Physics exercises_solution: Chapter 06

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Physics exercises_solution: Chapter 06

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 06

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  1. 6.1: a) (2.40 N) (1.5 m)  3.60 J b) (0.600 N)(1.50 m)  0.900 J c) 3.60 J  0.720 J  2.70 J . 6.2: a) “Pulling slowly” can be taken to mean that the bucket rises at constant speed, so the tension in the rope may be taken to be the bucket’s weight. In pulling a given length of rope, from Eq. (6.1), W  Fs  mgs  (6.75 kg ) (9.80 m / s 2 )(4.00 m)  264.6 J. b) Gravity is directed opposite to the direction of the bucket’s motion, so Eq. (6.2) gives the negative of the result of part (a), or  265 J . c) The net work done on the bucket is zero. 6.3: (25.0 N)(12.0 m)  300 J . 6.4: a) The friction force to be overcome is f   k n   k mg  (0.25)(30.0 kg)(9.80 m / s 2 )  73.5 N, or 74 N to two figures. b) From Eq. (6.1), Fs  (73.5 N)(4.5 m)  331 J . The work is positive, since the worker is pushing in the same direction as the crate’s motion. c) Since f and s are oppositely directed, Eq. (6.2) gives  fs  (73.5 N)(4.5 m)  331 J. d) Both the normal force and gravity act perpendicular to the direction of motion, so neither force does work. e) The net work done is zero.
  2. 6.5: a) See Exercise 5.37. The needed force is  k mg (0.25)(30 kg)(9.80 m / s 2 ) F   99.2 N, cos    k sin  cos 30  (0.25) sin 30 keeping extra figures. b) Fs cos   (99.2 N)(4.50 m) cos 30  386.5 J , again keeping an extra figure. c) The normal force is mg  F sin  , and so the work done by friction is  (4.50 m)(0.25)((30 kg)(9.80 m / s 2 )  (99.2 N) sin 30)  386.5 J . d) Both the normal force and gravity act perpendicular to the direction of motion, so neither force does work. e) The net work done is zero. 6.6: From Eq. (6.2), Fs cos   (180 N)(300 m) cos15.0  5.22  104 J. 6.7: 2 Fs cos   2(1.80  10 6 N)(0.75  10 3 m) cos14  2.62  10 9 J, or 2.6  10 9 J to two places. 6.8: The work you do is:   F  s  ((30 N)i  (40 N) ˆ)  ((9.0m)i  (3.0m) ˆ) ˆ j ˆ j  (30 N)(9.0 m)  (40 N)(3.0 m)  270 N  m  120 N  m  150 J 6.9: a) (i) Tension force is always perpendicular to the displacement and does no work. (ii) Work done by gravity is  mg ( y 2  y1 ). When y1  y 2 , Wmg  0 . b) (i) Tension does no work. (ii) Let l be the length of the string. Wmg   mg ( y 2  y1 )   mg (2l )  25.1 J The displacement is upward and the gravity force is downward, so it does negative work.
