PHYS 210 Chapter 7, Spring 2012

Chapter 7 Chapter 7 Kinetic Energy and Work Kinetic Energy and Work Forms of Energy Forms of Energy  Mechanical Mechanical  focus for now focus for now  chemical chemical  electromagnetic electromagnetic  nuclear nuclear Using Energy Considerations Using Energy Considerations  Energy can be transformed from one Energy can be transformed from one form to another form to another  Essential to the study of physics, Essential to the study of physics, chemistry, biology, geology, astronomy chemistry, biology, geology, astronomy  Can be used in place of Newton’s Can be used in place of Newton’s laws to solve certain problems more laws to solve certain problems more simply simply Work Done by a constant Force Work Done by a constant Force  Provides a link between force and Provides a link between force and energy energy  The work, The work, W W, done by a constant , done by a constant force on an object is defined as the force on an object is defined as the product of the component of the product of the component of the force along the direction of force along the direction of displacement and the magnitude of displacement and the magnitude of the displacement the displacement (Fcos ) x W    Work, cont. Work, cont.   F cos θ is the F cos θ is the component of the component of the force in the force in the direction of the direction of the displacement displacement  Δ x is the Δ x is the displacement displacement (Fcos ) x W    Work, cont. Work, cont.  This gives no information about This gives no information about  the time it took for the displacement to the time it took for the displacement to occur occur  the velocity or acceleration of the object the velocity or acceleration of the object Units of Work Units of Work  SI SI  Newton • meter = Joule Newton • meter = Joule  N • m = J N • m = J  US Customary US Customary  foot • pound foot • pound  ft • lb ft • lb – no special name no special name More About Work More About Work  Scalar quantity Scalar quantity  The work done by a force is zero when the The work done by a force is zero when the force is perpendicular to the displacement force is perpendicular to the displacement  cos 90° = 0 cos 90° = 0  If there are multiple forces acting on an If there are multiple forces acting on an object, the total work done is the algebraic object, the total work done is the algebraic sum of the amount of work done by each sum of the amount of work done by each force force More About Work, cont. More About Work, cont.  Work can be positive or negative Work can be positive or negative  Positive if the force and the Positive if the force and the displacement are in the same direction displacement are in the same direction  Negative if the force and the Negative if the force and the displacement are in the opposite displacement are in the opposite direction direction When Work is Zero When Work is Zero  Displacement is Displacement is horizontal horizontal  Force is vertical Force is vertical  cos 90° = 0 cos 90° = 0 Work Can Be Positive or Work Can Be Positive or Negative Negative  Work done by F is Work done by F is positive when positive when lifting the box lifting the box The Scalar Product of Two The Scalar Product of Two vectors vectors  Scalar product or Scalar product or dot product is a dot product is a mathematical tool. mathematical tool. A.B A.B = |A|| = |A|| B| B| cos( cos())  |A| and |B| are the magnitudes of |A| and |B| are the magnitudes of vectors vectors A A and and B B..   is the angle between is the angle between AA and and BB.. The Scalar Product of Two The Scalar Product of Two vectors vectors  Comparing Comparing A.B A.B = |A||B| = |A||B| cos( cos(), ), with with We see that the work is the dot We see that the work is the dot product between the force and the product between the force and the displacement: displacement: W = W = FF..rr r F W   cos ) (  The Scalar Product of Two The Scalar Product of Two vectors vectors  The Scalar product is commutative: The Scalar product is commutative: A.B A.B = = B.A B.A  The scalar Product obey the distributive The scalar Product obey the distributive law of Multiplication: law of Multiplication: A.(B + C) = A.B + A.C A.