12/04/32 Chapter 7. Kinetic Energy and Work 7.1. What is Physics? 7.2. What Is Energy? 7.3. Kinetic Energy 7.4. Work 7.5. Work and Kinetic Energy 7.6. Work Done by the Gravitational Force 7.7. Work Done by a Spring Force 7.8. Work Done by a General Variable Force 7.9. Power 1 What is Physics? 2 1 12/04/32 Chapter 7 Kinetic Energy and Work In this chapter we will introduce the following concepts: Kinetic energy of a moving object Work done by a force Power In addition, we will develop the work-kinetic energy theorem and apply it to solve a variety of problems. This approach is an alternative approach to mechanics. It uses scalars such as work and kinetic energy rather than vectors such as velocity and acceleration. Therefore it is simpler to apply. (7-1) 3 Kinetic Energy: We define a new physical parameter to describe the state of motion of an object of mass m and speed v. We define its kinetic energy K as mv 2 KпЂЅ . 2 Unit for Kinetic energy is: Kinetic energy is a scalar quantity. 4 (7-2) 2 12/04/32 Sample Problem 7-1 In 1896 in Waco, Texas, William Crush of the вЂњKatyвЂќ railroad 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 in front of 30,000 spectators. вЂў Hundreds of people were hurt by flying debris; several were killed. Assuming each locomotive weighed 1.2 x 106 N and its acceleration along the track was a constant 0.26 m/s2, what was the total kinetic energy of the two locomotives just before the collision? 5 SOLUTION: v 2 пЂЅ v 0 пЂ« 2a (x пЂ x 0 ) 2 v 2 пЂЅ 0 пЂ« 2 (0.26 m / s 2 ) (3.2 x103 m) v пЂЅ 40.8 m / s mпЂЅ 1.2 x 10 6 N пЂЅ 1.22 x 10 5 kg 2 9.8 m /s K пЂЅ 2 ( 12 mv 2 ) пЂЅ (1.22 x 105 kg) (40.8 m / s) 2 пЂЅ 2.0 x 108 J 6 3 12/04/32 Work Work W is energy transferred to or from an object by means of a force acting on the object. вЂў Energy transferred to the object is positive work, вЂў Energy transferred from the object is negative work. 7 m m Work: (symbol W) If a force F is applied to an object of mass m it can accelerate it and increase its speed v and kinetic energy K. Similarly F can decelerate m and decrease its kinetic energy. We account for these changes in K by saying that F has transferred energy W to or from the object. If energy is transferred to m (its K increases) we say that work was done by F on the object (W > 0). If on the other hand, energy is transferred from the object (its K decreases) we say that work was done by m (W < 0). 8 4 12/04/32 m m Finding an Expression for Work : Consider a bead of mass m that can move without friction along a straight wire along the x-axis. A constant force F applied at an angle пЃ¦ to the wire is acting on the bead. We apply Newton's second law: Fx пЂЅ max . We assume that the bead had an initial velocity v0 and after it has traveled a distance d its velocity is v . We apply the third equation of kinematics: v 2 пЂ v02 пЂЅ 2ax d . We multiply both sides by m / 2 п‚® m 2 m 2 m m F m v пЂ v0 пЂЅ 2ax d пЂЅ 2 x d пЂЅ Fx d пЂЅ F cos пЃ¦ d . K i пЂЅ v02 2 2 2 2 m 2 m 2 K f пЂЅ v п‚® The change in kinetic energy K f пЂ K i пЂЅ Fd cos пЃЄ. 2 Thus the work W done by the force on the bead is given by W пЂЅ Fx d пЂЅ Fd cos пЃЄ. W