Physics H. Instructor: Dr. Alaa Imam

Transcription

1 Physics H Instructor: Dr. Alaa Imam alaa_y_emam@hotmail.com

2 Chapter 7 KINETIC ENERGY AND WORK Sections 7-2, 7-3, 7-4, 7-5 What is Energy? Kinetic Energy Work Work and Kinetic Energy

3 Important skills from this lecture: 1. Describe the kinetic energy and its relation with velocity 2. Calculate kinetic energy 3. Define the unit of kinetic energy 4. Define work and its unit 5. Evaluate the work done by a constant force 6. Calculate the net work done by several forces 7. Identify the work-kinetic energy theorem

4 What Is Energy? Energy is a scalar quantity associated with the state (or condition) of one or more objects Energy is the ability to make things change A system that has energy has the ability to do work Living organisms need energy for growing & moving Some forms of energy: Thermal Energy Electrical Energy Chemical Energy Nuclear Energy Mechanical Energy (kinetic energy & potential energy)

5 Principle of energy conservation: Energy can be transformed from one type to another, and transferred from one object to another, but the total amount is always the same (energy is conserved) Properties of Energy Scalar quantity Conserved Transferred Measured with Joule In this chapter we focus only on one type of energy (kinetic energy) and one way of transferring energy (work)

6 Kinetic Energy Kinetic energy (K or K.E): the energy associated with the state of motion of an object The faster the object moves, the greater its K.E When the object is stationary, K.E = 0 For an object of mass m, and speed v (v < speed of light) K 1 2 mv2 (kinetic energy). Energy unit in SI is joule (J) 1 joule 1J 1 kg m 2 /s 2.

7 Sample Problem In 1896 in Waco, Texas, William Crush parked two locomotives at opposite ends of a 6.4-km-long track, fired them up, tied their throttles open, and then allowed them to crash head-on at full speed (Fig. 7-1) in front of 30,000 spectators. Hundreds of people were hurt by flying debris; several were killed. Assuming each locomotive weighed N and its acceleration was a constant 0.26 m/s 2,what was the total kinetic energy of the two locomotives just before the collision? 6.4km 3.2km

8 K.E = mv2 v 2 = v o 2 + 2a(x x 0 ) x x 0 = = 3.2km v 2 = 0 + 2(0.26m / s 2 )( ) = 40.8m / s m = N = kg 9.8m / s 2 K.E = mv2 = ( kg)(40.8m / s) 2 = J

9 Work If an object is accelerated à é K.E Energy is transferred to the object If an object is decelerated à ê K.E Energy is transferred from the object Energy transfer means that there is a work (W) done on the object Work W is energy transferred to or from an object by means of a force acting on the object. Energy transferred to the object is positive work, and energy transferred from the object is negative work. Work : the transferred energy doing work : the act of transferring the energy Work has the same units as energy It is a scalar quantity

10 7-5 Work and Kinetic Energy Finding an Expression for Work Let us find an expression Work for work & by Kinetic considering a bead Energy that can slide along africtionless wire that is stretched along a horizontal x axis (Fig. 7-2). A constant force F :, directed at an angle f to the wire, accelerates the bead along the wire. We can relate A bead the of force mass and m slides the tioacceleration along a frictionless with Newton s wire along second a horizontal law, written x for components axis. A along constant the x force axis: F :,, directed at an angle Φ to the wire, accelerates the bead along the wire F Applying Newton s 2 nd x ma x, (7-3) law along the x axis: F x ma x, (1) where m is the bead s mass. As the bead moves : through a displacement : d, φthe : The bead s velocity changes from v 0 to v. as the bead moves through a displacement : : : force changes the bead s velocity from an initial value v d eler, 0 to some other value v. Because the force is constant, we know that the acceleration is also constant. alu Thus, we can Because Eq. F 2-16 is constant to write, àfor a is components constant à we along can the use xthe axis, equation of motion: v 2 v 2 0 2a x d. (2) (7-4) Sub 1 into 2 gives: Solving this equation for a x,substituting 1 2 mv2 1 mv into Eq.7-3,and rearranging then give us F x d. 1 The 1 st term on the 2left mv2 1 side gives 2 mv 0 2 F the final x d. (7-5) kinetic energy K f The first term The on the 2 nd term left side on the of left the gives equation the initial is kinetic the kinetic energy energy K i K f of the bead at the end of the à the displacement left side gives d,and the change the ΔK.E second by the term forceis the kinetic energy K i of the bead at the The start right of side the of the displacement. equation shows Thus, that the change left side is equal of Eq. to F x d7-5 tells us The work W done on the bead by the force: the kinetic energy has been changed by the force, Wand the F x dright side tells us the change is equal to F x d.therefore,the work W done on the bead by the force halliday_c07_ hr.qxd :40 Page 143 F

