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  1. 1.) Does the work of a force on a body depend on the frame of reference in which it is calculated? Give some examples.
  2.  
  3. The work of a force on a body does NOT depend on the frame of reference in which it is calculated. The angle is always the angle between the two vectors.
  4.  
  5. 2.) Does your body do work (external or internal) when standing at rest? When walking steadily along a level road?
  6.  
  7. Work done is defined by force dot displacement, with this definition when the displacement is zero, the work done is also zero. When the body walks steadily along a
  8.  
  9. level road, work is done because a displacement is made and a force is given to counter friction.
  10.  
  11. 3.) Consider a pendulum swinging back and forth. During what part of the motion does the weight do positive work? Negative work?
  12.  
  13. Weight does positive work as the pendulum is moving down, and negative work while the pendulum is moving up.
  14.  
  15. 4.) Since v² = vx² + vy² + vz², Eq.(7.23) implies K = 1/2mv²x + 1/2mv²y + 1/2mv²z. Does this mean that the kinetic energy has x, y, and z components?
  16.  
  17. No, and because of this kinetic energy is a scalar value.
  18.  
  19. 5.) Consider a woman steadily climbing a flight of stairs. The external forces on the woman are her weight and the normal force of the stairs against her feet. During the
  20.  
  21. climb, the weight does negative work, while the normal force does no work. Under these conditions how can the kinetic energy of the woman remain constant? (Hint: The entire
  22.  
  23. woman cannot be regarded as a particle, since her legs are not rigid; but the upper part of thr body can be regarded as a particle, since it is rigid. What is the force of
  24.  
  25. her legs against the upper part of her body? Does this force do work?)
  26.  
  27. The kinetic energy of the woman can remain constant when her force equals the force of weight.
  28.  
  29. 6.) An automobile increases its speed from 80 to 88 km/h. What is the percentage of increase in kinetic energy? What is the percentage of reduction of travel time for a
  30.  
  31. given distance?
  32.  
  33. The percentage of increase is 10%. The percentage of reduction in travel time for a given distance is (1 - (80/88))*100%.
  34.  
  35. 7.) Two blocks in contact slide past one another and exert friction forces on one another. Can the friction force increase the kinetic energy of one block? Of both? Does
  36.  
  37. there exist a reference frame in which the friction force decreases the kinetic energy of both blocks?
  38.  
  39. If these blocks were heading in opposite directions, then both blocks would lose kinetic energy. If they are both travelling in one direction there is a possibility
  40.  
  41. of a block increasing its kinetic energy.
  42.  
  43. 8.) When an automobile with rear-wheel drive is accelerating on, say, a level road, the horizontal force of the road on the rear wheels does not give the automobile any
  44.  
  45. energy because the point of application of this force (point of contact of wheel on ground) is instantaneously at rest if the wheel is not slipping. What force gives the body
  46.  
  47. of the automobile energy? Where does this energy come from? (Hint: Consider the force that the rear axle exerts against its bearings.)
  48.  
  49. The energy comes from the force of friction.
  50.  
  51. 9.) Why do elevators have counterweights?
  52.  
  53. So less work is needed to be done to elevate the elevator.
  54.  
  55. 10.) A parachutist jumps out of an airplane, opens a parachute, and lands safely on the ground. Is the mechanical energy for this motion conserved?
  56.  
  57. No, as some of the energy was lost in the drag of the parachute.
  58.  
  59. 11.) If you release a tennis ball at some hight above a hard floor, it will bounce up and down severa l timees, with a gradually decreasing amplitude. Where does the ball
  60.  
  61. suffer a loss of mechanical energy?
  62.  
  63. Some energy is lost from the drag in the air but the majority of it is lost when the ball makes contact with the floor.
  64.  
  65. 12.) Two ramps, one steeper than the other, lead from the floor to a loading platform (Fig. 7.25). It takes more force to push a (frictionless) box up the steeper ramp.
  66.  
  67. Does this mean it takes more work to raise the box from the floor to the platform?
  68.  
  69. No, the same amount of work is needed for the box to go up both ramps.
  70.  
  71. 13.) Consider the two ramps described in the preceding question. Taking friction into account, which ramp requires less work for raising a box from the floor to the
  72.  
  73. platform?
  74.  
  75. The steeper ramp (with a shorter distance) will take less work for raising the box to the platform.
  76.  
  77. 14.) A stone is tied to a string. Can you whirl this stone in a vertical circle with constant speed? Can you whirl this stone with constant energy? For each of these two
  78.  
  79. cases describe how you must move your hand.
  80.  
  81. You can whirl this stone in a circle with constant speed, to do this you must ensure that the force you apply upwards increases to account for the force of gravity
  82.  
  83. and the force downward is in the opposite direction to also account for the force of gravity.
  84.  
  85. 7.) A record for stair climbing was achieved by a man who raced up the 1600 steps of the Empire State Building to a height of 320 m in 10 min 59 s. If his average mass
  86.  
