Chapter 4 Work, energy, and power

Outline Chapter 4 Work, energy, and power 4. Potential energy & Kinetic energy 4.3 Power By Liew Sau Poh 1 Objectives: (a) define the work done by a force dw = F ds (b) calculate the work done using a force displacement graph (c) calculate the work done in certain situations, including the work done in a spring (d) derive and use the formula: potential energy change = mgh near the surface of the Earth (e) derive and use the formula: kinetic energy = ½ mv Objectives: (f) state and use the work-energy theorem; (g) apply the principle of conservation of energy in situations involving kinetic energy and potential energy; (h) derive and use the formula P = Fv (i) use the concept of efficiency to solve problems. 3 4 Work is done when a force moves an object to a new place Work is done on an object when a force causes a displacement of the object. Work is done only when components of a force are parallel to a displacement. The result of force moving an object. Work is therefore done on the object. 5 6 3.1 Work If the object does not move, than no work has been done. You can try and push the wall for hours, use all that energy, and still not have done any work! Work is a transfer of energy. 7 F The image shows a box being pulled by a constant force along a horizontal surface and moved a displacement d. The force is applied parallel to the surface. The amount of word done is given by W = Fd, where W is the work, F is the force acting in the direction of the displacement, d W = (N)(m) = Joule d 8

The image shows a box being pulled by a constant force along a horizontal surface and moved a displacement d. The force is applied at an angle to the surface. Only the component of the force (F cos ) parallel to the displacement does work in the direction of the displacement. The amount of work done is given by W = F d cos F F x =F cos d 9 Here is an example of a situation where a force is applied and no work is done: A man holds a 50 N weight. No work is done (even though he must exert a force to hold the weight) because there is no displacement parallel to the direction of the weight. 10 Here is an example of a situation where there is a force applied and there is displacement and no work is done against the object's weight: A man walks around the room holding a 50 N weight. The object's weight acts down. For work to be done against the weight there has to be displacement in the direction of the weight (either up or down). There is none so no work is done against the weight. There is work done against friction as the man walks around the room. 11 It is important to specify whether you are talking about work done by an object or work done on an object. For example, work done by gravity depends upon the vertical height. You can push an object up an incline and the amount of work done by gravity is the same for all angles of the incline. 1 Conservative forces: Forces such as gravity, for which the work done does not depend upon the path taken are called conservative forces. Nonconservative forces: Forces such as friction, for which the work done does depend upon the path taken are called nonconservative forces. 13 Since work is scalar, a sign convention must be established for work. Positive work -the displacement and the force are in the same direction. Positive work the work done by a system is positive. Negative work - displacement and the force are in opposite directions. Friction does negative work (the frictional force acts in the direction opposite the motion). Negative work the work done on a system is negative 14 Net work: The net work done on an object determines its motion. If the net work is zero, the object moves at constant speed or is at rest. The object accelerates if the net work has a value other than zero. Orbiting Moon The moon orbiting the earth is an example of when a force is applied and there is no work done. Velocity vector Gravitational force 15 16

Orbiting Moon In the figure, the gravitational force acts inward (it is the source of the centripetal force) and the velocity of the moon is perpendicular to the gravitational force (or in a direction tangent to the circle or orbit). Velocity vector Gravitational force 17 Orbiting Moon The moon's displacement is in the direction of the velocity vector, perpendicular to the gravitational force. Thus, there is no component of the gravitational force parallel to the displacement and the work done by the gravitational force is zero. Since the net work done by gravity is zero, the moon moves at constant speed. 18 Force-displacement (F-s) graph F Force-displacement (F-s) graph F/N Area, A = total work 0 a b s 0 x s/m 19 0 Work done in a spring Equilibrium position, x 0 = 0 Work done, W = Fds = dk = -du; F = - du / ds Smooth surface For a spring with extension, x, due to force, F; -kx = - du / dx ; k = spring (elastic) constant Elastic Potential energy, U = 0 x kx dx = ½ kx Examples : Extension, x F F = -kx Sandy has a mass of 50.0 kg and climbs.0 m. How much work was done during the climb? What was the change in potential energy? w=fd (the force Sandy must supply is to overcome her weight) w=mgd w = (50.0 kg)(9.81 m/s )(.0 m) w = 981 J 1 Example 1: Burt has a mass of 15 kg and stands on a pogo stick. When Burt stands on a pogo stick, the spring compresses 0.050 m. How much work is performed on the spring? w = Fd w = mgd w = (15 kg)(9.81 m/s )(0.050 m) w = 61.3 J Examples 3: A block is pushed.5 m by a net force of 50.0 N in the direction of motion. How much work was done? W = Fd W = (50.0 N)(.5 m) W = 15 J 3 4