  3. 6.10: a) From Eq. (6.6), 2 1   1 m / s  K  (1600 kg) (50.0 km/h)     1.54  10 J. 5 2   3.6 km / h    b) Equation (6.5) gives the explicit dependence of kinetic energy on speed; doubling the speed of any object increases the kinetic energy by a factor of four. 1 m/s 6.11: For the T-Rex, K  1 (7000 kg )((4 km / hr ) 3.6 km/hr ) 2  4.32  103 J . The person’s 2 velocity would be v  2(4.32  10 3 J)/70 kg  11.1 m/s , or about 40 km/h. 6.12: (a) Estimate: v  1ms (walking) v  2 ms (running) m  70 kg Walking: KE  1 mv 2  1 (70 kg )(1 m / s) 2  35 J 2 2 Running: KE  1 (70 kg )(2 m / s) 2  140 J 2 (b) Estimate: v  60 mph  88 ft / s  30 m / s m  2000 kg KE  2 (2000 kg )(30 m / s) 2  9  10 5 J 1 (c) KE  Wgravity  mgh Estimate h  2 m KE  (1 kg)(9.8 m / s 2 )(2 m)  20 J 6.13: Let point 1 be at the bottom of the incline and let point 2 be at the skier. Wtot  K 2  K 1 1 2 K 1  mv0 , K 2  0 2 Work is done by gravity and friction, so Wtot  Wmg  W f . Wmg   mg ( y 2  y1 )   mgh W f   fs  (  k mg cos  )(h / sin  )    k mgh / tan  Substituting these expressions into the work-energy theorem and solving for v0 gives v0  2 gh(1   k / tan  )
  4. 6.14: (a) W  KE 1 1 2  mgh  mvf2  mv0 2 2 v0  vf2  2 gh  (25.0 m / s) 2  2 (9.80 m / s 2 )(15.0 m)  30.3 m / s (b) W  KE 1 1 2  mgh  mvf2  mv0 2 2 v  vf (30.3 m / s) 2  02 2 2 h 0  2g 2(9.80 ms / s 2 )  46.8 m 6.15: a) parallel to incline: force component  mg sin  , down incline; displacement  h sin  , down incline W||  (mg sin  )(h / sin  )  mgh perpendicular to incline: no displacement in this direction, so W  0 . Wmg  W||  W  mgh , same as falling height h. b) Wtot  K 2  K1 gives mgh  1 mv 2 and v  2 gh , same as if had been dropped 2 from height h. The work done by gravity depends only on the vertical displacement of the object. When the slope angle is small, there is a small force component in the direction of the displacement but a large displacement in this direction. When the slope angle is large, the force component in the direction of the displacement along the incline is larger but the displacement in this direction is smaller. c) h  15.0 m , so v  2 gh  17.1 s 6.16: Doubling the speed increases the kinetic energy, and hence the magnitude of the work done by friction, by a factor of four. With the stopping force given as being independent of speed, the distance must also increase by a factor of four. 6.17: Barring a balk, the initial kinetic energy of the ball is zero, and so W  (1 / 2)mv 2  (1 / 2)(0.145 kg)(32.0 m/s) 2  74.2 J.
  5. 6.18: As the example explains, the boats have the same kinetic energy K at the finish line, so (1 / 2)m A v A  (1 / 2)m B v B , or, with m B  2m A , v A  2v B . a) Solving for the ratio of the 2 2 2 2 speeds, v A / v B  2 . b) The boats are said to start from rest, so the elapsed time is the distance divided by the average speed. The ratio of the average speeds is the same as the ratio of the final speeds, so the ratio of the elapsed times is t B / t A  v A / v B  2 . 6.19: a) From Eq. (6.5), K 2  K 1 / 16 , and from Eq. (6.6), W  (15 / 16) K 1 . b) No; kinetic energies depend on the magnitudes of velocities only. 6.20: From Equations (6.1), (6.5) and (6.6), and solving for F, K 1 m(v 2  v1 ) 1 (8.00 kg )((6.00 m / s)  (4.00 m / s) ) 2 2 2 2 F  2  2  32.0 N. s s (2.50 m) K 1 (0.420 kg)((6.00 m / s) 2  (2.00 m / s) 2 ) 6.21: s  2  16.8 cm F (40.0 N) 6.22: a) If there is no work done by friction, the final kinetic energy is the work done by the applied force, and solving for the speed, 2W 2 Fs 2(36.0 N)(1.20 m) v    4.48 m / s. m m (4.30 kg ) b) The net work is Fs  f k s  ( F   k mg ) s , so 2( F   k mg ) s v m 2(36.0 N  (0.30)(4.30 kg)(9.80 m / s 2 ))(1.20 m)  (4.30 kg)  3.61 m / s. (Note that even though the coefficient of friction is known to only two places, the difference of the forces is still known to three places.)
