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PRINCIPLE OF WORK AND ENERGY (Sections 14.1-14.3)

PRINCIPLE OF WORK AND ENERGY (Sections 14.1-14.3). 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 : • Check homework, if any • Reading quiz

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PRINCIPLE OF WORK AND ENERGY (Sections 14.1-14.3)

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  1. PRINCIPLE OF WORK AND ENERGY (Sections 14.1-14.3) 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: • Check homework, if any • Reading quiz • Applications • Work of a force • Principle of work and energy • Concept quiz • Group problem solving • Attention quiz

  2. F 1. What is the work done by the force F ? A) F s B) –F s C) Zero D) None of the above. s1 s2 s 2. If a particle is moved from 1 to 2, the work done on the particle by the force, FR will be A) B) C) D) READING QUIZ

  3. 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, r) to control the forces experienced by the passengers?

  4. 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?

  5. 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 at = v (dv/ds) into Ft= mat, 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.

  6. r2 ò r1 By using the definition of the dot product and integrating, the total work can be written as U1-2 = F • dr 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 q, the increment of work dU done by the force is dU = F ds cos q

  7. If F is a function of position (a common case) this becomes s2 ò F cos q ds U1-2 = s1 WORK OF A FORCE (continued) If both F and q are constant (F = Fc), this equation further simplifies to U1-2 = Fc cos q (s2 - s1) 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.

  8. U1-2 = - W dy = - W (y2 - y1) = - W Dy y2 ò y1 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 The work of a weight is the product of the magnitude of the particle’s weight and its vertical displacement. If Dy is upward, the work is negative since the weight force always acts downward.

  9. The work of the spring force moving from position s1 to position s2 is s2 s2 = ò = ò U1-2 Fs ds k s ds = 0.5k(s2)2 - 0.5k(s1)2 s1 s1 WORK OF A SPRING FORCE When stretched, a linear elastic spring develops a force of magnitude Fs = ks, where k is the spring stiffness and s is the displacement from the unstretched position. If a particle is attached to the spring, the force Fs exerted on the particleisopposite to that exerted on the spring. Thus, the work done on the particle by the spring force will be negative or U1-2 = – [ 0.5k (s2)2 – 0.5k (s1)2 ] .

  10. 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., (ksi)(si). Beware, this is a trap that students often fall into! 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.

  11. PRINCIPLE OF WORK AND ENERGY By integrating the equation of motion,  Ft = mat = mv(dv/ds), the principle of work and energy can be written as  U1-2 = 0.5m(v2)2 – 0.5m(v1)2 or T1 +  U1-2 = T2 U1-2 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. T1 and T2 are the kinetic energies of the particle at the initial and final position, respectively. Thus, T1 = 0.5 m (v1)2 and T2 = 0.5 m (v2)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.

  12. PRINCIPLE OF WORK AND ENERGY (continued) Note that the principle of work and energy (T1 +  U1-2 = T2) 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. 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.

  13. 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 q = 135°. Plan: 1) Draw the FBD of the ball at q = 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.

  14. t N n 45° W => + Fn = man = m (v2/r) => W cos45° = m (v2/r) => (0.5)(9.81) cos 45° = (0.5/1.5)v2 => v = 3.2257 m/s EXAMPLE (continued) Solution: 1) Draw the FBD of the ball at q = 135°. 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 q = 135°, set N = 0.

  15. EXAMPLE (continued) 3) Apply the principle of work and energy between position 1 (q = 0) and position 2 (q = 135°). Note that the normal force (N) does no work since it is always perpendicular to the displacement direction. (Students: Draw a FBD to confirm the work forces). T1 + U1-2 = T2 0.5m (v1)2 – W Dy – (0.5k(s2)2 – 0.5k (s1)2) = 0.5m (v2)2 and v1 = 0, v2 = 3.2257 m/s s1 = s + 0.08 m, s2 = 0.08 m Dy = 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

  16. 2 kg 2 kg CONCEPT QUIZ 1. A spring with an unstretched length of 5 cm expands from a length of 2 cm to a length of 4 cm. The work done on the spring is _________ cm ·N. A) 0.5 k(2 cm)2 B) - [0.5 k(4 cm)2 - 0.5 k(2 cm)2] C) - [0.5 k(3 cm)2 - 0.5 k(1 cm)2] D) 0.5 k(3 cm)2 - 0.5 k(1 cm)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

  17. GROUP PROBLEM SOLVING Given: Block A has a weight of 60 N and block B has a weight of 10 N. The coefficient of kinetic friction between block A and the incline is mk = 0.2. Neglect the mass of the cord and pulleys. Find: The speed of block A after it moves 3 m 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.