(B + C) = A.B + A.C The Scalar Product of Two The Scalar Product of Two vectors vectors  If If AA is perpendicular to is perpendicular to BB, then: , then: A.B A.B = 0 = 0 It follows that the dot product between two It follows that the dot product between two units vectors is: units vectors is: Since the units vectors I, j and k are Since the units vectors I, j and k are mutually perpendicular one to another. mutually perpendicular one to another. Also we have: Also we have: 1 ˆ ˆ ˆ ˆ ˆ ˆ       k k j j i i 0 ˆ ˆ ˆ ˆ ˆ ˆ       k j i k i j The Scalar Product of Two The Scalar Product of Two vectors vectors A k A j A i z y x ˆ ˆ ˆ   A   B k B j B i B z y x ˆ ˆ ˆ         z z y y x x z y x z y x A B A B B A B k B j B i A k A j A i B A           ˆ ˆ ˆ ˆ ˆ ˆ   Consider the following vectors: Consider the following vectors: | cos( ) || |  A B A B A B A B A B z z y y x x          Example: Example:  The vectors The vectors AA and and BB are given by: are given by:  Determine the scalar product Determine the scalar product A.B A.B  Find the angle Find the angle  between between AA and and BB j i 3ˆ ˆ A 2   j i B  ˆ  2 ˆ   Solution: Solution: | cos( ) || | 4  B   A B A   j i 3ˆ ˆ A 2   j i B  ˆ  2 ˆ   1. 2. 4 3 2 )1 2 (    B    A   5.0 5 13 4 cos( )    13 3 2 | 2 2   A  |  5 2 )1 ( | 2 2    B  |   = 60.2° Example: Example:  A particle moving in the xy-plane undergo A particle moving in the xy-plane undergo a displacement in meters, a displacement in meters, as a constant force in Newtons as a constant force in Newtons acts on the particle. acts on the particle. 1. Calculate the work done by 1. Calculate the work done by FF.. 2. Calculate the magnitudes of 2. Calculate the magnitudes of FF and and rr.. W = W = FF..rr = 5 = 5 2 + 2 2 + 2 3 = 16 J 3 = 16 J j i r ˆ 0.3 ˆ 0.2    j i F ˆ 0.2 ˆ 0.5    m r 6.3 13 3 2 | | 2 2      N F 4.5 29 2 5 | | 2 2      Work Done by a varying Work Done by a varying Force Force  Consider a particle being Consider a particle being displaced along the x-axis displaced along the x-axis under the action of a force under the action of a force that varies with position. that varies with position.  In the small interval In the small interval x, x, FFx x can be considered to be can be considered to be constant and the work for constant and the work for the small displacement the small displacement xx:: W = F W = Fxx xx Work Done by a varying Work Done by a varying Force Force  If we imagine that the If we imagine that the FFxx versus versus xx is divided is divided into a large number of into a large number of intervals, the total intervals, the total work done for the work done for the displacement from displacement from xxii to to xxff is given by: is given by:    f i x x Fx x W Work Done by a varying Work Done by a varying Force Force  When the number of When the number of small intervals is very small intervals is very large we can large we can approximate the approximate the discrete sum ( discrete sum () into a ) into a continues integral: continues integral:     f i f i x x x x x x F dx x F lim x  0 Work Done by a varying Work Done by a varying Force Force  The work is the The work is the area under the area under the curve. curve.   f i x x Fxdx W Example: Example:  A force acting on A force acting on a particle varies a particle varies with x, as shown with x, as shown in the adjacent in the adjacent figure. Calculate figure. Calculate the work done by the work done by the force as the the force as the particle moves particle moves from x = 0 to from x = 0 to x=6.0 m. x=6.0 m. Solution: Solution:  The work is the area The work is the area under the curve = under the curve = Area of the Area of the trapezoid: trapezoid: 2 Height Large Base Small Base) Area   ( 25 J 2 5 6 4) W Area      ( Example Example The adjacent Figure a) shows two horizontal forces that act on a block that is sliding to the right across a frictionless floor. Figure b) shows three plots of the block’s kinetic energy K versus time t. Which of the plots best corresponds to the following three situations: (a) F1 = F2, (b) F1 > F2  , (c) F1 < F2       ? Example: Example: In the below figure, a greased pig has a choice of three frictionless slides along which to slide to the ground. Rank the slides according to how much work the gravitational force does on the pig during the descent, greatest first. Work done by a spring: Work done by a spring:  A block being pulled A block being pulled from from xxii = 0 to = 0 to xxff = = xxmax max on on a frictionless surface by a a frictionless surface by a force force