пЂЅ Fd cos пЃЄ FA W пЂЅ F пѓ—d W пЂЅ Fd cos пЃЄ FC 9 m m W пЂЅ F пѓ—d FB The unit of W is the same as that of K , i.e., joules. Note 1:The expressions for work we have developed apply when F is constant. Note 2:We have made the implicit assumption that the moving object is point-like. Note 3: W пЂѕ 0 if 0 пЂј пЃ¦ пЂј 90п‚°, W пЂј 0 if 90п‚° пЂј пЃ¦ пЂј 180п‚°. Net Work: If we have several forces acting on a body (say three as in the picture) there are two methods that can be used to calculate the net work W net . Method 1: First calculate the work done by each force: W A by force FA , W B by force FB , and W C by force FC . Then determine W net пЂЅW A пЂ«W B пЂ«W C . Method 2: Calculate first Fnet пЂЅ FA пЂ« FB пЂ« FC ; then determine W net пЂЅ F пѓ— d . 10 5 12/04/32 Properties of Work вЂў Only the force component along the objectвЂ™s displacement will contribute to work. вЂў The force component perpendicular to the displacement does zero work. вЂў A force does positive work when it has a vector component in the same direction displacement, вЂў A force does negative work when it has a vector component in the opposite direction. вЂў Work is a scalar quantity. 11 Work-Kinetic Energy Theorem We have seen earlier that K f пЂ K i пЂЅ Wnet . m m We define the change in kinetic energy as пЃ„K пЂЅ K f пЂ K i . The equation above becomes the work-kinetic energy theorem: пЃ„K пЂЅ K f пЂ Ki пЂЅ Wnet пѓ©Change in the kinetic пѓ№ пѓ©net work done on пѓ№ пѓЄenergy of a particle пѓє пЂЅ пѓЄ the particle пѓє пѓ« пѓ» пѓ« пѓ» The work-kinetic energy theorem holds for both positive and negative values of Wnet . If Wnet пЂѕ 0 п‚® K f пЂ Ki пЂѕ 0 п‚® K f пЂѕ Ki If Wnet пЂј 0 п‚® K f пЂ Ki пЂј 0 п‚® K f пЂј Ki 12 6 12/04/32 (a) decrease; (b) same; (c) negative, zero 13 Sample Problem 7-2 Figure 7-4a shows two industrial spies sliding anпЃІ initially stationary 225 kg floor safe a displacement d of magnitude 8.50 m, straight toward their truck. пЃІ вЂў The push F1 of Spy 001 is 12.0 N, directed atпЃІan angle of 30В° downward from the horizontal; the pull F2 of Spy 002 is 10.0 N, directed at 40В° above the horizontal. The magnitudes and directions of these forces do not change as the safe moves, and the floor and safe make 14 frictionless contact. 7 12/04/32 (a) What is the net work done on the safe пЃІby пЃІ пЃІ forces F1 and F2 during the displacement d ? SOLUTION: пЃІ Work done by F1 : W1 пЂЅ F1 d cos пЃ¦1 пЂЅ (12.0 N)(8.50 m)(cos 30пЃЇ ) пЂЅ 88.33 J пЃІ Work done by F2 : W2 пЂЅ F2 d cos пЃ¦2 пЂЅ (10.0 N)(8.50 m)(cos 40пЃЇ ) пЂЅ 65.11 J Total work done : W пЂЅ W1 пЂ« W2 пЂЅ 88.33 J пЂ« 65.11 J пЂЅ 153.4 J п‚» 153 J 15 (b) During the displacement, whatпЃІ is the work W g done on the safe by the gravitational force Fg and what is the work пЃІ WN done on the safe by the normal force N from the floor? SOLUTION: Wg пЂЅ mgd cos 90пЃЇ пЂЅ 0 WN пЂЅ Nd cos 90пЃЇ пЂЅ 0 16 8 12/04/32 (c) The safe is initially stationary. What is its speed vf at the end of the 8.50 m displacement? SOLUTION: Work done on object equals increase in kinetic energy : W пЂЅ K f пЂ K i пЂЅ 12 mv f пЂ 12 mv i 2 vf пЂЅ 2W пЂЅ m 2 2 (153.4 J) 225 kg пЂЅ 1.17 m / s We have assumed no frictional forces exist. 