  87. was 75 kg, how much work did he do against gravity? At what average rate (in J/s) did he do this work?
  88.  
  89. The work he did against gravity was Fg·D = |Fg||D|cos(Fg,D) = 735N * 320 m * 1 = 235200J. The rate at which he did this work was 235200J/(10min)*(60s/1min)+59s =
  90.  
  91. 356.904401 J/s.
  92.  
  93. 11.) A man pushes a heavy box up an inclined ramp making an angle of 30° with the horizontal. The mass of the box is 60 kg, and the coefficient of kinetic friction between
  94.  
  95. the box and the ramp is 0.45. How much work must the man do to push the box to a height of 2.5 m at constant speed? Assume that the man pushes on the box in a direction
  96.  
  97. parallel to the surface of the ramp.
  98.  
  99. Fpush = Ff + Fgsinθ
  100. Ff = μmgcosθ
  101. Ff = (0.45)(60 kg)(9.8 m/s²)(0.866025404)
  102. Ff = 229.150322 N
  103. Fgsinθ = (60 kg)(9.8 m/s²)(0.5)
  104. Fgsinθ = 294 N
  105. Fpush = 229.150322 N + 294 N = 523.150322 N
  106. Work Done = F·Dcos(F,D) = 523.150322 N * 2.5/sin 30 * 1 = 2615.7516 J
  107.  
  108. 17.) A man pulls a cart along a level road by means of a short rope stretched over his shoulder and attached to the front end of the cart. The friction force that oppposes
  109.  
  110. the motion of the cart is 250 N.
  111.  
  112. a.) If the rope is attached to the cart at shoulder height, how much work must the man do to pull the cart 50 m at constant speed?
  113.  
  114. Fpull = Ff
  115. Fpull = 250 N
  116. W = Fpull·Dcos(Fpull,D) = 12500 J
  117.  
  118. b.) If the rope is attached to the cart below shoulder height so it makes an angle of 30° with the horizontal, what is the tension in the rope? How much work must the man
  119.  
  120. now do to pull the cart 50 m? Assume that enough mass was added so the friction force is unchanged.
  121.  
  122. Fpull = Ff/cos30 = 288.675135 N
  123. W = Fpull·Dcos(Fpull,D) = 12500 J
  124.  
  125. 19.) An elevator consists of an elevator cage and a counterweight attached to the ends of the cable that runs over a pully (Fig. 7.27). The mass of the cage (with its
  126.  
  127. load) is 1200 kg, and the mass of the counterweight is 1000 kg. The elevator is driven by an electric motor attached to the pulley. Suppose that the elevator is initially at
  128.  
  129. rest on the first floor of the building and the motor makes the elevator accelerate upward at the rate of 1.5 m/s².
  130.  
  131. a.) What is the tension in the part of the cable attached to the elevator cage? What is the tension in the part of the cable attached to the counter weight?
  132.  
  133. Ftelevator cage = (1200 kg)(9.8 m/s² + 1.5 m/s²) = 13560 N
  134. Ftcounter weight = (1000 kg)(9.8 m/s² - 1.5 m/s²) = 8300 N
  135.  
  136. b.) The acceleration last exactly 1.0 s. How much work has the electric motor done in this interval? Ignore friction forces and ignore the mass of the pulley.
  137.  
  138. y = 0.75m/s²
  139. y = 0.75m
  140. Fmotor = (1200 kg - 1000 kg)(9.8 m/s²) = 2260 N
  141. W = Fmotor·ycos(Fmotor,y) = 2260 N * 0.75 m * 1 = 1695 J
  142.  
  143. c.) After the acceleration interval of 1.0 s, the motor pulls the elevator upward at constant speed until it reaches the third floor, exactly 10m above the first floor.
  144.  
  145. What is the total amount of work that the motor has done up to this point?
  146.  
  147. D = 10 m - 0.75 m = 9.25 m
  148. Fmotor = Fg
  149. Fmotor = (200 kg)(9.8 m/s²) = 1960 N
  150. W = Fmotor·Dcos(Fmotor,D) = 1960 N * 9.25 m * 1 = 18130 J
  151.  
  152. 21.) During a storm, a sailboat is anchored in a 10-m-deep harbor. The wind pushes against the boat with a steady horizontal force of 7000 N.
  153.  
  154. a.) The anchor rope that holds the boat in place is 50 m long and is stretched straight between the anchor and the boat (Fig. 7.28a). What is the tension in the rope?
  155.  
  156. Trope = Fwind/cosθ
  157. θ = sin−1(10/50) = 11.5369590328°
  158. Trope = 7000 N / cos 11.5369590328° = 7144 N
  159.  
  160. b.) How much work must the crew of the sailboat do to pull in 30 m of the achor rope, bringing the boat nearer to the anchor (Fig. 7.28b)? What is the tension in the rope
  161.  
  162. when the boat is in this new position?
  163.  