  6. 6.23: a) On the way up, gravity is opposed to the direction of motion, and so W   mgs  (0.145 kg)(9.80 m / s 2 )(20.0 m)  28.4 J . W 2(28.4 J ) b) v2  v12  2  (25.0 m / s) 2   15.26 m / s . m (0.145 kg ) c) No; in the absence of air resistance, the ball will have the same speed on the way down as on the way up. On the way down, gravity will have done both negative and positive work on the ball, but the net work will be the same. 6.24: a) Gravity acts in the same direction as the watermelon’s motion, so Eq. (6.1) gives W  Fs  mgs  (4.80 kg)(9.80 m / s 2 )(25.0 m)  1176 J. b) Since the melon is released from rest, K 1  0 , and Eq. (6.6) gives K  K 2  W  1176 J. 6.25: a) Combining Equations (6.5) and (6.6) and solving for v2 algebraically, Wtot 2(10.0 N)(3.0 m) v 2  v12  2  (4.00 m / s) 2   4.96 m / s. m (7.00 kg ) Keeping extra figures in the intermediate calculations, the acceleration is a  (10.0 kg  m / s 2 ) /(7.00 kg)  1.429 m / s 2 . From Eq. (2.13), with appropriate change in notation, v 2  v12  2as  (4.00 m / s) 2  2(1.429 m / s 2 )(3.0 m), 2 giving the same result. 6.26: The normal force does no work. The work-energy theorem, along with Eq. (6.5), gives 2K 2W v   2 gh  2 gL sin  , m m where h  L sin  is the vertical distance the block has dropped, and  is the angle the plane makes with the horizontal. Using the given numbers, v  2(9.80 m / s 2 )(0.75 m) sin 36.9  2.97 m / s.
  7. 6.27: a) The friction force is  k mg , which is directed against the car’s motion, so the net work done is   k mgs . The change in kinetic energy is K   K 1  (1 / 2) mv0 , and so 2 s  v 0 / 2  k g . b) From the result of part (a), the stopping distance is proportional to the 2 square of the initial speed, and so for an initial speed of 60 km/h, s  (91.2 m)(60.0 / 80.0) 2  51.3 m . (This method avoids the intermediate calculation of  k , which in this case is about 0.279.) 6.28: The intermediate calculation of the spring constant may be avoided by using Eq. (6.9) to see that the work is proportional to the square of the extension; the work needed to compress the spring 4.00 cm is (12.0 J )  3.00 cm 4.00 cm 2  21.3 J . 6.29: a) The magnitude of the force is proportional to the magnitude of the extension or compression; (160 N)(0.015 m / 0.050 m)  48 N, (160 N)(0.020 m / 0.050 m)  64 N. b) There are many equivalent ways to do the necessary algebra. One way is to note  169 N  that to stretch the spring the original 0.050 m requires 1  2   (0.050 m) 2  4 J ,  0.050 m   so that stretching 0.015 m requires (4 J )(0.015 / 0.050) 2  0.360 J and compressing 0.020 m requires (4 J )(0.020 / 0.050) 2  0.64 J . Another is to find the spring constant k  (160 N)  (0.050 m)  3.20  10 3 N / m , from which (1 / 2)(3.20  10 3 N / m)(0.015 m) 2  0.360 J and (1 / 2)(3.20  10 3 N / m )( 0.020 m ) 2  0.64 J . 6.30: The work can be found by finding the area under the graph, being careful of the sign of the force. The area under each triangle is 1/2 base  height . a) 1 / 2 (8 m)(10 N)  40 J . b) 1 / 2 (4 m)(10 N)  20 J . c) 1 / 2 (12 m)(10 N)  60 J . 6.31: Use the Work-Energy Theorem and the results of Problem 6.30. (2)(40 J ) a) v   2.83 m / s 10 kg b) At x  12 m , the 40 Joules of kinetic energy will have been increased by 20 J, so (2)(60 J) v  3.46 m / s . 10 kg
  8. 6.32: The work you do with your changing force is 6.9 6.9 6.9 N 0 F ( x) dx   0 (20.0 N) dx   0 3.0 m xdx N 2  (20.0 N) x | 0.9 (3.0 6 )( x / 2) | 0.9 6 m  138 N  m  71.4 N  m  209.4 J or  209 J The work is negative because the cow continues to advance as you vainly attempt to push her backward. 6.33: Wtot  K 2  K1 1 2 K 1  mv0 , K 2  0 2 Work is done by the spring force. Wtot   1 kx 2 , where x is the amount the spring is 2 compressed. 1 1 2  kx 2   mv0 and x  v 0 m / k  8.5 cm 2 2 6.34: a) The average force is (80.0 J) /(0.200 m)  400 N , and the force needed to hold the platform in place is twice this, or 800 N. b) From Eq. (6.9), doubling the distance quadruples the work so an extra 240 J of work must be done. The maximum force is quadrupled, 1600 N. Both parts may of course be done by solving for the spring constant k  2(80.0 J)  (0.200 m) 2  4.00  10 3 N/m , giving the same results.