11 w values displacement, for F x and we use d,we only can the use force this component equation to along calculate the object s the work displacement.the W the force bead component by the force. perpendicular to the displacement does zero work. To calculate the work a force does on an object as the object moves through some displacement, From Fig. 7-2, we use we only see the that force we can component write Falong the object s displacement. The x as F cos f, where f the angle W F x d ulate the work a force does on an object as the object moves through some To calculate the work a force does on an object as the object moves through some ment, force use component only the force perpendicular component to along the the displacement : between the directions of the displacement object s d and displacement. the does force zero F work. : The.Thus, displacement, we use only the force component along the object s displacement. The mponent perpendicular to the displacement does zero work. force component perpendicular to the displacement does zero work. From Fig. 7-2, Wwe Fd see cos that f we can (work write done by F x a as constant F cos force). f, where f is the (7-7) angle : Fig. between 7-2, the see directions that we can of the write displacement F x as = F cos f, d and where the fforce is the F : angle. Thus, : the Because directions the of right the displacement side of this equation d and the is force equivalent F :. Thus, to the scalar (dot) product F : : d,we can also write W Fd cos f (work done by a constant force). (3) (7-7) W Fd cos f (work done by a constant force). (7-7) Because the right side W of F : this : d equation is equivalent to the scalar (dot) product the F : right : (work done by a constant force), (4) (7-8) d,we side can of also this write equation is equivalent to the scalar (dot) product can also write where F is the magnitude of F :. (You may wish to review the discussion of scalar products in WFor Section F : using : 3-8.) d the W (work above Equation F : : d done equations 7-8 (work is especially by a constant to done force), calculate by a constant useful work force), done calculating (7-8) on an the work (7-8) : when F : and object d are by given a force in unit-vector notation. is the where magnitude F is the First, of magnitude F :. the (You force may of must wish F :. (You be to a review constant may wish the force discussion to review (no change of the scalar discussion in of scalar in Section products 3-8.) in Section Equation magnitude 3-8.) 7-8 or Equation is direction especially 7-8 as useful the is especially object for calculating moves) useful the for work calculating the work : : nd when d are F given : and in dsecond, unit-vector are given the in object notation. unit-vector must be notation. rigid (all its parts must move together, in the same direction) This component Small initial does no work. kinetic energy This component This component F Small initial

12 Signs for work: To find the sign of the work done by a force, consider the force vector component that is parallel to the displacement (Eq. 3 ) If 0 <Φ < 90 à cos Φ is +ve à W is + ve If 90 <Φ < 180 à cos Φ is ve à W is ve If Φ = 90 o à W=0 A force does positive work when it has a vector component in the same direction as the displacement, and it does negative work when it has a vector component in the opposite direction. It does zero work when it has no such vector component. F d If F, d are normal Ø=0 o Ø=180 o Ø=90 o

13 Units for work: Work has the SI unit of the joule, From Eq.3, an equivalent unit for joule is the newton.meter (N.m) The corresponding unit in the British system is the foot.pound (ft.lb). 1J 1 kg m 2 /s 2 1N m ft lb. Net work done by several forces: When two or more forces act on an object, W net = ΣW W net is calculated in two ways: 1. Find the work done by each force and then sum those works 2. Find F net of forces. Then use Eq. 3, substituting the magnitude F net for F and also the angle between the directions of F net and d for Φ

14 Work Kinetic Energy Theorem 1 2 mv2 1 2 mv 0 2 F x d. K K f K i W, change in the kinetic energy of a particle net work done on the particle K f K i W, kinetic energy after. the net work is done kinetic energy before the net work the net work done These statements are known as the work kinetic energy theorem for particles These statements are true for both +ve & ve work If W net is +ve à é K.E of the particle s by the work If W net is ve à ê K.E of the particle s by the work e.g., if k i = 5 J, and there is a net transfer of 2 J to the particle à (+ve W net )à k f = 7 J. If there is a net transfer of 2 J from the particle à ( ve W net )à k f = 3 J

15 CHECKPOINT 1 A particle moves along an x axis.does the kinetic energy of the particle increase,decrease, or remain the same if the particle s velocity changes (a) from 3m/s to 2 m/s and (b) from 2m/s to 2m/s? (c) In each situation,is the work done on the particle positive, negative, or zero? (a) K i = 1 2 mv i2 = 1 2 m( 3)2 = 4.5m (b) K i = 1 2 mv i2 = 1 2 m( 2)2 = 2m K f = 1 2 mv f2 = 1 2 m(2)2 = 2m K f = 1 2 mv f2 = 1 2 m(2)2 = 2m K i > K f K i = K f (c) W = K f K i = = 2.5 (c) W = K f K i = 2 2 = 0

