Tài liệu Physics exercises_solution: Chapter 06 docx

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 )(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 )  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 6.5: a) See Exercise 5.37 The needed force is  k mg (0.25)(30 kg)(9.80 m / s )  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 )  (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: Fs cos   2(1.80  10 N)(0.75  10 m) cos14  2.62  10 J, two places or 2.6  10 J to 6.8: The work you is:   ˆ ˆ F  s  ((30 N)i  (40 N) ˆ)  ((9.0m)i  (3.0m) ˆ) j 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  y1 ) When y1  y , Wmg  b) (i) Tension does no work (ii) Let l be the length of the string Wmg   mg ( y  y1 )   mg (2l )  25.1 J The displacement is upward and the gravity force is downward, so it does negative work 6.10: a) From Eq (6.6),   m / s  K  (1600 kg) (50.0 km/h)    1.54  10 J  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 m/s 6.11: For the T-Rex, K  (7000 kg )((4 km / hr ) 3.6 km/hr )  4.32  103 J The person’s velocity would be v  2(4.32  10 J)/70 kg  11.1 m/s , or about 40 km/h 6.12: (a) Estimate: v  1ms (walking) v  ms (running) m  70 kg Walking: KE  mv  (70 kg )(1 m / s)  35 J 2 Running: KE  (70 kg )(2 m / s)  140 J (b) Estimate: v  60 mph  88 ft / s  30 m / s m  2000 kg KE  (2000 kg )(30 m / s)   10 J (c) KE  Wgravity  mgh Estimate h  m KE  (1 kg)(9.8 m / s )(2 m)  20 J 6.13: Let point be at the bottom of the incline and let point be at the skier Wtot  K  K 1 K  mv0 , K  Work is done by gravity and friction, so Wtot  Wmg  W f Wmg   mg ( y  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  gh(1   k / tan  ) 6.14: (a) W  KE 1  mgh  mvf2  mv0 2 v0  vf2  gh  (25.0 m / s)  (9.80 m / s )(15.0 m)  30.3 m / s (b) W  KE 1  mgh  mvf2  mv0 2 2 v  vf (30.3 m / s)  02 h  2g 2(9.80 ms / s )  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  Wmg  W||  W  mgh , same as falling height h b) Wtot  K  K1 gives mgh  mv and v  gh , same as if had been dropped 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  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  (1 / 2)(0.145 kg)(32.0 m/s)  74.2 J 6.18: As the example explains, the boats have the same kinetic energy K at the finish line, 2 2 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 speeds, v A / v B  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  6.19: a) From Eq (6.5), K  K / 16 , and from Eq (6.6), W  (15 / 16) K 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, F 6.21: 2 2 K m(v  v1 ) (8.00 kg )((6.00 m / s)  (4.00 m / s) )    32.0 N (2.50 m) s s s K (0.420 kg)((6.00 m / s)  (2.00 m / s) )   16.8 cm (40.0 N) F 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 Fs 2(36.0 N)(1.20 m) v    4.48 m / s (4.30 kg ) m m b) The net work is Fs  f k s  ( F   k mg ) s , so v 2( F   k mg ) s m 2(36.0 N  (0.30)(4.30 kg)(9.80 m / s ))(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.)  6.23: a) On the way up, gravity is opposed to the direction of motion, and so W   mgs  (0.145 kg)(9.80 m / s )(20.0 m)  28.4 J 2(28.4 J ) W  (25.0 m / s)   15.26 m / s (0.145 kg ) m 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 b) v2  v12  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 )(25.0 m)  1176 J b) Since the melon is released from rest, K  , and Eq (6.6) gives K  K  W  1176 J 6.25: a) Combining Equations (6.5) and (6.6) and solving for v2 algebraically, v  v12  Wtot 2(10.0 N)(3.0 m)  (4.00 m / s)   4.96 m / s (7.00 kg ) m Keeping extra figures in the intermediate calculations, the acceleration is a  (10.0 kg  m / s ) /(7.00 kg)  1.429 m / s From Eq (2.13), with appropriate change in notation, v  v12  2as  (4.00 m / s)  2(1.429 m / s )(3.0 m), 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   gh  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 )(0.75 m) sin 36.9  2.97 m / s 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 / 2) mv0 , and so s  v /  k g b) From the result of part (a), the stopping distance is proportional to the square of the initial speed, and so for an initial speed of 60 km/h, s  (91.2 m)(60.0 / 80.0)  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 )   4.00 cm 3.00 cm  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 the necessary algebra One way is to note  169 N    (0.050 m)  J , that to stretch the spring the original 0.050 m requires  2 0.050 m    so that stretching 0.015 m requires (4 J )(0.015 / 0.050)  0.360 J and compressing 0.020 m requires (4 J )(0.020 / 0.050)  0.64 J Another is to find the spring constant k  (160 N)  (0.050 m)  3.20  10 N / m , from which (1 / 2)(3.20  10 N / m)(0.015 m)  0.360 J and (1 / 2)(3.20  10 N / m )( 0.020 m )  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) / (8 m)(10 N)  40 J b) / (4 m)(10 N)  20 J c) / (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 v (2)(60 J)  3.46 m / s 10 kg 6.32: The work you with your changing force is  6.9 F ( x) dx   6.9 (20.0 N) dx   6.9 3.0 N xdx m N )( x / 2) | 0.9 m  138 N  m  71.4 N  m  209.4 J or  209 J  (20.0 N) x | 0.9 (3.0 The work is negative because the cow continues to advance as you vainly attempt to push her backward 6.33: Wtot  K  K1 K  mv0 , K  Work is done by the spring force Wtot   kx , where x is the amount the spring is compressed 1  kx   mv0 and x  v m / k  8.5 cm 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)  4.00  10 N/m , giving the same results 6.35: a) The static friction force would need to be equal in magnitude to the spring force, m m s mg  kd or μs  ((020.0 N / )( 9)(80.086/ s 2))  1.76 , which is quite large (Keeping extra figures in 100 kg m the intermediate calculation for d gives a different answer.) b) In Example 6.6, the relation 1  k mgd  kd  mv12 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  gd k d m k   mg    mg    s   2 k g  s  m k   k  mg 2 (  s  2 s  k ) k  (0.10 kg)(9.80 m / s )   (20.0 N / m)    ((0.60)  2(0.60)(0.47)),   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  kx  (200 N / m)(0.025 m)  0.06 J 2 b) The work-energy theorem gives 2(0.06 J ) 2W v   0.18 m / s (4.0 kg ) m 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 6.38: a) K  4.0 J , so v  K m  (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  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  (1 / 2)mv , 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 ) , so the final speed is v k 2 ( x1  x )  m (4000 N / m ((0.375 m)  (0.200 m) )  2.40 m / s (70 kg) 6.40: a) From Eq (6.14), with dl  Rd , W  P2 0 F cos  dl  2wR  cos  d  2wR sin  P1    In an equivalent geometric treatment, when F is horizontal, F  dl  Fdx , and the total work is F  w times the horizontal distance, in this case (see Fig 6.20(a)) R sin θ0 , 2w giving the same result b) The ratio of the forces is w tan   cot θ0 2wR sin θ0 sin θ0 θ 2  cot (1  cos 0 ) wR(1  cos 0 ) c) 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 , 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 )(1.80 m) sin 40.0   5.7 cm (640 N / m) k 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 )(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 x 6.66: Let x be the distance past P when x  12.5 m ,  k  0.600 (a)  k  0.100  Ax A  0.500 / 12.5 m  0.0400 /m W  KE : Wf  KEf  KEi   μk mgdx   mvi2 xf g  (0.100  Ax)dx  vi2 2  x  g (0.100) xf  A f   vi2 2   x2  (9.80 m / s ) (0.100) xf  (0.0400 / m) f   (4.50 m / s) 2  Solve for xf : xf  5.11m (b)  k  0.100  (0.0400 / m)(5.11 m)  0.304 (c) Wf  KEf  KEi  μk mgx   mv12 (4.50 m / s) x  vi2 / μk g   10.3 m 2(0.100)(9.80 m / s ) 6.67: a) xa  (4.00 N m3 )(1.00 m)3  4.00 N b) xb  (4.00 N m3 )(2.00 m)3  32.0 N