FFapp app. The applied . The applied force is equal in force is equal in magnitude and opposite magnitude and opposite in direction to the spring in direction to the spring force F force Fss at all times. at all times. FFss = -kx = -kx   2 max 2 1 kx kxdx Fxdx       max f i x 0 x x s W Work done by a spring: Work done by a spring:  The work done by F The work done by Fapp app = -W = -Wss FFapp app = kx = kx   2max 2 1 kx kxdx Fxdx      max f i x 0 x x app W Kinetic Energy Kinetic Energy  Energy associated with the motion of Energy associated with the motion of an object an object   Scalar quantity with the same units as Scalar quantity with the same units as work work  Work is related to kinetic energy Work is related to kinetic energy 2 mv2 KE 1 Work-Kinetic Energy Work-Kinetic Energy Theorem Theorem  When work is done by a net force on When work is done by a net force on an object and the only change in the an object and the only change in the object is its speed, the work done is object is its speed, the work done is equal to the change in the object’s equal to the change in the object’s kinetic energy kinetic energy   Speed will increase if work is positive Speed will increase if work is positive  Speed will decrease if work is negative Speed will decrease if work is negative KE KE KE W i f net    Work and Kinetic Energy Work and Kinetic Energy  An object’s kinetic An object’s kinetic energy can also be energy can also be thought of as the thought of as the amount of work the amount of work the moving object could moving object could do in coming to rest do in coming to rest  The moving The moving hammer has kinetic hammer has kinetic energy and can do energy and can do work on the nail work on the nail Example: Example:  A 6.0 kg block initially A 6.0 kg block initially at rest is pulled to at rest is pulled to the right along a the right along a horizontal horizontal frictionless surface frictionless surface by a constant by a constant horizontal force of 12 horizontal force of 12 N. Find the speed of N. Find the speed of the block after it has the block after it has moved 3 m. moved 3 m. Solution: Solution:  Neither of the Neither of the vertically directed vertically directed forces (m forces (mgg and and nn) ) does work on the does work on the block, Since the block, Since the displacement is displacement is horizontal. Applying horizontal. Applying the work kinetic the work kinetic energy theorem: energy theorem:  W WFF = F. = F.x = x = KE KE   F. F.x = ½ m(v x = ½ m(vff))2 2 – ½ m(v – ½ m(vii))22 But v But vii = 0 = 0   3.5 m/s 6 12 3 2 m 2.F.Δ. f      v x Transferring Energy Transferring Energy  By Work By Work  By applying a force By applying a force  Produces a Produces a displacement of the displacement of the system system Transferring Energy Transferring Energy  Heat Heat  The process of The process of transferring heat by transferring heat by collisions between collisions between molecules molecules Transferring Energy Transferring Energy  Mechanical Waves Mechanical Waves  a disturbance a disturbance propagates propagates through a medium through a medium  Examples include Examples include sound, water, sound, water, seismic seismic Transferring Energy Transferring Energy  Electrical Electrical transmission transmission  transfer by means transfer by means of electrical current of electrical current Transferring Energy Transferring Energy  Electromagnetic Electromagnetic radiation radiation  any form of any form of electromagnetic electromagnetic waves waves  Light, microwaves, Light, microwaves, radio waves radio waves Notes About Conservation of Notes About Conservation of Energy Energy  We can neither create nor destroy We can neither create nor destroy energy energy  Another way of saying energy is Another way of saying energy is conserved conserved Power Power  Often also interested in the Often also interested in the rate rate at which at which the energy transfer takes place the energy transfer takes place  Power Power is defined as this rate of energy is defined as this rate of energy transfer transfer   SI units are Watts (W) SI units are Watts (W)  Fv t W P   2 2 s m kg s J W    Power, cont. Power, cont.  US Customary units are generally hp US Customary units are generally hp  need a conversion factor need a conversion factor 746 W s 550 ft lb 1hp   The instantaneous power is defined as: