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Chapter 7 Kinetic Energy and Work Kinetic Energy of a mass m is : K mv 1 2 2 Energy is a scalar quantity The SI unit of energy is 1 joule 1 J 1 kg m / s 2 2 This unit is named after an English scientist of the 1800s, James Prescott Joule. 1 Sample Problem 7-1 In 1896 in Waco, Texas, William Crush of the “Katy” 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. 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? 2 SOLUTION: v 2 v 0 2a (x x 0 ) 2 v 2 0 2 (0.26 m / s 2 ) (3.2 x103 m) v 40.8 m / s 1.2 x 10 6 N 5 m 1 . 22 x 10 kg 2 9.8 m /s K 2 ( 12 mv 2 ) (1.22 x 105 kg) (40.8 m / s) 2 2.0 x 10 J 8 3 Work Work W is positive if kinetic energy is transferred to an object by an external force. Work is negative if kinetic energy is transferred from an object. In other words, negative work is done if an object’s kinetic energy decreases. Work is a scalar and the SI unit is Joule. 4 Work done by an External Force W F d cos W F d To calculate the work done on an object by a force during a displacement, we use only the force component along the object's displacement. The force component 5 perpendicular to the displacement does zero work. 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. 1 J kg m / s 1 N m 0.738 ft lb 2 2 6 K K f Ki W change in the kinetic energy of a particle net work done on the particle Kf Ki W kinetic energy after kinetic energy the net the net work is done before the net work work done 7 Sample Problem 7-2 Figure 7-4a shows two industrial spies sliding an initially stationary 225 kg floor safe a displacement d of magnitude 8.50 m, straight toward their truck. The push F1 of Spy 001 is 12.0 N, directed atan angle of 30° downward from the horizontal; the pull F2 of Spy 002 is 10.0 N, directed at 40° above the horizontal. The magnitudes and directions of these forces do not change as the safe moves, and the floor and safe make frictionless contact. 8 (a) What is thenet work done on the safe by forces F1 and F2 during the displacement d ? SOLUTION: Work done by F1 : W1 F1 d cos 1 (12.0 N)(8.50 m)(cos 30 ) 88.33 J Work done by F2 : W2 F2 d cos 2 (10.0 N)(8.50 m)(cos 40 ) 65.11 J Total work done : W W1 W2 88.33 J 65.11 J 153.4 J 153 J 9 (b) During the displacement, what is the work Wg done on the safe by the gravitational force Fg and what is the work WN done on the safe by the normal force N from the floor? SOLUTION: Wg mgd cos 90 0 WN Nd cos 90 0 10 (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 K f K i mv f mv i 1 2 2 1 2 2 2W 2 (153.4 J) vf m 225 kg 1.17 m / s We have assumed no frictional forces exist. 11 Sample Problem 7-3 During a storm, a crate of crepe is sliding across a slick, oily parking lot through a displacement d while a steady wind pushes against the crate with a force F (20 N) î ( 6.0 N) ĵ. The situation and coordinate axes are shown in Fig. 7-5. 12 (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 : W F d (2.0 N) î ( 6.0 N) ĵ ( 3.0 m) i (2.0 N) ( 3.0 m) î î ( 6.0 N) ( 3.0 m) ĵ î ( 6.0 J) (1) 0 6.0 J The wind force does negative work, i.e. kinetic energy is taken out of the crate. 13 (b) If the crate has a kineticenergy of 10 J at the d , what is its kinetic beginning of displacement energy at the end of d ? SOLUTION: K f Ki W 10 J ( 6.0 J) 4.0 J 14 Work done by Gravitational Force Wg m g d cos Fg (down) The angle = is between and d . For the upward path, . 180 Wg (up path ) m g d cos180 m g d (1) m g d The work done by the gravitational force is negative for the upward path. The change of gravitational potential energy of the object equals the negative of the work done by the gravitational force. 15 After the object has reached its maximum height and is falling back down, the angle = 0o. The work done is Wg (down path ) m g d cos 0 m g d (1) m g d 16 Work done on Lifting and Lowering an Object against a Field at Constant Velocity An applied force of magnitude = mg pointed up can barely lift an object of weight mg. Assuming no change in the velocity, the work done by the applied force on the object in lifting it a distance d is : Wa mgd Wg (up path ) The work done by the applied force in lifting the object is positive. It has the opposite sign as the work done by the gravitational force. 17 The applied force of magnitude mg (pointed up) lowers the object with no change in velocity. The work done by the applied force on the object in lowering it a distance d is : Wa mgd The work done by the applied force in lowering the object is negative. The work done by the applied force equals the change in the gravitational potential energy of the object. 18 Sample Problem 7-4 Let us return to the lifting feats of Andrey Chemerkin and Paul Anderson. (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 gravitational force Fg acting on them? SOLUTION: The gravitational force points down and the displacement d points up : Wg (up path ) mgd cos (2548 N) (2.0 m) (cos 180 ) 5100 J The work done by the gravitational force is negative to the work done by the applied force. 19 (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 ) Wg (up ) 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. 20 (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 ) Wg (up path ) (27900 N) (0.010 m) 280 J 21 Sample Problem 7-5 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 Wg is done on the crate by the gravitational force Fg during the lift? 22 SOLUTION: The work done by the gravitational force during the lift is negative : Wg (up path ) mg sin d Since d sin = h, we have : Wg (up path ) mgh (15.0 kg) (9.8 m / s 2 ) (2.50 m) 368 J 23 (b) How much work WT is done on the crate by the force T from the cable during the lift? 