c) Equation 6.7 gives the work needed to move an object against the force; the work done by the force is the negative of this, x2    x 3dx   ( x2  x14 ) x1 With x1  xa  1.00 m and x2  xb  2.00 m , W  15.0 J , this work is negative 6.68: From Eq (6.7), with x1  , k b c x2  x2  x2 0 2 3  (50.0 N / m) x2  (233 N / m ) x2  (3000 N / m ) x2 x2 x2 W   Fdx   (kx  bx  cx3 ) dx  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 term is always in the  x -direction, and so the needed force, and hence the needed work, will be less when x2  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 ( 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 ( 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  K1  (1 / 2)m(v2  v12 )  (1 / 2)(0.120 kg)((2.80 m / s)  (0.70 m / s) )  0.441 J Note 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   (the force is repulsive), and x2  x1 , so the work done is again negative; 1 1 W       (2.12  10 26 N  m ((0.200 m 1 )  (1.25  109 m 1 )) x x     2.65  1017 J Note that x1 is so large compared to x that the term x1 is negligible Then, using Eq (6.13)) and solving for v2 , v2  v12  2W 2(2.65  1017 J)  (3.00  105 m / s)   2.41  105 m / s  27 (1.67  10 kg) m  b) With K  0, W   K1 Using W   x ,  2 2(2.12  1026 N  m )    2.82  1010 m K1 mv12 (1.67  10 27 kg)(3.00  105 m / s) 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 x2  6.71: The velocity and acceleration as functions of time are dx v(t )   2t  3t , a (t )  2  t dt a) v(t  4.00 s)  2(0.20 m / s )(4.00s)  3(0.02 m / s3 )(4.00s)  2.56 m / s b) ma  (6.00 kg)(2(0.20 m / s )  6(0.02 m / s )(4.00 s)  5.28 N c) W  K  K1  K  (1 / 2)(6.00 kg)(256 m / s)  19.7 J 6.72: In Eq (6.14), dl  dx and   31.0 is constant, and so P2 x2 P1 x1 W   F cos dl   F cos  dx 1.50 m  (5.00 N / m ) cos 31.0 1.00 m x dx  3.39 J The final speed of the object is then 2W 2(3.39 J ) v2  v12   (4.00 m / s)   6.57 m / s (0.250 kg ) m 6.73: a) K  K1  (1 / 2)m(v2  v12 )  (1 / 2)(80.0 kg)((1.50 m / s)  (5.00 m / s) )  910 J b) The work done by gravity is  mgh  (800 kg)(980 ms )(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 6.74: a) W    x0 b b dx  n x (n  1)) x n 1 1n Note that for this part, for n  1, x integral must be used,   x0 b n (n  1) x0 1  as x   b) When  n  , the improper  b  n n W  lim  ( x 1  x0 1 ), x2  ( n  1)   n 1 and because the exponent on the x 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  as x2   6.75: Setting the (negative) work done by the spring to the needed (negative) change in kinetic energy, kx  mv , and solving for the spring constant, 2 k mv0 (1200 kg )(0.65 m / s)   1.03  105 N / m (0.070 m) x2 6.76: a) Equating the work done by the spring to the gain in kinetic energy, 2 1 kx  mv , so v k 400 N / m x0  (0.060 m)  6.93 m / s m 0.0300 kg b) Wtot must now include friction, so mv  Wtot  kx0  fx0 , where f is the magnitude 2 of the friction force Then, k 2f v x0  x0 m m 400 N / m 2(6.00 N) (0.06 m)  (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 N/  Wtot  k ( x0  x )  f ( x0  x) , so the maximum speed is vmax   k 2f ( x0  x )  ( x0  x) m m 400 N / m 2(6.00 N) ((0.060 m)  (0.0150 m) )  (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 kx and the work done by friction is   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 kx   k mg ( x  L) Solving for L, (1 / 2)kx (1 / 2)(250 N / m)(0.250 m) L x  (0.250 m)  0.813 m, (0.30)(2.50 kg)(9.80 m / s )  k mg 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 )(2.00 m) sin 30  131.4 J The cat’s initial kinetic energy is v2  mv12  (7.00 kg)(2.40 m / s)  20.2 J , and 2( K1  W )  m 2(20.2 J  131.4 J )  6.58 m / s (7.00 kg ) 6.79: In terms of the bumper compression x and the initial speed v0 , the necessary relations are 2 kx  mv0 , kx  mg 2 Combining to eliminate k and then x, the two inequalties