4. Potential Energy & Kinetic Energy 4. Potential Energy & Kinetic Energy Three Forms Gravitational Energy Elastic Potential Energy Chemical Potential Energy 5 6 4. Potential Energy & Kinetic Energy Potential Energy is the energy associated with an object because of the position, shape, or condition of the object. Gravitational potential energy is the potential energy stored in the gravitational fields of interacting bodies. Gravitational potential energy depends on height from a zero level. 4. Potential Energy & Kinetic Energy Work must be done on an object to raise it to a higher level above the ground. So the object was given energy. W = PE = mgh gravitational PE = mass acceleration height free-fall 7 8 4. Potential Energy & Kinetic Energy W = PE = mgh its Gravitational Potential Energy. the Gravitational Potential Energy. doubling of the PE. Tripling the Height will increase PE by a factor of 3 Example 1: Work = change in Ep W = PE PE 1 For the same object at two heights W = mgh mgh 1 W = mg(h h 1 ) W = mg( h) NOTE: It is the difference in energy that is important. 9 30 Example : A 0.0 kg apple hangs 7.0 m above the ground from a tree. 1.Calculate the potential energy of the apple..what will be the change in potential energy if the apple falls and lands on a table located.0 m above the ground? Answer: PE = mgh PE = (0.0 kg)(9.81 m/s ) (7.0 m) Example : f PE o PE f = mgh f PE f = 0.0kg(9.80ms - )(.0m) PE f = 3.90 J - 13.7 J 10. J f PE o PE = 13.7J == 14 J 31 3 f -9.8 J mgh o f h o ) ) (.0 m 7.0 m) ) ( 5.0 m)

4. Potential Energy & Kinetic Energy The energy of an object that is due to the is called kinetic energy. Forms of Kinetic Energy Vibrational Due to vibrating Rotational Due to rotation Translational Motion from one place to another Note: When using KE, we are referring to Trans. KE 4. Potential Energy & Kinetic Energy Kinetic energy depends on speed and mass. KE = ½ mv Kinetic Energy = ½ mass x velocity 1 Joule = 1 kg x (m/s) KE is Measured in Joules 33 34 Work-Kinetic Energy Theorem The net work done by all the forces acting on an object is equal to the change in the kinetic energy. The net work done on a body equals its change in kinetic energy. W net KE net work = change in kinetic energy Work net = Change in Kinetic Energy W net = ½ m hammer (v f v i ) Since v f = 0 when the hammer stops, W net = ½ m hammer v i 35 36 Example 1: Calculate the KE of: a. A 10.0 kg ball traveling 5.00 m/s KE = ½ mv KE = ½ (10.0 kg)(5.00 m/s) KE = 15 J b. a 5.00 kg ball traveling 10.0 m/s KE = ½ mv KE = ½ (5.00 kg)(10.0 m/s) KE = 50. J 37 Example : If a 50. kg cart traveling along a horizontal surface 10.0 m/s slides to a halt: a. What is the change in kinetic energy of the cart? f - KE o - ½ mv o - ½ (50. kg)(10.0 m/s) -500 J 38 Example : b. How much work is performed on the cart by the horizontal surface? -500 J c. If the cart is brought to a halt over a distance of 5.00m, what is the frictional force acting on the cart? F = w/d F = -500 J / 5.00 m F = -5.0 X10 N The negative sign indicates that the force opposes the motion of the object Conservation of energy Total amount of energy in a closed system in constant. Initial energy = Final energy (½ mv + mgh) I = (½ mv + mgh) f 39 40