  9. 6.35: a) The static friction force would need to be equal in magnitude to the spring force, s mg  kd or μs  ((020.0 N / )( 9)(80.086/ s 2))  1.76 , which is quite large. (Keeping extra figures in m 0 .100 kg . m m the intermediate calculation for d gives a different answer.) b) In Example 6.6, the relation 1 1  k mgd  kd 2  mv12 2 2 was obtained, and d was found in terms of the known initial speed v1 . In this case, the condition on d is that the static friction force at maximum extension just balances the spring force, or kd   s mg . Solving for v12 and substituting, k v12  d 2  2 gd k d m 2 k   mg    mg    s   2 k g  s  m k   k  mg 2 2  (  s  2 s  k ) k  (0.10 kg)(9.80 m / s 2 ) 2    ((0.60) 2  2(0.60)(0.47)),   (20.0 N / m)  from which v1  0.67 m / s . 6.36: a) The spring is pushing on the block in its direction of motion, so the work is positive, and equal to the work done in compressing the spring. From either Eq. (6.9) or Eq. (6.10), W  1 kx 2  1 (200 N / m)(0.025 m) 2  0.06 J . 2 2 b) The work-energy theorem gives 2W 2(0.06 J ) v   0.18 m / s. m (4.0 kg ) 6.37: The work done in any interval is the area under the curve, easily calculated when the areas are unions of triangles and rectangles. a) The area under the trapezoid is 4.0 N  m  4.0 J . b) No force is applied in this interval, so the work done is zero. c) The area of the triangle is 1.0 N  m  1.0 J , and since the curve is below the axis ( Fx  0) , the work is negative, or  1.0 J . d) The net work is the sum of the results of parts (a), (b) and (c), 3.0 J. (e)  1.0 J  2.0 J  1.0 J .
  10. 6.38: a) K  4.0 J , so v  2 K m  2 (4.0 J ) (2.0 kg )  2.00 m / s . b) No work is done between x  3.0 m and x  4.0 m , so the speed is the same, 2.00 m/s. c) K  3.0 J , so v  2 K / m  2(3.0 J ) /( 2.0 kg )  1.73 m / s . 6.39: a) The spring does positive work on the sled and rider; (1 / 2)kx 2  (1 / 2)mv 2 , or v  x k / m  (0.375 m) (4000 N / m) /(70 kg )  2.83 m / s . b) The net work done by the spring is (1 / 2)k ( x12  x 2 ) , so the final speed is 2 k 2 (4000 N / m v ( x1  x 2 )  2 ((0.375 m) 2  (0.200 m) 2 )  2.40 m / s. m (70 kg) 6.40: a) From Eq. (6.14), with dl  Rd , P2 0 W  F cos  dl  2wR  cos  d  2wR sin  0 . P1 0    In an equivalent geometric treatment, when F is horizontal, F  dl  Fdx , and the total work is F  2 w times the horizontal distance, in this case (see Fig. 6.20(a)) R sin θ0 , giving the same result. b) The ratio of the forces is w tan  0  2 cot θ0 . 2w 2wR sin θ0 sin θ0 θ c) 2  2 cot 0 . wR(1  cos 0 ) (1  cos 0 ) 2 6.41: a) The initial and final (at the maximum distance) kinetic energy is zero, so the positive work done by the spring, (1 / 2)kx 2 , must be the opposite of the negative work done by gravity,  mgLsin θ , or 2mgL sin θ 2(0.0900 kg )(9.80 m / s 2 )(1.80 m) sin 40.0 x   5.7 cm. k (640 N / m) At this point the glider is no longer in contact with the spring. b) The intermediate calculation of the initial compression can be avoided by considering that between the point 0.80 m from the launch to the maximum distance, gravity does a negative amount of work given by  (0.0900 kg)(9.80 m / s 2 )(1.80 m  0.80 m) sin 40.0  0.567 J , and so the kinetic energy of the glider at this point is 0.567 J. At this point the glider is no longer in contact with the spring.