are v2 mg x and k  25 5g v a) Using the given numbers, (20.0 m / s) x  8.16 m, 5(9.80 m / s ) (1700 kg)(9.80 m / s )  1.02  10 N / m (20.0 m / s) b) A distance of m is not commonly available as space in which to stop a car k  25 6.80: The students positive work, and the force that they exert makes an angle of 30.0 with the direction of motion Gravity does negative work, and is at an angle of 60.0 with the chair’s motion, so the total work done is Wtot  ((600 N ) cos 30.0  (85.0 kg )(9.80 m / s ) cos 60.0)(2.50 m)  257.8 J , and so the speed at the top of the ramp is 2Wtot 2(257.8 J ) v2  v12   (2.00 m / s)   3.17 m / s (85.0 kg) m Note that extra figures were kept in the intermediate calculation to avoid roundoff error 6.81: a) At maximum compression, the spring (and hence the block) is not moving, so the block has no kinetic energy Therefore, the work done by the block is equal to its initial kinetic energy, and the maximum compression is found from kX  mv , or 2 m v k b) Solving for v in terms of a known X, k v X  m X  5.00 kg (6.00 m s)  0.600 m 500 N m 500 N m (0.150 m)  1.50 m s 5.00 kg 6.82: The total work done is the sum of that done by gravity (on the hanging block) and that done by friction (on the block on the table) The work done by gravity is (6.00 kg) gh and the work done by friction is  μk (8.00 kg) gh, so Wtot  (6.00 kg  (0.25)(8.00 kg) (9.80 m s ) (1.50 m)  58.8 J This work increases the kinetic energy of both blocks; Wtot  (m1  m2 )v , so 2(58.8 J) v  2.90 m s (14.00 kg) 6.83: See Problem 6.82 Gravity does positive work, while friction does negative work Setting the net (negative) work equal to the (negative) change in kinetic energy, (m1  μk m2 ) gh   (m1  m2 )v , and solving for  k gives μk  m1  (1 2) (m1  m2 )v gh m2 (6.00 kg)  (1 2) (14.00 kg) (0.900 m s) ((9.80 m s ) (2.00 m)) (8.00 kg)  0.79  6.84: The arrow will acquire the energy that was used in drawing the bow (i.e., the work done by the archer), which will be the area under the curve that represents the force as a function of distance One possible way of estimating this work is to approximate the F vs x curve as a parabola which goes to zero at x  and x  x0 , and has a maximum of F0 at x  x0 , so that F ( x)  F0 x0 x( x0  x) This may seem like a crude approximation to the figure, but it has the ultimate advantage of being easy to integrate; x0 x F0 F  x x3  Fdx   ( x0 x  x ) dx  20  x0    F0 x0  x0 x0  3   With F0  200 N and x0  0.75 m, W  100 J The speed of the arrow is then 2W m  (100 J) (0.025 kg)  89 m s Other ways of finding the area under the curve in Fig (6.28) should give similar results 6.85: f k  0.25 mg so W f  Wtot  (0.25 mg ) s, where s is the length of the rough patch Wtot  K  K1  2 2 K1  mv0 , K  mv2  m(0.45v0 )  0.2025 mv0 2 2  The work-energy relation gives  (0.25mg ) s  (0.2025  1) mv0 The mass divides out and solving gives s  1.5 m 6.86: Your friend’s average acceleration is v  v0 6.00 m/s a   2.00 m/s 3.00 s t Since there are no other horizontal forces acting, the force you exert on her is given by Fnet  ma  (65.0 kg)(2.00 m/s )  130 N Her average velocity during your pull is 3.00 m/s, and the distance she travels is thus 9.00 m The work you is Fx  (130 N)(9.00 m)  1170 J , and the average power is therefore 1170 J/3.00 s  390 W The work can also be calculated as the change in the kinetic energy 6.87: a) (800 kg)(9.80 m/s )(14.0 m)  1.098  105 J, or 1.10  105 J to three figures b) (1 / 2)(800 kg)(18.0 m/s2 )  1.30  105 J c) 6.88: 1.10  10 J  1.30  10 J  3.99 kW 60 s P  Fv  mav  m(2  βt )(2t  βt )  m(4 2t  18βt  18 β 2t )  (0.96 N/s)t  (0.43 N/s )t  (0.043 N/s )t At t  400 s, the power output is 13.5 W 6.89: Let t equal the number of seconds she walks every day Then, (280 J/s)t  (100 J/s)(86400 s  t )  1.1  10 J Solving for t, t  13,111 s  