KE=½ mv ; PE= mgh When an object falls it is converting potential energy into kinetic energy. At the bottom of the fall all the potential energy has been changed into kinetic energy. Determine 1) Position(s) of instantaneous rest? ) Position(s) of max velocity? 3) Position(s) of max KE? 4) Position(s) of max PE? 5) Position(s) of min KE? 6) Position(s) of min PE? A B C D E 41 4 KE=½ mv ; PE= mgh Determine 7) Position(s) after which KE increases? 8) Position(s) after which PE increases? 9) Position(s) after which KE decreases? 10) Position(s) after which PE decreases? A B C D E 43 KE=½ mv ; PE= mgh Determine 1) Position(s) of instantaneous rest? ) Position(s) of maximum velocity? 3) Position(s) of maximum KE? 4) Position(s) of maximum PE? 5) Position(s) of minimum KE? B E A D C h O h 44 KE=½ mv ; PE= mgh Determine 6) Position(s) of minimum PE? 7) Position(s) after which KE increases? 8) Position(s) after which PE increases? 9) Position(s) after which KE decreases? 10) Position(s) after which PE decreases? B E A D C h O h 45 Example: 00. kg rock is pushed off of a 00. m cliff. What will be the height of the rock when it falls at a rate of 10.0 m/s? TE o = TE f (Method 1) KE o + PE o = KE f + PE f PE f = PE o KE f mgh f = mgh o ½ mv f h f = (gh o ½ v f ) g h f = (9.80 m/s (00. m)) ½ (10.0 m/s) 9.80 m/s h f = 195 m 46 (Method ) PE o = mgh o PE o = 00. kg(9.80 m/s )(00. m) PE o = 39000 J KE f = ½ mv f KE f = ½ (00. kg)(10.0 m/s) KE f = 10000 J 4.3 Power PE f = PE o KE f PE f = 39000 J 10000 J mgh f = 38000 J h f = 38000 J / mg h f = 38000 J / (00.kg(9.80 m/s)) h f = 195 m 47 48

4.3 Power Power Sources Power is the rate at which energy is used. Power = work / time P = W / t = Fd/t P = Joules / sec = Watt P = ft-lbs / sec = horse power 1kW = 1000W 1 horsepower = 0.75 kw So an engine rated at 134 hp = 100kW 49 50 Example An elevator whose mass is 1,195 kg can carry passengers with up to 935 kg mass. A 4,55 N constant friction force is opposite to the upward movement. a) What is the minimum power required by the engine elevator to make sure to lift the elevator at a constant speed of 4.5 m/s? b) What is the required power when the elevator is designed to develop an upward acceleration of 1.75 m/s and the speed is 4.5 m/s? Solution: a) The free body diagram indicates the engine must develop a force T to rise the elevator. f represents the constant friction force and mg, the elevator weight. Constant speed means zero acceleration, a = 0. Second Newton' Law states: F y = T - f - mg = ma = 0 51 5 Where m is the total mass equal to,130 kg (elevator plus passengers). Then T = f + mg T = 4.55x10 3 N + (.13x10 3 kg)(9.8m/s ) T =.51x10 4 N The power P is: P = Tv = (.51x10 4 N)(4.5 m/s) = 11.3x10 4 W b) Now the engine must do additional work for accelerating the elevator. The only change in the setting of the problem is that a 0 Fy = T - f - mg = ma T = m(a + g) + f = (.13x10 3 kg)(1.75 + 9.8)m/s + 4,55 N =.89x10 4 N The necessary power P is: P = (.89x10 4 N)(4.5 m/s) = 13.0x10 4 Watts 53 54 Efficiency Compares work output to work input Work output / Work input Can never be greater than 100% Machines can never give out more work than is put in Friction reduces efficiency Expressed as a percent Example: 1. A box is slid up an incline with a force of 50N. The length of the incline is 7 meters, and its height is 5 meters. The box weighs 70N. What is the efficiency? 55 56

Solution Work output = FR x dr Work input FE x de = 70N x 5m 50N x 7m = 350 350 = 1 = 100% (FRICTIONLESS) Example:. A box is slid up an incline with a force of 100N. The length of the incline is 7 meters, and its height is 5 meters. The box weighs 70N. What is the efficiency? F R = 70N d R = 5m 57 58 Solution Work output = FR x dr Work input FE x de = 70N x 5m 100N x 7m = 350 700 = 0.5 = 50% 59 Summary Work W = Fs Power P = Fv P = W/t Work, Energy & Power Conservation of Energy / Transformation of Energy Energy Efficiency Kinetic Energy K=1/ mv Potential Energy U= mgh 60