16 (W Fd cos f) or changed Eq. 7-8 when (W energy is transferred F d) to it by calculate F 1 and F 2. those works. Since we know the magnitudes and directions of the forces, we choose Eq (12.0 N)(8.50 m)(cos 30.0 ) is Sample Problem 2 (10.0 N)(8.50 m)(cos 40.0 ) J J 153 J. (Answer) cement, therefore, the spies transkinetic energy of the safe. Calculations: We relate the speed to the work done by combining Eqs and 7-1: Calculations: From Eq. 7-7 and the free-body diagram for the safe in Fig. 7-4b,the work done by is W K f K i 1 2 mv f mv i 2. Figure 7-4a shows two industrial spies sliding an initially : stationary kg J, floor safe a displacement d of magnitude 8.50 m, straight toward their truck. The push F : F : 1 of spy 001 is 2 is data, we find that 12.0 N, directed at an angle of 30.0 downward from the horizontal; the pull F : of spy 002 2W 2 v is 10.0 N, directed at 40.0 f above the horizontal. The A A magnitudes m A 2(153.4 J) J. A 225 and kg directions of these forces do not change 1.17 as m/s. the safe moves, and the floor W W 1 W J J and safe make frictionless contact. W 1 F 1 d cos f 1 (12.0 N)(8.50 m)(cos 30.0 ) and the work done by The initial speed v i is zero, and we now know that the work done is J. Solving for v f and then substituting known W 2 F 2 d cos f 2 (10.0 N)(8.50 m)(cos 40.0 ) Thus, the net work W is J 153 J. During the 8.50 m displacement, Spy 002 therefore, the spies transfer 153 J of energy Spy to 001 the kinetic energy O Only force of components the safe. pa do F : 1 parallel to the displacement do work. (Answer) (Answer) (c) end The cha Cal com The don data (a) d d Safe F N F g F (b) F 1 Spy 001 Fig. 7-4 (a) Two spies move a floor safe through a : displacement d. (b) A free-body diagram for the safe. Additional examples, video, and practice available at WileyPLUS Additional examples, video, and

17 Fig. 7-4b,the izontal; work done the by pull F 1 Fis 2 of spy 002 is 10.0 N, directed at 40.0 A A above the horizontal. The magnitudes The and speed directions of the safe of changes because its kinetic energy is KEY IDEA 1 F 1 d cos f these 1 (12.0 N)(8.50 m)(cos 30.0 ) forces do not change as the safe moves, changed and when the (W energy floor is transferred F : : d ) to it by F : and F : J, and safe make frictionless contact. Because these forces are constant in rk done by F : forces, we choose Eq. Calculations: 7-7. We relate (a) What is the net work done on the safe by forces F : rection, the speed we to can the find work the work done they by do 2 is and Spy 002 : Spy 001 O the F : Calculations: From combining Eq. 7-7 Eqs. and and the 7-1: free-body diagram fo 2 F 2 d cos f2 2 during (10.0 the N)(8.50 safe displacement m)(cos in Fig ) d? 7-4b,the work done by Calculations: F : pa ner W K Thus, with mg as the f K i J. tational mv is 2 f 2 1 force, mv 2 i 2. W do Calculations: From Eq F and the 1 d cos f free-body 1 (12.0 N)(8.50 m)(cos we write 30.0 ) The initial diagram speed for v the safe in Fig. 7-4b,the KEY work done IDEAS by F : i is zero, and we now know that the work et work W is J, 1 is done is J. Solving for v W g mgd cos 90 m f and then substituting known and the work done by W 1 W(1) 2 The net WJ 1 work F 1 dw cos done J f 1 on (12.0 the safe N)(8.50 by data, F : the m)(cos two we 2 isfind 30.0 ) forces that is the and W N F N d cos 90 F e a ody sum of the works they W J 153 J. J, do 2 individually. F (Answer) 2 d cos (2) fbecause 2 (10.0 we can N)(8.50 m)(cos 40.0 ) 2W We should have known d this result. B treat the safe as a particle and the forces are constant v f in and the work done by F : (a) ddit 8.50 m displacement, therefore, the perpendicular to the displacement o both magnitude and direction, 2 isspies transenergy to the kinetic energy of the safe. work on the safe and do not transfer A m 2(153.4 J) J. A 225 kg Thus, the net work we can W use is either Eq. 7-7 (W Fd Wcos 2 f) F 2 dor cos Eq. f (10.0 (W N)(8.50 F : : 1.17 m/s. (Answer) W W 1 d ) m)(cos Wto 2 calculate 40.0 ) thosej J works. Since we know J. the magnitudes and directions of the (c) The safe is initially stationary. W J 153 J. (Answer forces, we choose Eq end of the 8.50 m displacement? During the 8.50 m displacement, Spy 002 therefore, the spies trans Thus, the fer net 153 work W J is of energy Spy to 001 the kinetic energy Only force of components the safe. Calculations: W W From 1 W Eq and J the free-body J diagram for parallel to the displacement the safe in Fig. 7-4b,the work done by F : KEY IDEA 1 is do work J 153 J. (Answer) The speed F of the safe changes beca W 1 F 1 d cos f 1 (12.0 N)(8.50 m)(cos 30.0 ) N F changed when 2 energy is transferred Safe J, 30.0 and the work done by F : Calculations: We relate the speed. 7-4 (a) Two spies move a floor safe through 2 is a d F g F 1 : combining Eqs and 7-1: placement d. (b) A free-body diagram for the safe. (a) (b) W 2 F 2 d cos f 2 (10.0 N)(8.50 m)(cos 40.0 ) W K f K i J. mv f 2 Fig. 7-4 (a) Two spies move a floor safe through a : displacement d. (b) A free-body diagram for the safe. Additional examples, video, and practice available at The WileyPLUS initial speed v i is zero, and we n Thus, the net work W is done is J. Solving for v f and t treat the safe as a particle and the forces are constant in both magnitude and direction, we can use either Eq. 7-7 (W Fd cos f) or Eq. 7-8 to calculate those works. Since we know the magnitudes and directions of the