3.6 hours 6.90: a) The hummingbird produces energy at a rate of 0.7 J/s to 1.75 J/s At 10 beats/s, the bird must expend between 0.07 J/beat and 0.175 J/beat b) The steady output of the athlete is 500 W/70 kg  W/kg, which is below the 10 W/kg necessary to stay aloft Though the athlete can expend 1400 W/70 kg  20 W/kg for short periods of time, no human-powered aircraft could stay aloft for very long Movies of early attempts at human-powered flight bear out this observation d d 6.91: From the chain rule, P  dt W  dt (mgh)  dm gh, for ideal efficiency Expressing dt the mass rate in terms of the volume rate and solving gives (2000  106 W) m3  1.30  103 (0.92)(9.80 m/s )(170 m)(1000 kg/m ) s 6.92: a) The power P is related to the speed by Pt  K  mv , so v  b) a) a dv d  dt dt Pt  m x  x0   v dt  2P d t m dt 2P  m t Pt m P 2mt P 12 2P 8P t dt  t2  t2 m  m 9m 6.93: a) (7500  103 kg )(1.05  103 kg/m3 )(9.80 m/s )(1.63 m)  1.26  105 J b) (1.26  105 J)/(86,400 s)  1.46 W 6.94: a) The number of cars is the total power available divided by the power needed per car, 13.4  106 W  177, (2.8  103 N)(27 m/s) rounding down to the nearest integer b) To accelerate a total mass M at an acceleration a and speed v, the extra power needed is Mav To climb a hill of angle  , the extra power needed is Mg sin v These will be nearly the same if a ~ g sin  ; if g sin  ~ g tan  ~ 0.10 m/s , the power is about the same as that needed to accelerate at 0.10 m/s2 c) (1.10  106 kg)(9.80 m/s )(0.010)(27 m/s)  2.9 MW d) The power per car needed is that used in part (a), plus that found in part (c) with M being the mass of a single car The total number of cars is then 13.4  106 W  2.9  106 W  36, (2.8  103 N  (8.2  104 kg)(9.80 m/s )(0.010))(27 m/s) rounding to the nearest integer 6.95: a) P0  Fv  (53  103 N)(45 m/s)  2.4 MW b) P  mav  (9.1  105 kg)(1.5 m/s )(45 m/s)  61 MW c) Approximating sin  , by tan  , and using the component of gravity down the incline as mg sin  , P2  (mg sin  )v  (9.1  105 kg)(9.80 m/s )(0.015)(45 m/s)  6.0 MW 6.96: a) Along this path, y is constant, and the displacement is parallel to the force, so W  αy  xdx  (2.50 N/m )(3.00 m) (2.002m)  15.0 J b) Since the force has no y-component, no work is done moving in the y-direction c) Along this path, y varies with position along the path, given by y  1.5 x, so Fx   (1.5 x) x  1.5x , and W   Fx dx  1.5  x dx  1.5(2.50 N/m ) (200 m)3  10.0 J 6.97: a) P  Fv  ( Froll  Fair )v  ((0.0045)(62.0 kg)(9.80 m/s )  (1 / 2)(1.00)(0.463 m (1.2 kg/m3 )(12.0 m/s) )(12.0 m/s)  513 W 6.98: a) 28.0  103 W P F   1.68  103 N v (60.0 km/h)((1 m/s)/(3.6 km/h)) b) The speed is lowered by a factor of one-half, and the resisting force is lowered by a factor of (0.65  0.35 / 4), and so the power at the lower speed is (28.0 kW)(0.50)(0.65  0.35/4)  10.3 kW  13.8 hp c) Similarly, at the higher speed, (28.0 kW)(2.0)(0.65  0.35  4)  114.8 kW  154 hp 6.99: a) (8.00 hp)(746 W/hp)  358 N (60.0 km/h)((1 m/s)/(3.6 km/h)) b) The extra power needed is mgv||  (1800 kg)(9.80 m/s ) 60.6 km/h sin(arctan(1 / 10))  29.3 kW  39.2 hp, 3.6 km/h m/s so the total power is 47.2 hp (Note: If the sine of the angle is approximated by the tangent, the third place will be different.) c) Similarly, mgv||  (1800 kg)(9.80 m/s ) 60.0 km/h sin(arctan(0.010))  2.94 kW  3.94 hp, 3.6 km/h m/s This is the rate at which work is done on the car by gravity The engine must work on the car at a rate of 4.06 hp d) In this case, approximating the sine of the slope by the tangent is appropriate, and the grade is (8.00 hp)(746 W/hp)  0.0203, (1800 kg)(9.80 m/s2 )(60.0 km/h)((1 m/s)/(3.6 km/h)) very close to a 2% grade 6.100: Use the Work–Energy Theorem, W  KE , and integrate to find the work x KE   mv0 and W   ( mg sin   μmg cos  )dx Then, x   Ax W   mg  (sin   Ax cos  )dx, W   