, the force F s exerted on the particle is opposite to that exerted on the spring. Thus, the work done on the particle by the spring force will be negative or U 1-2 = [ 0.5k (s 2 ) 2 0.5k (s 1 ) 2 ]. Engr222 Spring 2004 Chapter 14 5
Spring Forces It is important to note the following about spring forces: 1. The equations just shown are for linear springs only. Recall that a linear spring develops a force according to F = ks (essentially the equation of a line). 2. The work of a spring is not just spring force times distance at some point, i.e., (ks i )(s i ). 3. Always double check the sign of the spring work after calculating it. It is positive work if the force put on the object by the spring and the movement are in the same direction. Principle of Work and Energy By integrating the equation of motion, F t = ma t = mv(dv/ds), the principle of work and energy can be written as U 1-2 = 0.5m(v 2 ) 2 0.5m(v 1 ) 2 or T 1 + U 1-2 = T 2 U 1-2 is the work done by all the forces acting on the particle as it moves from point 1 to point 2. Work can be either a positive or negative scalar. T 1 and T 2 are the kinetic energies of the particle at the initial and final position, respectively. Thus, T 1 = 0.5 m (v 1 ) 2 and T 2 = 0.5 m (v 2 ) 2. The kinetic energy is always a positive scalar (velocity is squared!). So, the particle s initial kinetic energy plus the work done by all the forces acting on the particle as it moves from its initial to final position is equal to the particle s final kinetic energy. Engr222 Spring 2004 Chapter 14 6
Principle of Work and Energy - continued Note that the principle of work and energy (T 1 + U 1-2 = T 2 ) is not a vector equation. Each term results in a scalar value. Both kinetic energy and work have the same units, that of energy. In the SI system, the unit for energy is called a joule (J), where 1 J = 1 N m. In the FPS system, units are ft lb. The principle of work and energy cannot be used, in general, to determine forces directed normal to the path, since these forces do no work. The principle of work and energy can also be applied to a system of particles by summing the kinetic energies of all particles in the system and the work due to all forces acting on the system. Example Given:A 0.5 kg ball of negligible size is fired up a vertical track of radius 1.5 m using a spring plunger with k = 500 N/m. The plunger keeps the spring compressed 0.08 m when s = 0. Find: The distance s the plunger must be pulled back and released so the ball will begin to leave the track when θ = 135. Plan: 1) Draw the FBD of the ball at θ = 135. 2) Apply the equation of motion in the n-direction to determine the speed of the ball when it leaves the track. 3) Apply the principle of work and energy to determine s. Engr222 Spring 2004 Chapter 14 7
Solution: 1) Draw the FBD of the ball at θ = 135. n t 45 W N Example - continued The weight (W) acts downward through the center of the ball. The normal force exerted by the track is perpendicular to the surface. The friction force between the ball and the track has no component in the n-direction. 2) Apply the equation of motion in the n-direction. Since the ball leaves the track at θ = 135, set N = 0. => + F n = ma n = m (v 2 /ρ) => W cos45 = m (v 2 /ρ) => (0.5)(9.81) cos 45 = (0.5/1.5)v 2 => v = 3.2257 m/s Example - continued 3) Apply the principle of work and energy between position 1 (θ = 0) and position 2 (θ = 135 ). Note that the normal force (N) does no work since it is always perpendicular to the displacement direction. T 1 + U 1-2 = T 2 0.5m (v 1 ) 2 W y (0.5k(s 2 ) 2 0.5k (s 1 ) 2 ) = 0.5m (v 2 ) 2 and v 1 = 0, v 2 = 3.2257 m/s s 1 = s + 0.08 m, s 2 = 0.08 m y = 1.5 + 1.5 sin 45 = 2.5607 m => 0 (0.5)(9.81)(2.5607) [0.5(500)(0.08) 2 0.5(500)(5 + 0.08) 2 ] = 0.5(0.5)(3.2257) 2 => s = 0.179 m = 179 mm Engr222 Spring 2004 Chapter 14 8
Concept Quiz 1. A spring with an unstretched length of 5 in. expands from a length of 2 in to a length of 4 in. The work done on the spring is in lb. A) 0.5 k (2 in) 2 B) - [0.5 k(4 in) 2-0.5 k(2 in) 2 ] C) - [0.5 k(3 in) 2-0.5 k(1 in) 2 ] D) 0.5 k(3 in) 2-0.5 k(1 in) 2 2. Two blocks are initially at rest. How many equations would be needed to determine the velocity of block A after block B moves 4 m horizontally on the smooth surface? A) One B) Two C) Three D) Four 2 kg 2 kg Group Problem Solving Given: Block A has a weight of 60 lb and block B has a weight of 10 lb. The coefficient of kinetic friction between block A and the incline is µ k = 0.2. Neglect the mass of the cord and pulleys. Find: The speed of block A after it moves 3 ft down the plane, starting from rest. Plan: 1) Define the kinematic relationships between the blocks. 2) Draw the FBD of each block. 3) Apply the principle of work and energy to the system of blocks. Engr222 Spring 2004 Chapter 14 9
Solution: Group Problem Solving - continued 1) The kinematic relationships can be determined by defining position coordinates s A and s B, and then differentiating. s A s B Since the cable length is constant: 2s A + s B = l 2 s A + s B = 0 s A = 3ft => s B = -6 ft and 2v A + v B = 0 => v B = -2v A Note that, by this definition of s A and s B, positive motion for each block is defined as downwards. 2) Draw the FBD of each block. y Group Problem Solving - continued 5 x 4 3 W A A µn A N A 2T Sum forces in the y-direction for block A (note that there is no motion in this direction): T B W B F y = 0: N A (4/5)W A = 0 => N A = (4/5)W A Engr222 Spring 2004 Chapter 14 10
Group Problem Solving - continued 3) Apply the principle of work and energy to the system (the blocks start from rest). T 1 + U 1-2 = T 2 (0.5m A (v A1 ) 2 +.5m B (v B1 ) 2 ) + ((3/5)W A 2T µn A ) s A + (W B T) s B = (0.5m A (v A2 ) 2 + 0.5m B (v B2 ) 2 ) v A1 = v B1 = 0, s A = 3ft, s B = -6 ft, v B = -2v A, N A = (4/5)W A => 0 + 0 + (3/5)(60)(3) 2T(3) (0.2)(0.8)(60)(3) + (10)(-6) T(-6) = 0.5(60/32.2)(v A2 ) 2 + 0.5(10/32.2)(-2v A2 ) 2 => v A2 = 3.52 ft/s Note that the work due to the cable tension force on each block cancels out. Attention Quiz 1. What is the work done by the normal force N if a 10 lb box is moved from A to B? A) - 1.24 lb ft B) 0 lb ft C) 1.24 lb ft D) 2.48 lb ft N B 2. If a spring force is F = 5s 3 N/m and the spring is compressed by s = 0.5 m, the work done on a particle attached to the spring will be A) 0.625 N m B) 0.625 N m C) 0.0781 N m D) 0.0781 N m Engr222 Spring 2004 Chapter 14 11
Textbook Problem 14- Announcements Engr222 Spring 2004 Chapter 14 12
Power and Efficiency - Section 14.4 Today s Objectives: Students will be able to: a) Determine the power generated by a machine, engine, or motor. b) Calculate the mechanical efficiency of a machine. In-Class Activities: Reading quiz Applications Define power Define efficiency Concept quiz Group problem solving Attention quiz Reading Quiz 1. The formula definition of power is. A) du / dt B) F v C) F dr/dt D) All of the above. 2. Kinetic energy results from. A) displacement B) velocity C) gravity D) friction Engr222 Spring 2004 Chapter 14 13
Applications Engines and motors are often rated in terms of their power output. The power requirements of the motor lifting this elevator depend on the vertical force F that acts on the elevator, causing it to move upwards. Given the desired lift velocity for the elevator, how can we determine the power requirement of the motor? Applications - continued The speed at which a vehicle can climb a hill depends in part on the power output of the engine and the angle of inclination of the hill. For a given angle, how can we determine the speed of this jeep, knowing the power transmitted by the engine to the wheels? Engr222 Spring 2004 Chapter 14 14
Power Power is defined as the amount of work performed per unit of time. If a machine or engine performs a certain amount of work, du, within a given time interval, dt, the power generated can be calculated as P = du/dt Since the work can be expressed as du = F dr, the power can be written P = du/dt = (F dr)/dt = F (dr/dt) = F v Thus, power is a scalar defined as the product of the force and velocity components acting in the same direction. Power - continued Using scalar notation, power can be written P = F v = F v cos θ where θ is the angle between the force and velocity vectors. So if the velocity of a body acted on by a force F is known, the power can be determined by calculating the dot product or by multiplying force and velocity components. The unit of power in the SI system is the watt (W) where 1 W = 1 J/s = 1 (N m)/s. In the FPS system, power is usually expressed in units of horsepower (hp) where 1 hp = 550 (ft lb)/s = 746 W. Engr222 Spring 2004 Chapter 14 15