18 (b) During the displacement, what is the work W g done on the safe by the gravitational force F : g and what is the work W N done on the safe by the normal force F : N from the floor? W g mgd cos 90 mgd(0) 0 (Answer) and W N F N d cos 90 F N d(0) 0. (Answer) (c) The safe is initially stationary. What is its speed v f at the end of the 8.50 m displacement? W K f K i 1 2 mv f mv i 2. The initial speed v i is zero, and we now know that the work done is J. Solving for v f and then substituting known data, we find that v f A 2W m 1.17 m/s. A 2(153.4 J) 225 kg (Answer)

19 Sample Problem During a storm, a crate of crepe is sliding across a slick, : oily parking lot through a displacement d ( 3.0 m)î while a steady wind pushes against the crate with a force F : (2.0 N)î ( 6.0 N)ĵ. The situation and coordinate axes are shown in Fig (a) How much work does this force do on the crate during the displacement? The parallel force component does negative work, slowing the crate. y d F x W F : d : [(2.0 N)î ( 6.0 N)ĵ] [( 3.0 m)î]. W (2.0 N)( 3.0 m)î î ( 6.0 N)( 3.0 m)ĵˆ îˆ ˆ ˆ ( 6.0 J)(1) J. (Answer) The force does a negative 6.0 J of work on the crate, transferring 6.0 J of energy from the kinetic energy of the crate

20 (b) If the crate has a kinetic energy of 10 J at the beginning : : of displacement d, what is its kinetic energy at the end of d? K f K i W 10 J ( 6.0 J) 4.0 J.

21 Examples: Q.1: 5kg block moves with a speed of 72km/h. Its kinetic energy is: (a) 900kg.m 2 /s 2 (b) 1000kg.m 2 /s 2 (c) 1200kg.m 2 /s 2 (d) 50kg.m 2 /s 2 v = 72km / h = = 20m / s K = 1 2 mv2 = 1 2 (5)(20)2 = 1000J Q.2: A 5kg block moves with velocity v=6i + 8j m/s. Its kinetic energy is: (a) 250J (b) 400J (c) 540J (d) 180J v = = 10m / s K = 1 2 mv2 = 1 2 (5)(10)2 = 250J Q.3: 1 joule is equal to: (a) kg.m 2 /s (b) kg.m/s 3 (c) kg.m/s 2 (d) kg.m 2 /s 2

22 Q.4: A particle moves 5m in the positive x-direction while being acted upon by a constant force F = 2i + 2j. The work done on the particle by this force is: (a) 20J (b) 10J (c) 30J (d) 15J W = F d = (2î + 2 ĵ) 5î = 10J Q.5: A force acts on a 3kg particle in such away that the position of the object is x = 3t 4t 2 + t 3, where x in meters and t in second. Find the work done on the object by the force from t = 0 to t = 4s (a) 528J (b) 10J (c) 50J (d) 180J v(t) = dx 2 = 3 8t + 3t dt v(0) = 3m / s, v(4) = 3 8(4) + 3(16) = 19m / s W = 1 2 m(v 2 2 v 2 1 ) = 3 (361 9) = 528J 2

23 Q.6: Force F acts on a particle of mass m making a displacement D. If F = 7i + 3j 1.5k N, and D = 2i 3j + 2.5k m. The work done by the force is: (a) 9.25J (b) 7.25J (c) 5.25J (d) 1.25J W = F d = (7î + 3ĵ 1.5 ˆk) (2î 3ĵ ˆk) = = 1.25J Q.7: A 5kg cart is moving horizontally at 6m/s. In order to change its speed to 10m/s, the net work done on the cart must be: (a) 40J (b) 90J (c) 160J (d) 400J v 1 = 6m / s, v 2 = 10m / s W = 1 2 m(v 2 2 v 2 1 ) = 5 (100 36) = 160J 2