  18. sA sB GROUP PROBLEM SOLVING (continued) Solution: 1) The kinematic relationships can be determined by defining position coordinates sA and sB, and then differentiating. Since the cable length is constant: 2sA + sB = l 2DsA + DsB = 0 DsA = 3m => DsB = -6 m and 2vA + vB = 0 => vB = -2vA Note that, by this definition of sA and sB, positive motion for each block is defined as downwards.

  19. T WA 2T y x B A mNA 5 3 NA 4 WB GROUP PROBLEM SOLVING (continued) 2) Draw the FBD of each block. Sum forces in the y-direction for block A (note that there is no motion in this direction): Fy = 0: NA –(4/5)WA = 0 => NA = (4/5)WA

  20. GROUP PROBLEM SOLVING (continued) 3) Apply the principle of work and energy to the system (the blocks start from rest). T1 + U1-2 = T2 (0.5mA(vA1)2 + .5mB(vB1)2) + ((3/5)WA – 2T – mNA)DsA + (WB – T)DsB = (0.5mA(vA2)2 + 0.5mB(vB2)2) vA1 = vB1 = 0, DsA = 3 m, DsB = -6 m, vB = -2vA, NA = (4/5)WA => 0 + 0 + (3/5)(60)(3) – 2T(3) – (0.2)(0.8)(60)(3) + (10)(-6) – T(-6) = 0.5(60/9.81)(vA2)2 + 0.5(10/9.81)(-2vA2)2 => vA2 = 1.94 m/s Note that the work due to the cable tension force on each block cancels out.

  21. 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: • Check homework, if any • Reading quiz • Applications • Define power • Define efficiency • Concept quiz • Group problem solving • Attention quiz

  22. 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

  23. 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?

  24. 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?

  25. 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.

  26. POWER (continued) Using scalar notation, power can be written P = F • v = F v cos q where q 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 .

  27. 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 e = (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 e = (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.

  28. Solving Problems • Find the resultant externalforce 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 q). • 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.

  29. EXAMPLE Given: A sports car has a mass of 2 Mg and an engine efficiency of e = 0.65. Moving forward, the wind creates a drag resistance on the car of FD = 1.2v2 N, where v is the velocity in m/s. The car accelerates at 5 m/s2, 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.

  30. + Fx = max => Fc – 1.2v2 = (2000)(5) => Fc = (10,000 + 1.2v2) N EXAMPLE (continued) Solution: 1) Draw the FBD of the car. The drag force and weight are known forces. The normal force Nc and frictional force Fc 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 ax = 5 m/s2:

  31. EXAMPLE (continued) 3) The constant acceleration equations can be used to determine the car’s velocity. vx = vxo + axt = 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: Po = (Fc)(vx ) = [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. Pi = Po/e = 209.6/0.65 = 322 kW

  32. 1. A motor pulls a 10 N block up a smooth incline at a constant velocity of 4 m/s. Find the power supplied by the motor. A) 8.4 N.m/s B) 20 N.m/s C) 34.6 N.m/s D) 40 N.m/s 30º CONCEPT QUIZ 2. A twin engine jet aircraft is climbing at a 10 degree angle at 264 m/s. The thrust developed by a jet engine is 1000 N. The power developed by the engines is A) (1000 N)(140 m/s) B) (2000 N)(140 m/s) cos 10 C) (1000 N)(140 m/s) cos 10 D) (2000 N)(140 m/s)

  33. GROUP PROBLEM SOLVING Given: A 50-N (= 50-kg) 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 N 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 m. 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 meters. 3) Calculate the power supplied by the motor and to the motor.