Efficiency The mechanical efficiency of a machine is the ratio of the useful power produced (output power) to the power supplied to the machine (input power) or ε = (power output)/(power input) If energy input and removal occur at the same time, efficiency may also be expressed in terms of the ratio of output energy to input energy or ε = (energy output)/(energy input) Machines will always have frictional forces. Since frictional forces dissipate energy, additional power will be required to overcome these forces. Consequently, the efficiency of a machine is always less than 1. Solving Problems Find the resultant external force acting on the body causing its motion. It may be necessary to draw a free-body diagram. Determine the velocity of the point on the body at which the force is applied. Energy methods or the equation of motion and appropriate kinematic relations, may be necessary. Multiply the force magnitude by the component of velocity acting in the direction of F to determine the power supplied to the body (P = F v cos θ). In some cases, power may be found by calculating the work done per unit of time (P = du/dt). If the mechanical efficiency of a machine is known, either the power input or output can be determined. Engr222 Spring 2004 Chapter 14 16
Example Given:A sports car has a mass of 2 Mg and an engine efficiency of ε = 0.65. Moving forward, the wind creates a drag resistance on the car of F D = 1.2v 2 N, where v is the velocity in m/s. The car accelerates at 5 m/s 2, starting from rest. Find: The engine s input power when t = 4 s. Plan: 1) Draw a free body diagram of the car. 2) Apply the equation of motion and kinematic equations to find the car s velocity at t = 4 s. 3) Determine the power required for this motion. 4) Use the engine s efficiency to determine input power. Solution: 1) Draw the FBD of the car. Example - continued The drag force and weight are known forces. The normal force N c and frictional force F c represent the resultant forces of all four wheels. The frictional force between the wheels and road pushes the car forward. 2) The equation of motion can be applied in the x-direction, with a x = 5 m/s 2 : + F x = ma x => F c 1.2v 2 = (2000)(5) => F c = (10,000 + 1.2v 2 ) N Engr222 Spring 2004 Chapter 14 17
Example - continued 3) The constant acceleration equations can be used to determine the car s velocity. v x = v xo + a x t = 0 + (5)(4) = 20 m/s 4) The power output of the car is calculated by multiplying the driving (frictional) force and the car s velocity: P o = (F c )(v x ) = [10,000 + (1.2)(20) 2 ](20) = 209.6 kw 5) The power developed by the engine (prior to its frictional losses) is obtained using the efficiency equation. P i = P o /ε = 209.6/0.65 = 322 kw Concept Quiz 1. A motor pulls a 10 lb block up a smooth incline at a constant velocity of 4 ft/s. Find the power supplied by the motor. 30º A) 8.4 ft lb/s B) 20 ft lb/s C) 34.6 ft lb/s D) 40 ft lb/s 2. A twin engine jet aircraft is climbing at a 10 degree angle at 264 ft/s. The thrust developed by a jet engine is 1000 lb. The power developed by the engines is A) (1000 lb)(140 ft/s) B) (2000 lb)(140 ft/s) cos 10 C) (1000 lb)(140 ft/s) cos 10 D) (2000 lb)(140 ft/s) Engr222 Spring 2004 Chapter 14 18
Group Problem Solving Given: A 50-lb load (B) is hoisted by the pulley system and motor M. The motor has an efficiency of 0.76 and exerts a constant force of 30 lb on the cable. Neglect the mass of the pulleys and cable. Find: The power supplied to the motor when the load has been hoisted 10 ft. The block started from rest. Plan: 1) Relate the cable and block velocities by defining position coordinates. Draw a FBD of the block. 2) Use the equation of motion or energy methods to determine the block s velocity at 10 feet. 3) Calculate the power supplied by the motor and to the motor. Solution: Group Problem Solving - continued 1) Define position coordinates to relate velocities. s m s B Here s m is defined to a point on the cable. Also s B is defined only to the lower pulley, since the block moves with the pulley. From kinematics, s m + 2s B = l => v m + 2v B = 0 => v m = -2v B Draw the FBD of the block: 2T Since the pulley has no mass, a force balance requires that the tension in the B lower cable is twice the tension in the upper W B = 50 lb cable. Engr222 Spring 2004 Chapter 14 19
Group Problem Solving - continued 2) The velocity of the block can be obtained by applying the principle of work and energy to the block (recall that the block starts from rest). + T 1 + U 1-2 = T 2 0.5m(v 1 ) 2 + [2T(s) w B (s)] = 0.5m (v 2 ) 2 0 + [2(30)(10) - (50)(10)] = 0.5(50/32.2)(v 2 ) 2 => v 2 = v B = 11.35 ft/s Since this velocity is upwards, it is a negative velocity in terms of the kinematic equation coordinates. The velocity of the cable coming into the motor (v m ) is calculated from the kinematic equation. v m = - 2v B = - (2)(-11.35) = 22.70 ft/s Group Problem Solving - continued 3) The power supplied by the motor is the product of the force applied to the cable and the velocity of the cable: P o = F v = (30)(22.70) = 681 (ft lb)/s The power supplied to the motor is determined using the motor s efficiency and the basic efficiency equation. P i = P o /ε = 681/0.76 = 896 (ft lb)/s Converting to horsepower P i = 896/550 = 1.63 hp Engr222 Spring 2004 Chapter 14 20
Attention Quiz 1. The power supplied by a machine will always be the power supplied to the machine. A) less than B) equal to C) greater than D) A or B 2. A car is traveling a level road at 88 ft/s. The power being supplied to the wheels is 52,800 ft lb/s. Find the combined friction force on the tires. A) 8.82 lb B) 400 lb C) 600 lb D) 4.64 x 10 6 lb Textbook Problem 14- Engr222 Spring 2004 Chapter 14 21
Announcements Potential Energy and Conservation of Energy Sections 14.5-6 Today s Objectives: Students will be able to: a) Understand the concept of conservative forces and determine the potential energy of such forces. b) Apply the principle of conservation of energy. In-Class Activities: Reading quiz Applications Conservative force Potential energy Conservation of energy Concept quiz Group problem solving Attention quiz Engr222 Spring 2004 Chapter 14 22
Reading Quiz 1. The potential energy of a spring is A) always negative. B) always positive. C) positive or negative. D) equal to ks. 2. When the potential energy of a conservative system increases, the kinetic energy A) always decreases. B) always increases. C) could decrease or D) does not change. increase. Applications The weight of the sacks resting on this platform causes potential energy to be stored in the supporting springs. If the sacks weigh 100 lb and the equivalent spring constant is k = 500 lb/ft, what is the energy stored in the springs? Engr222 Spring 2004 Chapter 14 23
Applications - continued When a ball of weight W is dropped (from rest) from a height h above the ground, the potential energy stored in the ball is converted to kinetic energy as the ball drops. What is the velocity of the ball when it hits the ground? Does the weight of the ball affect the final velocity? Conservative Force A force F is said to be conservative if the work done is independent of the path followed by the force acting on a particle as it moves from A to B. In other words, the work done by the force F in a closed path (i.e., from A to B and then back to A) equals zero. F dr = 0 z F B This means the work is conserved. A conservative force depends only on the position of the particle, and is independent of its velocity or acceleration. x A y Engr222 Spring 2004 Chapter 14 24
Conservative Force - continued A more rigorous definition of a conservative force makes use of a potential function (V) and partial differential calculus, as explained in the textbook. However, even without the use of the these mathematical relationships, much can be understood and accomplished. The conservative potential energy of a particle/system is typically written using the potential function V. There are two major components to V commonly encountered in mechanical systems, the potential energy from gravity and the potential energy from springs or other elastic elements. V total = V gravity + V springs Potential Energy Potential energy is a measure of the amount of work a conservative force will do when it changes position. In general, for any conservative force system, we can define the potential function (V) as a function of position. The work done by conservative forces as the particle moves equals the change in the value of the potential function (the sum of V gravity and V springs ). It is important to become familiar with the two types of potential energy and how to calculate their magnitudes. Engr222 Spring 2004 Chapter 14 25