24 Chapter 7 KINETIC ENERGY AND WORK Sections 7-6, 7-7, 7-9 Work Done by the Gravitational Force The Work Done by a Spring Force Power

25 Important skills from this lecture: 1. Calculate the amount of work done by gravitational force in both raising and falling object 2. Define the spring force and its relationship with displacement of the spring 3. Calculate the spring force from Hook s law 4. Define the power and its unit 5. Calculate the average and instantaneous power 6. Calculate the power in terms of force exerted on a body and its velocity

26 6.0 N)( 3.0 m)ĵ î tor notation, we choose Eq K f K i W 10 J ( 6.0 J) 4.0 J. (Answer) KEY IDEA 7-6 Work Done Less by the Gravitational Force Because kinetic the energy force means does negative that the work crate on has the been crate, slowed. it reduces work the done crate s We next Work examine the Done by on kinetic an the object energy. Gravitational by the gravitational force Force acting on J. (Answer) ( 6.0 N)ĵ] [( 3.0 m)î]. ditional examples, it. Figure 7-6 video, shows and a particle-like practice available tomato of at mass WileyPLUS m that is thrown upward with ot products, only î î, ĵ ĵ, and Calculation: Using the work kinetic energy initial If tomato speed of v 0 mass and thus m is with thrown initial upward kinetic with energy initial K i speed 1 theorem in. vas the tomato o rises, it is slowed by a gravitational force F : 2 mv2 0 ix E). Here we obtain the form of Eq. 7-11, we have ; that is, the tomato s kinetic energy K f decreases Its speed because is slowed F : g î ( 6.0 N)( 3.0 m)ĵ î does K f to work K v i W by on the the 10 J gravitational tomato ( 6.0as J) rises. 4.0 J. force Because (Answer) F g à F : g can treat 6.0 J. the tomato (Answer) g does as a work particle, Less on kinetic we the can tomato energy use means Eq. 7-7 that (Win the the crate Fd kinetic cos has been f) energy to slowed. express of the t the work gative 7-6done Work àduring Done 0 The a tomato s a displacement by the Gravitational : K.E decreases d. For the force from magnitude Force K to K f ( 1 i ( 1 F, we use mg as 2 mv2 0) 2 mv2 d ). the magnitude examples, of F : g.thus,the video, and work practice W g done available by the at gravitational WileyPLUS force F : g is peed orce Additional does negative. We next examine work done on an object by the gravitational force acting on, decreasing it. Figure speed The 7-6 gravitational shows work is: inetic energy. W a particle-like g mgd cos f tomato of mass m that is thrown upward with (work done by gravitational force). (7-12) initial speed v 0 and W thus g mgd with cos initial f kinetic (work energy done by Kgravitational i 1. As force). the tomato For a rising object, force F : : rises, it is slowed by a gravitational g is force directed F : 2 mv2 0 ; that opposite is, the the tomato s displacement kinetic energy d, as indicated in because Fig. 7-6.Thus, F : g K i decreases 7-6 Work Done by the Gravitational Force g does f work 180 on and the tomato For a rising object, the direction of F : as it rises. Because we can treat gis opposite to the displacement the We tomato next examine as a particle, the work àφ = 180 àw we g can done mgduse an cos Eq. object (W by the mgd( 1) Fd gravitational cos 0 a mgd. f) to force express acting the on work ive it. Figure 7-6 shows a particle-like : tomato of mass m that is thrown upward with (7-13) done during a displacement d. For the force magnitude F, we use mg as the magnitude of F : v 0 and thus with initial kinetic energy K gravitational force acting al edforce The initial minus speed sign tells W g that mgd during cos the 180 object s mgd( 1) rise, the mgd. of mass g.thus,the work W g done by the gravitational i 1. force F : As the tomato on rises, the it object is slowed The transfers by a gravitational sign means: energy F : g is g transfers in the force amount F : 2 mv2 0 ; that energy mgd is, the in the from tomato s amount the kinetic energy mgd from energy the of object the : he ity gravitational v 0 to decreases object.this force because F : g g does work on the tomato as it rises. Because we can treat à ve is Wconsistent work à slowing with 0 athe of slowing the object of the as object rises as it rises. e-like kinetic tomato the of mass tomato as a g particle, mgdwe cos can f use (work Eq. 7-7 done (Wby gravitational Fd cos f) to force). express the work (7-12) negative After the object has reached : its maximum height and is falling back down, : ws change from velocity done v during to a displacement d : : the angle For 0 a ffalling between object, and in displacement the same direction d is zero. of Thus, from For a rising object, force F :. For the force magnitude F, we use mg as the : magnitude kinetic of F : g.thus,the work W g done is directed by the gravitational opposite the force displacement F : d, as indi- g speed the displacement lacement d. A g gy. es the resulting cated change in àφ Fig. = 7-6.Thus, 0 à f W g 180 mgd andcos 0 mgd( 1) mgd. (7-14) W of the tomato, from g mgd cos f (work done by gravitational force). (7-12) W 1 g mgd cos 180 mgd( 1) mgd. (7-13) 2 mv2 ). For The a rising + sign object, means: force F : : g is transfers directed energy opposite in the the displacement amount mgd d, as to indicated minus à sign Fig. +ve 7-6.Thus, tells work that àf speeding during 180 andthe up object s of the object rise, the as gravitational falls force acting the object orce The ass on the object transfers energy in the amount mgd from the kinetic energy of the : W v 0 to g mgd cos 180 mgd( 1) mgd. (7-13) object.this is consistent with the slowing of the object as it rises. netic nal force The minus sign tells us that during the object s rise, the gravitational force acting F g F g F g v v 0