  11. 6.42: The initial and final kinetic energies of the brick are both zero, so the net work done on the brick by the spring and gravity is zero, so (1 2)kd 2  mgh  0 , or d  2mgh / k  2(1.80 kg )(9.80 m / s 2 )(3.6 m) /(450 N / m)  0.53 m. The spring will provide an upward force while the spring and the brick are in contact. When this force goes to zero, the spring is at its uncompressed length. 6.43: Energy  (power )( time)  (100 W )(3600 s)  3.6  10 5 J 1 K  mv 2 so v  2 K/m  100 s for m  70 kg. 2 6.44: Set time to stop: F  ma :  k mg  ma a  μk g  (0.200)(9.80 m / s 2 )  1.96 m / s 2 v  v0  at 0  8.00 m / s  (1.96 m / s 2 )t t  4.08 s 2 KE 1 mv P  2 t t 1 (20.0 kg )(8.00 m / s 2 )  2  157 W 4.08 s 6.45: The total power is (165 N)(9.00 m / s)  1.485  10 3 W , so the power per rider is 742.5 W, or about 1.0 hp (which is a very large output, and cannot be sustained for long periods). (1.0  1019 J / yr ) 6.46: a)  3.2  1011 W. (3.16  10 s / yr) 7 3.2  1011 W b)  1.2 kW/person . 2.6  108 folks 3.2  1011 W c)  8.0  10 8 m 2  800 km 2 . (0.40)1.0  10 W / m 3 2
  12. 6.47: The power is P  F  v . F is the weight, mg, so P  (700 kg) (9.8 m s 2 ) (2.5 m s)  17.15 kW. So, 17.15 kW 75 kW.  0.23, or about 23% of the engine power is used in climbing. 6.48: a) The number per minute would be the average power divided by the work (mgh) required to lift one box, (0.50 hp) (746 W hp)  1.41 s, (30 kg) (9.80 m s 2 ) (0.90 m) or 84.6 min. b) Similarly, (100 W)  0.378 s, (30 kg) (9.80 m s 2 ) (0.90 m) or 22.7 min. 6.49: The total mass that can be raised is (40.0 hp) (746 W hp)(16.0 s)  2436 kg, (9.80 m s 2 ) (20.0 m) so the maximum number of passengers is 1836 kg 65.0 kg  28. 6.50: From any of Equations (6.15), (6.16), (6.18) or (6.19), Wh (3800 N) (2.80 m) P   2.66  103 W  3.57 hp. t (4.00 s) (0.70) Pave (0.70) (280,000 hp)(746 W hp) 6.51: F    8.1  106 N. v (65 km h) ((1 km h) (3.6 m s)) 6.52: Here, Eq. (6.19) is the most direct. Gravity is doing negative work, so the rope  must do positive work to lift the skiers. The force F is gravity, and F  Nmg , where N is the number of skiers on the rope. The power is then P  ( Nmg ) (v) cos   1m s   (50) (70 kg) (9.80 m s 2 ) (12.0 km h)   3.6 km h  cos (90.0  15.0)     2.96  10 W. 4 Note that Eq. (1.18) uses  as the angle between the force and velocity vectors; in this case, the force is vertical, but the angle 15.0 is measured from the horizontal, so   90.0  15.0 is used.
  13. 6.53: a) In terms of the acceleration a and the time t since the force was applied, the speed is v  at and the force is ma, so the power is P  Fv  (ma) (at )  ma 2t. b) The power at a given time is proportional to the square of the acceleration, tripling the acceleration would mean increasing the power by a factor of nine. c) If the magnitude of the net force is the same, the acceleration will be the same, and the needed power is proportional to the time. At t  15.0 s , the needed power is three times that at 5.0 s, or 108 W. 6.54: dK d  1 2    mv  dt dt  2  dv  mv dt  mva  mav  Fv  P. 6.55: Work done in each stroke is W  Fs and Pav  W t  100 Fs t t  1.00 s, F  2mg and s  0.010 m. Pav  0.20 W.