17 Sample Problem 7-3 вЂў During a storm, a crate of crepe is sliding пЃІ across a slick, oily parking lot through a displacement d while a steady wind pushes against the crate with a force пЃІ F пЂЅ (20 N) Л†i пЂ« (пЂ 6.0 N) Л†j. The situation and coordinate axes are shown in Fig. 7-5. 18 9 12/04/32 (a) How much work does this force from the wind do on the crate during the displacement? SOLUTION: Work done by the wind force on crate : пЃ› пЃќ пЃІ пЃІ W пЂЅ F пѓ— d пЂЅ (2.0 N) Л†i пЂ« ( пЂ 6.0 N) Л†j пѓ— пЃ› (пЂ 3.0 m) i пЃќ пЂЅ (2.0 N) (пЂ 3.0 m) Л†i пѓ— Л†i пЂ« ( пЂ 6.0 N) ( пЂ 3.0 m) Л†j пѓ— Л†i пЂЅ ( пЂ 6.0 J) (1) пЂ« 0 пЂЅ пЂ 6.0 J The wind force does negative work, i.e. kinetic energy is taken out of the crate. 19 (b) If the crate has a kineticпЃІenergy of 10 J at the beginning of displacement d , what is its kinetic пЃІ energy at the end of d ? SOLUTION: K f пЂЅ K i пЂ« W пЂЅ 10 J пЂ« ( пЂ 6.0 J) пЂЅ 4.0 J 20 10 12/04/32 Work Done by the Gravitational Force : Consider a tomato of mass m that is thrown upward at point A with initial speed v0 . As the tomato rises, it slows down by the gravitational force Fg so that at point B it has a smaller speed v. The work Wg пЂЁ A п‚® B пЂ© done by the gravitational force on the tomato as it travels from point A to point B is Wg пЂЁ A п‚® B пЂ© пЂЅ mgd cos180п‚° пЂЅ пЂ mgd . The work Wg пЂЁ B п‚® A пЂ© done by the gravitational force on the tomato as it travels from point B to point A is Wg пЂЁ B п‚® A пЂ© пЂЅ mgd cos 0п‚° пЂЅ mgd . B A 21 Work Done by a Force in Lifting an Object : The work-kinetic energy theorem states that пЃ„K пЂЅ K f пЂ K i пЂЅW net . We also have that K i пЂЅ K f п‚® пЃ„K пЂЅ 0 п‚®W net пЂЅ 0. There are two forces acting on the object: The gravitat ional force Fg and the applied force F that lifts the object. W net пЂЅW a пЂЁ A п‚® B пЂ© пЂ«W g пЂЁ A п‚® B пЂ© пЂЅ 0 п‚® W a пЂЁ A п‚® B пЂ© пЂЅ пЂW g пЂЁ A п‚® B пЂ© . W g пЂЁ A п‚® B пЂ© пЂЅ mgd cos180п‚° пЂЅ -mgd п‚®W a пЂЁ A п‚® B пЂ© пЂЅ mgd . Work Done by a Force in Lowering an Object : In this case the object moves from B to A . W g пЂЁ B п‚® A пЂ© пЂЅ mgd cos 0п‚° пЂЅ mgd W a пЂЁ B п‚® A пЂ© пЂЅ пЂW g пЂЁ B п‚® A пЂ© = пЂ mgd B . A m 22 11 12/04/32 Sample Problem 7-4 Let us return to the lifting feats of Andrey Chemerkin and Paul Anderson. вЂў (a) Chemerkin made his record-breaking lift with rigidly connected objects (a barbell and disk weights) having a total mass m = 260.0 kg; he lifted them a distance of 2.0 m. During the lift, how much work was done on the objects by the пЃІ gravitational force Fg acting on them? SOLUTION: The gravitational force points down and the displacement d points up : Wg (up path) пЂЅ mgd cos пЃ¦ пЂЅ (2548 N) (2.0 m) (cos 180пЃЇ ) пЂЅ пЂ 5100 J The work done by the gravitational force is negative 23 to the work done by the applied force. (b) How much work was done on the objects by Chemerkin's force during the lift? SOLUTION: The work done by the applied force is : WAC (up) пЂЅ пЂ Wg (up) пЂЅ пЂ« 5100 J (c) While Chemerkin held the objects stationary above his head, how much work was done on them by his force? Since the displacement d is zero, the work done is zero. 24 12 12/04/32 (d) How much work was done by the force Paul Anderson applied to lift objects with a total weight of 27 900 N, a distance of 1.0 cm? SOLUTION: The work done by the applied force of Paul Anderson is : WPA (up path) пЂЅ пЂ Wg (up path) пЂЅ (27900 N) (0.010 m) пЂЅ 280 J 25 Sample Problem 7-5 вЂў An initially stationary 15.0 kg crate of cheese wheels is pulled, via a cable, a distance L = 5.70 m up a frictionless ramp, to a height h of 2.50 m, where it stops (Fig. 7-8a). (a) How much work W g is done on the crate by пЃІ the gravitational force Fg during the lift? 26 13 12/04/32 SOLUTION: The work done by the gravitational force during the lift is negative : Wg (up path) пЂЅ пЂ mg sin пЃ± d Since d sin пЃ± = h, we have : Wg (up path) пЂЅ пЂ mgh пЂЅ пЂ (15.0 kg) (9.8 m / s 2 ) (2.50 m) пЂЅ пЂ 368 J 27 (b) HowпЃІ much work WT is done on the crate by the force from the cable during the lift? T SOLUTION: Since the crate has zero velocity before and after the lift, the work done by the applied force must be equal and opposite to the work done by the gravitational force. WT пЂЅ пЂ Wg (up path) пЂЅ 368 J 28 14 12/04/32 Sample Problem 7-6 вЂў An elevator cab of mass m = 500 kg is descending with speed vi = 4.0 m/s when its supporting cable begins to slip, allowing it to fall with constant пЃІ пЃІ acceleration a = g /5. вЂў (a) During the fall through a distance d = 12 m, what is the work W g done on the cab byпЃІ the gravitational force Fg ? 29 SOLUTION: During the fall, the work done on the cab by the gravitational force is positive. Wg (down path) пЂЅ mgd cos 0пЃЇ пЂЅ (500 kg) (9.8 m / s 2 ) (12 m) (1) пЂЅ 5.88 x 104 J п‚» 59 kJ 30 15 12/04/32 (b) During the 12 m fall, what is the work WT done пЃІ on the cab by the upward pullT of the elevator cable? SOLUTION: mg пЂ T пЂЅ ma (positive direction is down) T пЂЅ m (g пЂ a ) пЂЅ 54 mg This gives the magnitude of T, the direction of T is up The work done on the cab by T during the fall is negative. WT (down path) пЂЅ пЂ 54 mg d пЂЅ пЂ 4.70 x 10 4 J п‚» пЂ 47 kJ 31 (c) What is the net work W done on the cab during the fall? SOLUTION: W пЂЅ Wg пЂ« WT пЂЅ 5.88 x 10 4 J пЂ 4.70 x 10 4 J пЂЅ 1.18 x 10 4 J п‚» 12 kJ (d) What is the cab's kinetic energy at the end of the 12 m fall? SOLUTION: K f пЂЅ K i пЂ« W пЂЅ 12 mv i пЂ« W 2 пЂЅ 12 (500 kg) (4.0 m / s) 2 пЂ« 1.18 x 10 4 J пЂЅ 1.58 x 10 4 J п‚» 16 kJ 16 32 12/04/32 The Spring Force : Fig. a shows a spring in its relaxed state. In fig. b we pull one end of the spring and stretch it by an amount d . The spring resists by exerting a force F on our hand in the opposi te direction. In fig. c we push one end of the spring and compress it by an amount d . Again the spring resists by exerting a force F on our hand in the opposite direction. The force F exerted by the spring on whatever agent (in the picture it is our hand) is trying to change its natural length either by extending or by compressing it is given by the equation F пЂЅ пЂkx. Here x is the amount by which the spring has been extended or compressed. This equation is known