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 Wg (up path ) 368 J 24 Sample Problem 7-6 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 acceleration a = g /5. (a) During the fall through a distance d = 12 m, what is the work Wg done on the cab by the gravitational force Fg ? 25 SOLUTION: During the fall, the work done on the cab by the gravitational force is positive. Wg (down path ) mgd cos 0 (500 kg) (9.8 m / s 2 ) (12 m) (1) 5.88 x 10 4 J 59 kJ 26 (b) During the 12 m fall, what is the work WT done on the cab by the upward pull T of the elevator cable? SOLUTION: mg T ma (positive direction is down) T m (g a ) 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 ) 54 mg d 4.70 x 10 4 J 47 kJ 27 (c) What is the net work W done on the cab during the fall? SOLUTION: W Wg WT 5.88 x 10 4 J 4.70 x 10 4 J 1.18 x 10 4 J 12 kJ (d) What is the cab's kinetic energy at the end of the 12 m fall? SOLUTION: K f K i W mv i W 1 2 2 12 (500 kg) (4.0 m / s) 2 1.18 x 10 4 J 1.58 x 10 4 J 16 kJ 28 The Spring Force When a spring of length L0 is either compressed or extended, a restoring force acts opposite to d . This restoring force is known as the spring force and is given by : FS k d The constant k is a scalar and is called the spring constant or force constant. The SI unit is N/m. The negative sign indicates that the spring force is always opposite in direction to the displacement. Hooke’s law is named after Robert 29 Hooke of the late 1600s. Work done by the Spring Force The work done by the spring force as the block is moved from position xi to xf is : xf WS F dx xi xf xf xi xi WS ( kx ) dx k x dx ( 12 k )[ x 2 ]xx if ( 12 k ) ( x f x i ) 2 2 We have assumed that all the mass is at the block and the spring itself is massless. 30 The work done by the spring force on the block is negative if xf > xi. If we set at L0 , xi = 0, then the work done by the spring force on the block for extension x is WS k x 1 2 2 Change in potential energy of spring = - work done by the spring force. P.E. change kx 1 2 2 The work done by the spring force is to increase P.E. and to decrease K.E. 31 Work Done on the Spring by an Applied Force If a block is stationary before and after a displacement, the work done on the block by an applied force is negative to the work done on it by the spring force. Therefore, for an extension x, the work done by an applied force on the block is : WA k x 1 2 2 The work done by the applied force equals to elastic potential energy of the spring. The work done by the applied force is to increase the P.E. of the spring. 32 Sample Problem 7-7 A package of spicy Cajun pralines lies on a frictionless floor, attached to the free end of a spring in the arrangement of Fig. 7-10a. An applied force of magnitude Fa = 4.9 N would be needed to hold the package stationary at x1 = 12 mm. (a) How much work does the spring force do on the package if the package is pulled rightward from x0 = 0 to x2 = 17 mm? 33 SOLUTION: FS k x FS 4.9 N k 408 N / m 3 x1 12 x 10 m Work done by the spring is : 3 Ws kx 2 (408 N / m) (17 x 10 m) 1 2 2 1 2 2 0.059 J 34 (b) Next, the package is moved leftward to x3 = -12 mm. How much work does the spring force do on the package during this displacement? Explain the sign of this work. SOLUTION: Ws 12 kx f 12 kx i 2 2 12 (408 N / m) ( 12 x 10 3 m) 2 (17 x 10 3 m) 2 0.030 J 30 mJ The spring force does positive work as the block moves from +17mm to its relaxed position and negative work as the block moves from the relaxed position to -12mm. The former work is larger resulting in WS being positive. 35 Sample Problem 7-8 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 SOLUTION: We assume the spring is massless. Work done by the spring on the canister is negative. This work is : WS 12 kd 2 Kinetic energy change of the canister is : k f k i 12 mv 2 Therefore, 12 kd 2 12 mv 2 m 0.40 kg dv (0.50 m / s) k 750 N / m 1.2 x 10 2 m 1.2 cm 37 Work done by a General Variable Force W F d r ( work by var iable force ) 38 Power The instantaneous power P is dW P dt Power is a scalar and the SI unit is J / s which is known as the Watt (W), named after James Watt. 1 watt 1 W 1 J / s 0.738 ft lb / s 1 horsepower 1 hp 550 ft lb / s 746 W 1 kiwowatt hour 1kW h (103 W) (3600 s) 3.60 x 10 6 J 3.60 MJ Note that the kilowatt-hr is a unit of work 39 dx d W F cos d x P F cos dt dt dt P F v cos The angle is between the force and velocity. Therefore, P F v 40 Sample Problem 7-10 Figure 7-14 shows constant forces F1 and F2 acting on a box as the box slides rightward across a frictionless floor. Force F1 is horizontal, with magnitude 2.0 N; force F2 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. 41 (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 F1 v cos 180 (2.0 N) (3.0 m / s) cos 180 6.0 W P2 F2 v cos 60 (4.0 N) (3.0 m / s) cos 60 6.0 W Pnet P1 P2 0 The kinetic energy of the box is not changing. The speed of the box remains at 3 m/s. The net power does not change. 42 (b) If the magnitude of F2 is, instead, 6.0 N, what now is the net power, and is it changing? SOLUTION: P2 F2 v cos 60 (6.0 N) (3.0 m / s) cos 60 9.0 W Pnet P1 P2 6.0 W 9.0 W 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. 43 Homework (due Oct 18) 5P 7E 11P 17P 19P 21E 23P 31E 33P 44