27 Work Done by the Gravitational Force Rising: 1. Object s kinetic energy decreases 2. Object deceleration 3. The gravitational force dose ve work 4. Φ = W g mgd cos 180 mgd( 1) mgd. Falling: 1. Object kinetic energy increases 2. Object acceleration 3. The gravitational force dose positive work on it 4. Φ = 0 5. W g mgd cos 0 mgd( 1) mgd.

28 Sample Problem One of the lifts of Paul Anderson in the 1950s remains a record: Anderson stooped beneath a reinforced wood platform, placed his hands on a short stool to brace himself, and then pushed upward on the platform with his back, lifting the platform straight up by 1 cm. The platform held automobile parts and safe filled with lead, with a total weight of 27900N. (a) As Anderson lifted the load, how much work was done on it by the gravitational force F g? mg = 27900N, d = 1cm = 0.01m, φ = 180 W g = mgd cosφ = (27900)(0.01)cos180 = 289J

29 The Spring Force A spring force: is the force from a spring, It is variable force Many forces in nature have the same mathematical form as the spring force When an object (block) is attached to the spring free end, and a force acts on it à 3 states of a spring: x = 0 F x = 0 Block attached to spring d F s x negative F x positive x positive F x negative F s d Relaxed (neither compressed nor extended) x = 0 is called Equilibrium position 0 x x 0 Compressed (c) By pushing the block to the left, the spring now pushes on the block toward the right x stretched By pulling the block to the right, the spring pulls on the block to the left (Restoring force) 0 x x

30 Figure 7-9a shows a spring in its relaxed state th extended. One end is fixed, and a particle-like obje fo The spring force F : is proportional to the other, to free the end. displacement If we stretch : the of spring by p s d the free end from posi its equilibrium in Fig. 7-9b, position the spring pulls on the block towar ing force acts to restore the relaxed state, it is sometim The spring force is given If bywe F : scompress kd : the (Hooke s spring law), by pushing the block spring now pushes on the block toward the right. F x kx (Hooke s law), To a good approximation for many springs, the The sign means: F : : sis in portional the opposite to the direction displacement of d of the free end fro posi is in the relaxed state.the spring force is given by The constant k is called the spring constant (or force constant) F : s kd : (Hooke s la k measures the stiffness of the spring The larger k is, the stiffer which the spring is known as Hooke s law after Robert Hoo à the stronger the spring s late pull 1600s. or push The minus for a given sign in Eq indicates th The SI unit for k is the N/m force is always opposite the direction of the displac If the spring is stretched The toward constant the right k is called the spring constant (or for à x is +ve à F x is ve (it of is the a pull stiffness toward of the left) spring.the larger k is, the stiff If the spring is compressed k is, toward the stronger the left the spring s pull or push for a give à x is ve à F x is +ve (it is a push toward the right) k is the newton per meter. A spring force is a variable force because it is a function of x, à F x is wrote as F(x) In Fig. 7-9 an x axis has been placed parallel to Hooke s law is a linear relationship the origin (x between 0) at Fthe x and position x of the free end w state. For this common arrangement, we can write E

31 The Work Done by a Spring Force To find the work done by the spring force we assume: The spring is massless The spring is an ideal spring (obeys Hooke s law) The contact between the block and the floor is frictionless It is not possible to find the work by using W = Fd cos Φ? Because F x is not constant (variable force) To find the work done by the spring, we use calculus: 1. If the block s initial position is x i & later position is x f 2. The distance between those two positions is Divided into many segments of tiny length Δx 3. The force magnitude (F x1 in the 1 st segment, F x2 in the 2 ed segment, and so on) is constant within each segment 4. à the work done within each segment is found using the relation W = Fd cos Φ, Φ = 180 à cos Φ = 1

32 5. The work done is F x1 Δx in segment 1, F x2 Δx in segment 2, and so on 6. The net work W s done by the spring, from x i to x f, is the sum of all these works: W s F xj x, 7. When Δx à 0 à W s x f 8. If x i = 0 and if we call the final position x, x i W s x f x i F x dx. elds W s 1 2 kx i kx f 2 kx dx k xf x i x dx (work by a spring force). Work W s is positive if the block ends up closer to the relaxed position (x 0) than it was initially. It is negative if the block ends up farther away from x 0. It is zero if the block ends up at the same distance from x 0. W s 1 2 kx2 ( 1 2 k)[x2 x ] f xi ( 1 2 k)(x f 2 x i2 ). (work by a spring force).

33 CHECKPOINT 2 For three situations, the initial and final positions, respectively, along the x axis for the block in Fig. 7-9 are (a) 3cm,2cm;(b) 2cm,3cm;and (c) 2cm,2cm.In each situation, is the work done by the spring force on the block positive, negative, or zero? x = 0 F x = 0 Block attached to spring (a) W = 1 2 Kx 2 i 1 2 Kx 2 f 0 (a) x W = 1 2 K ( 3)2 2 W = 1 2 K [ 9 4 ] W = +ve ( ) 2 x positive F x negative d d F s x 0 (b) F s x negative F x positive x W = 1 2 Kx 2 i 1 2 Kx 2 (b) f W = 1 2 K ( 2)2 3 W = 1 2 K [ 4 9 ] W = ve ( ) 2 (c) W W W W = = = = 1 Kx 2 1 K 2 1 K i [ ( 2) ( 2) ] [ 4 4] Kx 2 f x 0 (c) x

34 Sample Problem A package of spicy Cajun pralines lies on a frictionless floor, attached to the free end of a spring in the arrangement of Fig.7-9a. A rightward applied force of magnitude F a = 4.9 N would be needed to hold the package at x 1 = 12mm. (a) How much work does the spring force do on the package if the package is pulled rightward from x 0 = 0 to x 2 = 17 mm? x = 0 F x = 0 Block attached to spring F S = k x k = F S = 4.9 N x m = 408 N / m 0 (a) x W s = 1 2 kx 2 2 = 1 2 (408 N / m)(17 x 10 3 m) 2 = J x positive F x negative F s d (b) Next, the package is moved leftward to x 3 = 12 mm. How much work does the spring force do on the package during this displacement? Explain the sign of this work. d 0 (b) F s x x negative F x positive x W s = 1 2 kx f kx i 2 = 1 2 (408 N / m) ( m) 2 + ( m) 2 x 0 (c) x = J = 30 mj

35 Power Power (P): the time rate at which work is done by a force If a force does a work W in a time interval Δt à the average power due to the force during that time interval is: P avg W t (average power). The instantaneous power P is: P dw dt (instantaneous power). The SI unit of power is: J/s, and called watt (W) 1 watt 1W 1J/s ft lb/s 1 horsepower 1 hp 550 ft lb/s 746 W. Work can be expressed as power multiplied by time, as in the unit kilowatt-hour 1 kilowatt-hour 1 kw h (10 3 W)(3600 s) J 3.60 MJ.

36 For a particle moves along a straight line on an x axis, and a constant force F acted on it with an angle Φ P dw dt P Fv cos f. F cos dx dt F cos dx dt, P F : v : (instantaneous power). CHECKPOINT 3 A block moves with uniform circular motion because a cord tied to the block is anchored at the center of a circle.is the power due to the force on the block from the cord positive, negative, or zero? φ = 90 P = F.v = Fvcos90 = 0 v F

37 Sample Problem Figure 7-14 shows constant forces F : and F : 1 2 acting on a box as the box slides rightward across a frictionless floor. Force F : is horizontal, with magnitude 2.0 N; force F : 1 2 is angled upward by 60 to the floor and has magnitude 4.0 N. The speed v of the box at a certain instant is 3.0 m/s. What is the power due to each force acting on the box at that instant, and what is the net power? Is the net power changing at that instant? Negative power. (This force is removing energy.) Frictionless F 1 F 2 60 Positive power. (This force is supplying energy.) v

38 A block moves Calculation: We use Eq for each force. For force F : with uniform 1, chored at the center of a circle at angle f to velocity v:, we have positive, negative, or zero? P 1 F 1 v cos f 1 (2.0 N)(3.0 m/s) cos 180 force. For force F : 6.0 W. (Answer) This negative result tells us that force F : 1 is transferring en- ergy from the box at the rate of 6.0 J/s. : For force, at angle f 2 60 to velocity v:, we have F : 2 CHECKPOINT 3 P 2 F 2 v cos f 2 (4.0 N)(3.0 m/s) cos W. (Answer) This positive result tells us that force F : 2 is transferring energy to the box at the rate of 6.0 J/s. The net power is the sum of the individual powers: P net P 1 P W 6.0 W 0, (Answer) which tells us that the net rate of transfer of energy to or from the box is zero.thus, the kinetic energy (K 1 2 mv2 ) of the box is not changing, and so the speed of the box will remain at 3.0 m/s. With neither the forces F : and F : 1 2 nor the velocity v: changing, force. we see from For Eq that force P and P aref :

39 Examples: Q.1: A horizontal force of 180N used to pull a 50kg box on a rough horizontal surface to the right with a distance of 8m. If the box moves at a constant speed, find: (a) The work done by the horizontal force F W = Fd cosφ = (180)(8)cos0 = 1440J F N (b) The work done by the friction force f k F Constant speed à a x = 0 F net,x = ma x = 0 F f k = 0 F = f k = 180N W = f k d cos180 = 180(8)( 1) = 1440N mg (b) The work done by the force of gravity W g = mgd cos90 = 0 (c) The work don by the normal force W N = mgd cos90 = 0

40 Q.2 A watt is equal to: (a) kg.m 2 /s 2 (b) kg.m 2 /s 3 (c) kg.m/s 3 (d) kg.m 3 /s 2 P = W t = kg m2 s 2 s = kg m2 s 3 Q.3: Horse-power (hp)= (a) 1000W (b) 100W (c) 749W (d) 476W Q.4: Which of the following group does not contain a scalar quantity? (a) velocity, force, power (b) displacement, acceleration, force (c) acceleration, speed, work (d) energy, work, distance Q.5: An object of mass 1kg moves in a horizontal circle of radius 0.5m at a constant speed of 2m/s. The power done on the object during on revolution is: (a) 1J (b) 2J (c) 4J (d) zero φ = 90 P = F.v = Fvcos90 = 0

41 Q.6: A block of mass 0.5kg is dropped from a height of 45m above the ground. The work done by the gravitational force is: (a) 5J (b) 40J (c) 10J (d) 220.5J W g = mgd cosφ = 0.5(9.8)(45)cos0 = 220J motion Q.7: In the previous question, the average power during the time interval of 10s is: (a) 20W (b) 22W (c) 10W (d) 5W P = W Δt = = 22W F g Q.8: A spring of k = 300N/m initially at x =0, and forced to move to x = 10cm. The work done by the spring force is: (a) 1.5 J (b) 5.5 J (c) 1 J(d) 1.5 J W = 1 2 kx2 = 1 2 (300)(0.1)2 = 1.5J

42 Q.9: A mass of 100kg is pushed by a horizontal force across rough horizontal floor at a constant speed of 5m/s. If μ k =0.2, at what rate is work being done by the horizontal force F? (a) 50W (b) 980W (c) 392W (d) 400W Constant speed à a x = 0 F net,x = ma x = 0 F f k = 0 F = f k = µ k F N = µ k mg F = 0.2(100)(9.8) = 196N P = Fvcosφ = 196(5)(1) = 980W f k F N mg F

43 Problems: 15 Figure 7-27 shows three forces applied to a trunk that moves leftward by 3.00 m over a frictionless floor. The force magnitudes are F N, F N, and F N, and the indicated angle is u During the displacement, (a) what is the net work done on the trunk by the three forces and (b) does the kinetic energy of the trunk increase or decrease? F 2 F 1 θ F 3 Fig Problem SSM ILW A 100 kg block is pulled at a constant speed of 5.0 m/s across a horizontal floor by an applied force of 122 N directed 37 above the horizontal. What is the rate at which the force does work on the block?

44 Problems for chapter 7 15 Figure 7-27 shows three forces applied to a trunk that moves leftward by 3.00 m over a frictionless floor. The force magnitudes are F N, F N, and F N, and the indi- F 1 cated angle is u During the displacement, (a) what is the net work done on the trunk by the three forces and (b) does the kinetic F energy of the trunk increase or decrease? (a) The forces are constant, so the work done by any one of them is given by W F d, where d is the displacement. Force F1 is in the direction of the displacement, so W Fdcos (5.00 N)(3.00 m)cos J Force F 2 makes an angle of 120 with the displacement, so ϕ 2 = 180 o 60 o =120 o F 2 θ Motion direction W F dcos (9.00 N)(3.00m)cos J Force F 3 is perpendicular to the displacement, so The net work done by the three forces is W 3 = F 3 d cos 3 = 0 since cos 90 = 0. W W1 W2 W J 13.5 J J. (b) If no other forces do work on the box, its kinetic energy increases by 1.50 J during the displacement.

45 45 SSM ILW A 100 kg block is pulled at a constant speed of 5.0 m/s across a horizontal floor by an applied force of 122 N directed 37 above the horizontal. What is the rate at which the force does work on the block? 45. The power associated with force F is given by P F v, where v is the velocity of the object on which the force acts. Thus, P F v Fv 2 cos (122 N)(5.0 m/s)cos W.