as "Hooke's law" and k is known as the "spring constant" F пЂЅ пЂkx 33 x O xi (a) x O (b) xf x O (c) Work Done by a Spring Force : Consider the relaxed spring of spring constant k shown in (a). By applying an external force we change the spring's length from xi (see b) to x f (see c). We will calculate the work Ws done by the spring on the external agent (in this case our hand) that changed the spring length. We assume that the spring is massless and that it obeys Hooke's law. We will use the expression Ws пЂЅ xf xf xf xf xi xi xi пѓІ F ( x) dx пЂЅ пѓІ пЂkxdx пЂЅ пЂk пѓІ xdx. пѓ© x2 пѓ№ kx 2 kx Ws пЂЅ пЂk пѓЄ пѓє пЂЅ i пЂ . Quite often we start with a relaxed 2 2 пѓ« 2 пѓ» xi 2 f spring (xi пЂЅ 0) and we either stretch or compress the spring by an amount x ( x f пЂЅ п‚± x). In this case Ws пЂЅ пЂ 17 kx 2 . 2 34 12/04/32 (a) positive; (b) negative; (c) zero 35 Sample Problem 7-8 вЂў In Fig. 7-11, a cumin canister of mass m = 0.40 kg slides across a horizontal frictionless counter with speed v = 0.50 m/s. It then runs into and compresses a spring of spring constant k = 750 N/m. When the canister is momentarily stopped by the spring, by what distance d is the spring compressed? 36 18 12/04/32 SOLUTION: We assume the spring is massless. Work done by the spring on the canister is negative. This work is : WS пЂЅ пЂ 12 kd 2 Kinetic energy change of the canister is : k f пЂ k i пЂЅ пЂ 12 mv 2 Therefore, пЂ 12 kd 2 пЂЅ пЂ 12 mv 2 dпЂЅv m 0.40 kg пЂЅ (0.50 m / s) k 750 N / m пЂЅ 1.2 x 10 пЂ 2 m пЂЅ 1.2 cm 37 Work Done by a Variable Force F( x ) Acting Along the x - Axis : A force F that is not constant but instead varies as a function of x is shown in fig. a. We wish to calculate the work W that F does on an object it moves from position xi to position x f . We partition the interval пЂЁ xi , x f пЂ© into N "elements" of length пЃ„x each, as is shown in fig. b. The work done by F in the jth interval is пЃ„W j пЂЅ Fj ,avg пЃ„x, where Fj ,avg is the average value of F N over the j -th element. W пЂЅ пѓҐ Fj ,avg пЃ„x. We then take the limit of j пЂЅ1 the sum as пЃ„x п‚® 0, (or equivalently N п‚® п‚Ґ). N xf j пЂЅ1 xi W пЂЅ lim пѓҐ Fj ,avg пЃ„x пЂЅ пѓІ F ( x)dx. Geometrically, W is the area between the F ( x) curve and the x-axis, between xi and x f (shaded blue in fig. d). xf WпЂЅ пѓІ F ( x)dx xi 19 38 12/04/32 Three - Dimensional Analysis : In the general case the force F acts in three-dimensional space and moves an object on a three-dimensional path from an initial point A to a final point B. Л† The force has the form F пЂЅ F пЂЁ x, y, z пЂ© Л†i пЂ« F пЂЁ x, y, z пЂ© Л†j пЂ« F пЂЁ x, y, z пЂ© k. x y z Points A and B have coordinates пЂЁ xi , yi , zi пЂ© and пЂЁ x f , y f , z f пЂ© , respectively. dW пЂЅ F пѓ— dr пЂЅ Fx dx пЂ« Fy dy пЂ« Fz dz B xf yf A xi yi W пЂЅ пѓІ dW пЂЅ пѓІ Fx dx пЂ« xf WпЂЅ пѓІ zf Fy dy пЂ« пѓІ Fz dz zi yf z zf пѓІ F dx пЂ« пѓІ F dy пЂ« пѓІ F dz x xi y yi B z O x zi path y A 39 Work - Kinetic Energy Theorem with a Variable Force : Consider a variable force F ( x) that moves an object of mass m from point A ( x пЂЅ xi ) dv . We t hen dt dv multiply both sides of the last equation with dx