Potential Energy Due to Gravity The potential function (formula) for a gravitational force, e.g., weight (W = mg), is the force multiplied by its elevation from a datum. The datum can be defined at any convenient location. V g = _ + W y V g is positive if y is above the datum and negative if y is below the datum. Remember, YOU get to set the datum. Elastic Potential Energy Recall that the force of an elastic spring is F = ks. It is important to realize that the potential energy of a spring, while it looks similar, is a different formula. V e (where e denotes an elastic spring) has the distance s raised to a power (the result of an integration) or Ve = 1 ks 2 Notice that the potential function V e always yields positive energy. 2 Engr222 Spring 2004 Chapter 14 26
Conservation of Energy When a particle is acted upon by a system of conservative forces, the work done by these forces is conserved and the sum of kinetic energy and potential energy remains constant. In other words, as the particle moves, kinetic energy is converted to potential energy and vice versa. This principle is called the principle of conservation of energy and is expressed as T = + + = Constant 1 V1 T2 V2 T 1 stands for the kinetic energy at state 1 and V 1 is the potential energy function for state 1. T 2 and V 2 represent these energy states at state 2. Recall, the kinetic energy is defined as T = ½ mv 2. Example Given: The girl and bicycle weigh 125 lbs. She moves from point A to B. Find: The velocity and the normal force at B if the velocity at A is 10 ft/s and she stops pedaling at A. Plan: Note that only kinetic energy and potential energy due to gravity (V g ) are involved. Determine the velocity at B using the conservation of energy equation and then apply equilibrium equations to find the normal force. Engr222 Spring 2004 Chapter 14 27
Solution: Example - continued Placing the datum at B: T + V = T + V 1 ( 2 A 125 32.2 )(10) A 2 B V ft B = 45. 1 s B + 125(30) = 1 ( 2 125 32.2 ) v 2 B Equation of motion applied at B: 2 v Fn = man = m ρ N B 125 = 125 32.2 N B = 283lb (45.1) 50 2 Concept Quiz 1. If the work done by a conservative force on a particle as it moves between two positions is 10 ft-lb, the change in its potential energy is A) 0 ft-lb. B) -10 ft-lb. C) +10 ft-lb. D) None of the above. 2. Recall that the work of a spring is U 1-2 = -½ k(s 22 s 12 ) and can be either positive or negative. The potential energy of a spring is V = ½ ks 2. Its value is A) always negative. B) either positive or negative. C) always positive. D) an imaginary number! Engr222 Spring 2004 Chapter 14 28
Group Problem Solving Given: The mass of the collar is 2 kg and the spring constant is 60 N/m. The collar has no velocity at A and the spring is un-deformed at A. Find: The maximum distance y the collar drops before it stops at Point C. Plan: Apply the conservation of energy equation between A and C. Set the gravitational potential energy datum at point A or point C (in this example, choose point A). Group Problem Solving - continued Solution: Notice that the potential energy at C has two parts (T c = 0). V c = (V c ) e + (V c ) g Placing the datum for gravitational potential at A yields a conservation of energy equation with the left side all zeros. Since T c equals zero at points A and C, the equation becomes 0 + 0 = 60 0 + [ ( 2 (.75).75) 2(9.81) y] Note that (V c ) g is negative since point C is below the datum. 2 + y Since the equation is nonlinear, a numerical solver can be used to find the solution or root of the equation. This solving routine can be done with a calculator or a program like Excel. The solution yields y = 1.61 m. 2 2 Engr222 Spring 2004 Chapter 14 29
Group Problem Solving - continued Also notice that since the velocities at A and C are zero, the velocity must reach a maximum somewhere between A and C. Since energy is conserved, the point of maximum kinetic energy (maximum velocity) corresponds to the point of minimum potential energy. By expressing the potential energy at any given position as a function of y and then differentiating, we can determine the position at which the velocity is maximum (since dv/dy = 0 at this position). The derivative yields another nonlinear equation which could be solved using a numerical solver. Attention Quiz 1. The principle of conservation of energy is usually to apply than the principle of work & energy. A) harder B) easier C) the same amount of work D) Don t pick this one. 2. If the pendulum is released from the horizontal position, the velocity of its bob in the vertical position is A) 3.8 m/s. B) 6.9 m/s. C) 14.7 m/s. D) 21 m/s. Engr222 Spring 2004 Chapter 14 30
Textbook Problem 14- Engr222 Spring 2004 Chapter 14 31