  164. x1 = sqrt(r² - y²) = 49 m
  165. x2 = sqrt(r² - y²) = 17.3 m
  166. DeltaX = 31.7 m
  167. W = Fpull·DeltaXcos(0) = 7000 N * 31.7 m * 1 = 221900 J
  168.  
  169. Trope = Fwind/cosθ
  170. θ = sin−1(10/20) = 30°
  171. Trope = 7000 N / cos 30° = 8082 N
  172.  
  173. 35.) Calculate the kinetic energy that the Earth has owing to its motion around the Sun.
  174.  
  175. Kinetic energy is given by 1/2mv²
  176. Kinetic energy = (1/2)(5.9742*10^24 kg)(29786 m/s)² = 2.65*10^33 J
  177.  
  178. 45.) A mass of 150 g is held by a horizontal spring of spring constant 20 N/m. It is displaced from its equilibrium position and released from rest. As it passes trhough
  179.  
  180. equilibrium, its speed is 5.0 m/s. For the motion from the release position to the equilibrium position, what is the work done by the spring?
  181.  
  182. The work done by the spring is = (1/2)(0.150 kg)(5 m/s)² = 1.875 J
  183.  
  184. 53.) A box of mass 40 kg is initially at rest on a flat floor. The coefficient of kinetic friction between the box and the floor is μk = 0.6. A woman pushes horizontally
  185.  
  186. against the box with a force of 250 N until the box attains a speed of 2.0 m/s.
  187.  
  188. a.) What is the change of kinetic energy of the box?
  189.  
  190. (1/2)(40 kg)(2 m/s)² = 80 J
  191.  
  192. b.) What is the work done by the friction force on the box?
  193.  
  194. Ff = (0.6)(40 kg)(9.8 m/s²) = 235.2 N
  195. Fnet = 250 N - 235.2 N = 14.8 N
  196. a = 14.8 N / 40 kg = 0.37 m / s²
  197. t = v/a = 5.4 s
  198. x = (1/2)(0.37 (5.4)²) = 5.4 m
  199. W = Ff·xcos(Ff,x) = 235.2 N * 5.4 m * -1 = -1270 J
  200.  
  201. c.) What is the work done by the woman on the box?
  202.  
  203. W = Fpush·xcos(Fpush,x) = 250 N * 5.4 m * 1 = 1350 J
  204.  
  205. 59.) In pole vaulting, the jumper achieves great height by converting her kinetic energy of running into gravitational potential enrgy (Fig. 7.32). The pole plays an
  206.  
  207. intermediate role in this process. When the jumper leaves the ground, part of her translational kinetic energy has been converted into kinetic energy of rotation (with the
  208.  
  209. foot of the pole as the center of rotation) and part has been converted into elastic potential energy of deformation of the pole. When the jumper reaches her highest point,
  210.  
  211. all of this energy has been converted into gravitational potential energy. Suppose that a jumper runs at a speed of 10 m/s. If the jumper converts all of the corresponding
  212.  
  213. kinetic energy into gravitational potential energy, how high will her center of mass rise? The actual height reached by pole vaulters is 5.7 m (measured from the ground). Is
  214.  
  215. this consistent with your calculation?
  216.  
  217. t = v/a = 1.02 s
  218. ymax = (1/2)(9.8 (1.02)^2)
  219. ymax = 5.1 m
  220.  
  221. This result is close enough to 5.7 m to be considered acceptable.
  222.  
  223. 63.) A block released from rest slides down to the bottom of a plane of incline 15° from a height of 1.5 m; the block attains a speed of 3.5 m/s at the bottom. By
  224.  
  225. considering the vork done by gravity and the frictional force, determine the coefficient of friction.
  226.  
  227. U = mgh = m(9.8 m/s²)(1.5 m) = 14.7m J
  228. E = 1/2mv² = 1/2m(3.5 m/s)² = 6.125m J
  229. 14.7m J - 6.125m J = 8.575m J
  230.  
  231. Friction caused a 8.575m J loss of energy.
  232.  
  233. 1.5 m / sin 15 = 5.8 m
  234. -8.575m J/5.8 m = -1.48m N
  235. -1.48m N/m = -1.48 m/s²
  236. Ff = μmgcosθ
  237. af = μmgcosθ / m
  238. af = μgcosθ
  239. af/gcosθ = μ
  240. 1.48 m/s²/9.8 m/s²cos 15 = μ
  241. 0.16 = μ
  242.  
  243. 65.) A 2.5-g Ping-Pong bal is dropped from a window and strikes the ground 20 m below with a speed of 9.0 m/s. What fraction of its initial potential energy was lost to air friction?
  244.  
  245. U = mgh = (0.0025 g)(9.8 m/s²)(20 m) = 0.49 J
  246. E = 1/2mv² = (1/2)(0.0025 g)(9 m/s)² = 0.10125 J
  247. Difference in energy = 0.49 J - 0.10125 J = 0.38875 J
  248.  
  249. The fraction of initial potential energy that was lost is 0.38875 J/0.49 J
  250.  
  251. 67.) A skateboarder starts from rest and descends a ramp through a vertical
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