  34. sm Here sm is defined to a point on the cable. Also sB is defined only to the lower pulley, since the block moves with the pulley. From kinematics, sm + 2sB = l => vm + 2vB = 0 => vm = -2vB sB 2T Since the pulley has no mass, a force balance requires that the tension in the lower cable is twice the tension in the upper cable. B WB = 50 N GROUP PROBLEM SOLVING (continued) Solution: 1) Define position coordinates to relate velocities. Draw the FBD of the block:

  35. + T1 + U1-2 = T2 0.5m(v1)2 + [2T(s) – wB(s)] = 0.5m (v2)2 0 + [2(30)(10) - (50)(10)] = 0.5(50/9.81)(v2)2 => v2 = vB = 6.26 m/s Since this velocity is upwards, it is a negative velocity in terms of the kinematic equation coordinates. vm = - 2vB = - (2)(-6.26) = 12.52 m/s 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). The velocity of the cable coming into the motor (vm) is calculated from the kinematic equation.

  36. 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: Po = F • v = (30)(12.52) = 375.6 (N · m)/s The power supplied to the motor is determined using the motor’s efficiency and the basic efficiency equation. Pi = Po/e = 375.6/0.76 = 494.2 (N · m )/s

  37. 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 m/s. The power being supplied to the wheels is 52,800 N · m /s. Find the combined friction force on the tires. A) 8.82 N B) 400 N C) 600 N D) 4.64 x 106 N

  38. POTENTIAL ENERGY AND CONSERVATION OF ENERGY (Sections 14.5-14.6) • Today’s Objectives: • Students will be able to: • Understand the concept of conservative forces and determine the potential energy of such forces. • Apply the principle of conservation of energy. • In-Class Activities: • Check homework, if any • Reading quiz • Applications • Conservative force • Potential energy • Conservation of energy • Concept quiz • Group problem solving • Attention quiz

  39. 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.

  40. 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 N and the equivalent spring constant is k = 500 N/m, what is the energy stored in the springs?

  41. 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?

  42. 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. This means the work is conserved. z B F A ò = F d r 0 · y x CONSERVATIVE FORCE A conservative force depends only on the position of the particle, and is independent of its velocity or acceleration.

  43. 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 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 texts. However, even without the use of the these mathematical relationships, much can be understood and accomplished.

  44. 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 Vgravity and Vsprings). It is important to become familiar with the two types of potential energy and how to calculate their magnitudes.

  45. 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 + W y = _ g POTENTIAL ENERGY DUE TO GRAVITY Vg is positive if y is above the datum and negative if y is below the datum. Remember, YOU get to set the datum.

  46. Ve (where ‘e’ denotes an elastic spring) has the distance “s” raised to a power (the result of an integration) or 1 = 2 V ks 2 e 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. Notice that the potential function Ve always yields positive energy.

  47. + = + T V T V 1 1 2 2 = Constant 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 T1 stands for the kinetic energy at state 1 and V1 is the potential energy function for state 1. T2 and V2 represent these energy states at state 2. Recall, the kinetic energy is defined as T = ½ mv2.

  48. CONCEPT QUIZ 1. If the work done by a conservative force on a particle as it moves between two positions is –10 N.m, the change in its potential energy is A) 0 N.m. B) -10 N.m. C) +10 N.m. D) None of the above. 2. Recall that the work of a spring is U1-2 = -½ k(s22 – s12) and can be either positive or negative. The potential energy of a spring is V = ½ ks2 . Its value is A) always negative. B) either positive or negative. C) always positive. D) an imaginary number!

  49. 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).

  50. Placing the datum for gravitational potential at A yields a conservation of energy equation with the left side all zeros. Since Tc equals zero at points A and C, the equation becomes Note that (Vc)g is negative since point C is below the datum. 60 + = + + - - 2 2 2 0 0 0 [ ( (. 75 ) y . 75 ) 2 ( 9 . 81 ) y ] 2 GROUP PROBLEM SOLVING (continued) Solution: Notice that the potential energy at C has two parts (Tc = 0). Vc = (Vc)e + (Vc)g 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.

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