and get Fdx пЂЅ m dx. dt to point B ( x пЂЅ x f ). We apply Newton's second law: F пЂЅ ma пЂЅ m We integrate both sides over dx from xi to x f xf xf xi xi dv пѓІ Fdx пЂЅ пѓІ m dt dx . dv dv dx dv dv dx пЂЅ п‚® dx пЂЅ dx пЂЅ vdv. Thus the integral becomes dt dx dt dt dx dt xf W пЂЅ m пѓІ vdv пЂЅ xi 2 m 2 x f mv f mvi2 пѓ©пѓ«v пѓ№пѓ» пЂЅ пЂ пЂЅ K f пЂ K i пЂЅ пЃ„K . xi 2 2 2 Note: The work-kinetic energy theorem has exactly the same form as in the case when F is constant! W пЂЅ K f пЂ Ki пЂЅ пЃ„K O m F(x) .A x 20 dx . B x-axis 40 12/04/32 41 42 21 12/04/32 43 44 22 12/04/32 Power We define "power" P as the rate at which work is done by a force F. If F does work W in a time interval пЃ„t then we define the average power as Pavg пЂЅ W пЃ„t The instantaneous power is defined as PпЂЅ dW dt Unit of P : The SI unit of power is the watt. It is defined as the power of an engine that does work W = 1 J in a time t = 1 second. A commonly used non-SI power unit is the horsepower (hp), defined as 1 hp = 746 W. The kilowatt-hour The kilowatt-hour (kWh) is a unit of work. It is defined as the work performed by an engine of power P = 1000 W in a time t = 1 hour , W пЂЅ Pt пЂЅ 1000 п‚ґ 3600 пЂЅ 3.60 п‚ґ10 6 J . The kWh is used by electrical utility companies (check your latest electric bill). 45 Consider a force F acting on a particle at an angle пЃ¦ to the motion. The rate at which F does work is given by P пЂЅ dW F cos пЃ¦ dx dx пЂЅ пЂЅ F cos пЃ¦ пЂЅ Fv cos пЃ¦. dt dt dt P пЂЅ Fv cos пЃ¦ пЂЅ F пѓ— v v 46 23 12/04/32 zero 47 Sample Problem 7-11 пЃІ пЃІ вЂў Figure 7-14 shows constant forces F1 and F2 acting on a box as the box slides пЃІrightward across a frictionless floor. Force F1 is horizontal, пЃІ with magnitude 2.0 N; force F2 is angled upward by 60В° to the floor and has magnitude 4.0 N. The speed v of the box at a certain instant is 3.0 m/s. 48 24 12/04/32 (a) 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? SOLUTION: P1 пЂЅ F1 v cos 180 пЃЇ пЂЅ (2.0 N) (3.0 m / s) cos 180 пЃЇ пЂЅ пЂ 6.0 W P2 пЂЅ F2 v cos 60 пЃЇ пЂЅ (4.0 N) (3.0 m / s) cos 60 пЃЇ пЂЅ 6.0 W Pnet пЂЅ P1 пЂ« P2 пЂЅ0 The kinetic energy of the box is not changing. The speed of the box remains at 3 m/s. The49net power does not change. пЃІ (b) If the magnitude of F2 is, instead, 6.0 N, what now is the net power, and is it changing? SOLUTION: P2 пЂЅ F2 v cos 60 пЃЇ пЂЅ (6.0 N) (3.0 m / s) cos 60 пЃЇ пЂЅ 9.0 W Pnet пЂЅ P1 пЂ« P2 пЂЅ пЂ 6.0 W пЂ« 9.0 W пЂЅ 3.0 W There is a net rate of transfer of energy to the box. The kinetic energy of the box increases. The net power also increases. 25 50 12/04/32 Homework # 3 (due March 28) вЂ« Ш±ШЁЩЉШ№ Ш«Ш§Щ†ЩЉвЂ¬23 вЂ«ШЄШіЩ„ЩЉЩ… Ш§Щ„Щ€Ш§Ш¬ШЁ Щ€Ш§Щ„ЩѓЩ€ЩЉШІ Ш§Ш§Щ„Ш«Щ†ЩЉЩ†вЂ¬ вЂў Question 7 пЃ± Question 8 пЃ± 10 пЃ± 20 пЃ± 27 вЂў 37 вЂў 39 вЂў 43 вЂў 67 вЂў 71 51 Q. # Ans. 7 -------------- 8 (a) 3 m; (b) 3 m; (c) 0 and 6 m; (d) negative x 10 5.0 п‚ґ103 J 20 (a) +1.31 J. (b) 0.935 m/s. 27 a) 7.2 J; (b) 7.2 J; (c) 0; (d) - 25 J 37 5.3Г—102 J 39 (a) 42 J; (b) 30 J; (c) 12 J; (d) 6.5m/s, positive x direction; (e) 5.5m/s, positive x direction; 43 4.9Г—102 W 67 (a) 1.20 J; (b) 1.10 m/s 71 (a) 23 mm; (b) 45 N 52 26