mg sin x  cos     Set W  KE   Ax  mv0   mg sin x  cos   2   To eliminate x, note that the box comes to a rest when the force of static friction balances the component of the weight directed down the plane So, mg sin   Ax mg cos ; solve this for x and substitute into the previous equation sin  x A cos  Then,    sin    A     sin  sin    A cos   cos  , v0   g   A cos        and upon canceling factors and collecting terms, v0  stationary whenever v0  3g sin  A cos  3g sin  Or the box will remain A cos  6.101: a) Denote the position of a piece of the spring by l; l  is the fixed point and l  L is the moving end of the spring Then the velocity of the point corresponding to l, denoted u, is u (l )  v (when the spring is moving, l will be a function of time, and so u L is an implicit function of time) The mass of a piece of length dl is dm  M dl , and so L dK  1 Mv 2 dmu  l dl , 2 L3 and L Mv 2 Mv K   dK  l dl  L3  b) kx  mv , so v  (k m) x  (3200 N m) (0.053 kg) (2.50  10 2 m)  6.1 m s 2 c) With the mass of the spring included, the work that the spring does goes into the kinetic energies of both the ball and the spring, so kx  mv  Mv Solving for v, 2 v k (3200 N m) x (2.50  10  m)  3.9 m s mM (0.053 kg)  (0.243 kg) d) Algebraically, (1 2)kx mv   0.40 J and (1  M 3m) (1 2)kx Mv   0.60 J (1  3m M ) 6.102: In both cases, a given amount of fuel represents a given amount of work W0 that the engine does in moving the plane forward against the resisting force In terms of the range R and the (presumed) constant speed v,   W0  RF  Rv   v   In terms of the time of flight T , R  vt , so β  W0  vTF  T v   v  a) Rather than solve for R as a function of v, differentiate the first of these relations with dW respect to v, setting dv0  to obtain dR F  R dF  For the maximum range, dR  0, so dv dv dv dF dv  Performing the differentiation, dF dv  2v  β v  0, which is solved for 14 14  3.5  105 N  m s  β v    0.30 N  s m   32.9 m s  118 km h      d b) Similarly, the maximum time is found by setting dv ( Fv)  0; performing the differentiation, 3v  β v  0, which is solved for 14  β  v   3  14  3.5  105 N  m s    3(0.30 N  s m      25 m s  90 km h 6.103: a) The walk will take one-fifth of an hour, 12 From the graph, the oxygen consumption rate appears to be about 12 cm3 kg  min, and so the total energy is (12 cm3 kg  min) (70 kg) (12 min) (20 J cm3 )  2.0  105 J b) The run will take Using an estimation of the rate from the graph of about 33 cm3 kg  gives an energy consumption of about 2.8  105 J c) The run takes min, and with an estimated rate of about 50 cm3 kg  min, the energy used is about 2.8  105 J d) Walking is the most efficient way to go In general, the point where the slope of the line from the origin to the point on the graph is the smallest is the most efficient speed; about km h   6.104: From F  ma , Fx  max , Fy  ma y and Fz  maz The generalization of Eq (6.11) is then dv dv dv ax  vx x , a y  v y y , az  vz z dx dy dz The total work is then Wtot   ( x2 , y , z ) ( x1 , y1 , z1 ) Fx dx  Fy dy  Fz dz y2 z2 dv  x dv  dv  m  vx x dx   v y y dy   vz z dz   x1 y1 z1 dx dy dz    vx2 vy2 vz  m  vx dvx   v y dv y   vz dvz   v  v y1 vz  x1  2 2  m(vx  vx1  v y  v y1  vz22  vz21 2  mv2  mv12 2 ... spring From either Eq (6.9) or Eq (6.10), W  kx  (200 N / m)(0.025 m)  0 .06 J 2 b) The work-energy theorem gives 2(0 .06 J ) 2W v   0.18 m / s (4.0 kg ) m 6.37: The work done in any interval... is the magnitude 2 of the friction force Then, k 2f v x0  x0 m m 400 N / m 2(6.00 N) (0 .06 m)  (0 .06 m)  4.90 m / s 0.0300 kg (0.0300 kg) c) The greatest speed occurs when the acceleration... maximum speed is vmax   k 2f ( x0  x )  ( x0  x) m m 400 N / m 2(6.00 N) ((0 .060 m)  (0.0150 m) )  (0 .060 m  0.0150 m) (0.0300 kg) (0.0300 kg)  5.20 m/s, which is larger than the result