  14. 6.56: Let M  total mass and T  time for one revolution 1 KE   (dm)v 2 2 M dm  dx L 2πx v T L 2 1  M   2πx  KE    dx    0 2 L  T    4π L 2 2 1M    2    x dx  2 L  T 0   4π   L3  2 2 2 1M    2      π ML2 T 2  3  3 2 L  T   5 revolutions in 3 seconds  T  3 5 s 2 KE  π 2 (12.0 kg) (2.00 m) 2 (3 5 s) 2  877 J. 3 6.57: a) (140 N) (3.80 m)  532 J b) (20.0 kg) (9.80 m s 2 ) (3.80 m) ( sin 25)  315 J c) The normal force does no work. d) W f   f k s   μk ns   μk mgs cos θ  (0.30) (20.0 kg) (9.80 m s 2 ) (3.80 m) cos 25  203 J e) 532 J  315 J  203 J  15 J (14.7 J to three figures). f) The result of part (e) is the kinetic energy at the top of the ramp, so the speed is v  2 K m  2(14.7 J) (20.0 kg)  1.21 m s .
  15. 6.58: The work per unit mass is (W m)  gh. a) The man does work, (9.8 N kg) (0.4 m)  3.92 J kg. b) (3.92 J kg) (70 J kg)  100  5.6%. c) The child does work, (9.8 N kg) (0.2 m)  1.96 J kg. (1.96 J kg) (70 J kg)  100  2.8%. d) If both the man and the child can do work at the rate of 70 J kg, and if the child only needs to use 1.96 J kg instead of 3.92 J kg, the child should be able to do more pull ups. 6.59: a) Moving a distance L along the ramp, sin  L, sout  L sin α, so IMA  sin α . 1 b) If AMA  IMA, ( Fout Fin )  ( sin sout ) and so ( Fout ) ( sout )  ( Fin ) ( sin ) , or Wout  Win . c) d) Wout ( Fout )( sout ) Fout Fin AMA E    . Win ( Fin )( sin ) sin sout IMA w  Wg s (7.35  103 J) 6.60: a) m     41.7 kg. g g (9.80 m s 2 ) (18.0 m) Wn 8.25  103 J b) n    458 N. s 18.0 m c) The weight is mg  Wsg  408 N, so the acceleration is the net force divided by the 458 N  408 N mass,  1.2 m s 2 . 41.7 kg 6.61: a) 2 2 1 2 1  2πR  1  2π (6.66  106 m)  mv  m   (86,400 kg)   (90.1 min) (60 s min )   2.59  1012 J.  2 2  T  2   b) (1 2) mv  (1 2) (86,400 kg) ((1.00 m) (3.00 s))  4.80  103 J. 2 2
  16. 6.62: a) W f   f k s   μk mg cos θs  (0.31) (5.00 kg) (9.80 m s 2 ) cos 12.0(1.50 m)  22.3 J (keeping an extra figure) b) (5.00 kg) (9.80 m s 2 ) sin 12.0 (1.50 m)  15.3 J. c) The normal force does no work. d) 15.3 J  22.3 J  7.0 J. e) K 2  K1  W  (1 2) (5.00 kg) (2.2 m s) 2  7.0 J  5.1 J, and so v2  2(5.1 J ) /(5.00 kg )  1.4 m / s . 6.63: See Problem 6.62: The work done is negative, and is proportional to the distance s that the package slides along the ramp, W  mg (sin θ  μk cos θ ) s . Setting this equal to the (negative) change in kinetic energy and solving for s gives (1 / 2)mv12 v12 s  mg (sin θ  μk cos θ ) 2 g (sin θ  μk cos θ ) (2.2 m/s) 2   2.6 m. 2(9.80 m / s 2 )(sin 12  (0.31) cos12) As a check of the result of Problem 6.62, (2.2 m / s) 1  (1.5 m) /( 2.6 m)  1.4 m / s . 6.64: a) From Eq. (6.7), x x2 dx x2  1 2 1 1 W   Fx dx  k  2  k    k   . x  x1 x x1  x  x1  2 x1  The force is given to be attractive, so Fx  0 , and k must be positive. If x2  x1 , 1 x2  1 x1 , and W  0 . b) Taking “slowly” to be constant speed, the net force on the object is zero, so the force applied by the hand is opposite Fx , and the work done is negative of that   found in part (a), or k x11  x12 , which is positive if x2  x1 . c) The answers have the same magnitude but opposite signs; this is to be expected, in that the net work done is zero. 6.65: F  mg ( RE / r ) 2 RE  mgR  2 2 W    Fds     2 E  dr   mgRE ((1 / r ) | E )  mgRE 2 R   r  1   Wtot  K 2  K1 , K1  0 This gives K 2  mgRE  1.25  1012 J K 2  1 mv2 so v2  2 K 2 / m  11,000 m / s 2 2
  17. 6.66: Let x be the distance past P.  k  0.100  Ax when x  12.5 m ,  k  0.600 A  0.500 / 12.5 m  0.0400 /m (a) W  KE : Wf  KEf  KEi 1   μk mgdx  0  mvi2 2 xf 1 g  (0.100  Ax)dx  vi2 0 2  x  1 2 g (0.100) xf  A f   vi2  2 2  x2  1 (9.80 m / s 2 ) (0.100) xf  (0.0400 / m) f   (4.50 m / s) 2  2 2 Solve for xf : xf  5.11m (b)  k  0.100  (0.0400 / m)(5.11 m)  0.304 (c) Wf  KEf  KEi 1  μk mgx  0  mv12 2 (4.50 m / s) 2 x  vi2 / 2 μk g   10.3 m 2(0.100)(9.80 m / s 2 ) 6.67: a) xa  (4.00 N m3 )(1.00 m)3  4.00 N. 3 b) xb  (4.00 N m3 )(2.00 m)3  32.0 N. c) Equation 6.7 gives the work needed to 3 move an object against the force; the work done by the force is the negative of this, x2  4   x 3dx   ( x2  x14 ). x1 4 With x1  xa  1.00 m and x2  xb  2.00 m , W  15.0 J , this work is negative.
  18. 6.68: From Eq. (6.7), with x1  0 , x2 x2 k 2 b 3 c 4 W   Fdx   (kx  bx 2  cx3 ) dx  x2  x2  x2 0 0 2 3 4  (50.0 N / m) x2  (233 N / m ) x2  (3000 N / m ) x2 2 2 3 3 4 a) When x2  0.050 m , W  0.115 J , or 0.12 J to two figures. b) When x2  0.050 m, W  0.173 J , or 0.17 J to two figures. c) It’s easier to stretch the spring; the quadratic  bx 2 term is always in the  x -direction, and so the needed force, and hence the needed work, will be less when x2  0 . 2 6.69: a) T  marad  m vR  (0.120kg) (0.70 m/s)  0.147 N , or 0.15 N to two figures. b) At 2 ( 0 .40 m) 2 the later radius and speed, the tension is (0.120kg) (2.80 m/s)  9.41 N , or 9.4 N to two ( 0 .10 m) figures. c) The surface is frictionless and horizontal, so the net work is the work done by the cord. For a massless and frictionless cord, this is the same as the work done by the person, and is equal to the change in the block’s kinetic energy, K 2  K1  (1 / 2)m(v2  v12 )  (1 / 2)(0.120 kg)((2.80 m / s) 2  (0.70 m / s) 2 )  0.441 J . Note 2 that in this case, the tension cannot be perpendicular to the block’s velocity at all times; the cord is in the radial direction, and for the radius to change, the block must have some non-zero component of velocity in the radial direction. 6.70: a) This is similar to Problem 6.64, but here   0 (the force is repulsive), and x2  x1 , so the work done is again negative; 1 1 W       (2.12  10 26 N  m 2 ((0.200 m 1 )  (1.25  109 m 1 )) x x   1 2   2.65  1017 J. Note that x1 is so large compared to x 2 that the term 1 x1 is negligible. Then, using Eq. (6.13)) and solving for v2 , 2W 2(2.65  1017 J) v2  v12   (3.00  105 m / s) 2   27  2.41  105 m / s. m (1.67  10 kg) b) With K 2  0, W   K1 . Using W   x 2 ,   2 2(2.12  1026 N  m 2 ) x2     2.82  1010 m. K1 mv12 (1.67  10 27 kg)(3.00  105 m / s) 2 c) The repulsive force has done no net work, so the kinetic energy and hence the speed of the proton have their original values, and the speed is 3.00  105 m / s .
  19. 6.71: The velocity and acceleration as functions of time are dx v(t )   2t  3t 2 , a (t )  2  6 t dt a) v(t  4.00 s)  2(0.20 m / s )(4.00s)  3(0.02 m / s3 )(4.00s) 2  2.56 m / s. 2 b) ma  (6.00 kg)(2(0.20 m / s 2 )  6(0.02 m / s 3 )(4.00 s)  5.28 N. c) W  K 2  K1  K 2  (1 / 2)(6.00 kg)(256 m / s) 2  19.7 J. 6.72: In Eq. (6.14), dl  dx and   31.0 is constant, and so P2 x2 W   F cos dl   F cos  dx P1 x1 1.50 m  (5.00 N / m 2 ) cos 31.0 x 2 dx  3.39 J. 1.00 m The final speed of the object is then 2W 2(3.39 J ) v2  v12   (4.00 m / s) 2   6.57 m / s. m (0.250 kg ) 6.73: a) K 2  K1  (1 / 2)m(v2  v12 ) 2  (1 / 2)(80.0 kg)((1.50 m / s) 2  (5.00 m / s) 2 )  910 J. b) The work done by gravity is  mgh  (800 kg)(980 ms 2 )(520 m)  408  103 J , so the work done by the rider is 910 J  (408 103 J)  317 103 J .   b b b 6.74: a) W   dx   . x0 x n (n  1)) x n 1 x0 (n  1) x0 1 n 1n Note that for this part, for n  1, x  0 as x   . b) When 0  n  1 , the improper integral must be used,  b  W  lim  ( x 2 1  x0 1 ), n n x2  ( n  1)   n 1 and because the exponent on the x 2 is positive, the limit does not exist, and the integral diverges. This is interpreted as the force F doing an infinite amount of work, even though F  0 as x2  . 6.75: Setting the (negative) work done by the spring to the needed (negative) change in kinetic energy, 1 kx 2  1 mv 0 , and solving for the spring constant, 2 2 2 2 mv0 (1200 kg )(0.65 m / s) 2 k   1.03  105 N / m. x2 (0.070 m) 2
  20. 6.76: a) Equating the work done by the spring to the gain in kinetic energy, 2 kx 0  2 mv , so 1 2 1 2 k 400 N / m v x0  (0.060 m)  6.93 m / s. m 0.0300 kg b) Wtot must now include friction, so 1 mv 2  Wtot  1 kx0  fx0 , where f is the magnitude 2 2 2 of the friction force. Then, k 2 2f v x0  x0 m m 400 N / m 2(6.00 N)  (0.06 m) 2  (0.06 m)  4.90 m / s. 0.0300 kg (0.0300 kg) c) The greatest speed occurs when the acceleration (and the net force) are zero, or kx  f , x  kf  40000 Nm  0.0150 m . To find the speed, the net work is 6. N/ Wtot  1 k ( x0  x 2 )  f ( x0  x) , so the maximum speed is 2 2 k 2 2f vmax  ( x0  x 2 )  ( x0  x) m m 400 N / m 2(6.00 N)  ((0.060 m) 2  (0.0150 m) 2 )  (0.060 m  0.0150 m) (0.0300 kg) (0.0300 kg)  5.20 m/s, which is larger than the result of part (b) but smaller than the result of part (a). 6.77: Denote the initial compression of the spring by x and the distance from the initial position by L. Then, the work done by the spring is 1 kx 2 and the work done by friction is 2   k mg ( x  L) ; this form takes into account the fact that while the spring is compressed, the frictional force is still present (see Problem 6.76). The initial and final kinetic energies are both zero, so the net work done is zero, and 1 kx 2   k mg ( x  L) . Solving for 2 L, (1 / 2)kx 2 (1 / 2)(250 N / m)(0.250 m) 2 L x  (0.250 m)  0.813 m,  k mg (0.30)(2.50 kg)(9.80 m / s 2 ) or 0.81 m to two figures. Thus the book moves .81 m  .25 m  1.06 m , or about 1.1 m. 6.78: The work done by gravity is Wg   mgL sin θ (negative since the cat is moving up), and the work done by the applied force is FL, where F is the magnitude of the applied force. The total work is Wtot  (100 N)(2.00 m)  (7.00 kg )(9.80 m / s 2 )(2.00 m) sin 30  131.4 J. The cat’s initial kinetic energy is 1 2 mv12  1 (7.00 kg)(2.40 m / s) 2  20.2 J , and 2 2( K1  W ) 2(20.2 J  131.4 J ) v2    6.58 m / s. m (7.00 kg )
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