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KB>100 Module 4.6-4.12 Unfinished

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  1. Module 4.06                                                                     Karnbir Sandhu
  2.                                                                                 Physics
  3.  
  4. 1. Evan is a championship runner of the school's track team. During one of his trials, his speed was measured at 10.1 m/s. Find Evan's kinetic energy during that trial if his mass is 58.6 kg.
  5. Givens:
  6. M=58.6 kg
  7. V=10.1 m/s
  8. KE=?                   
  9. KE=1/2mv^2                     
  10.                
  11. Work:
  12. KE=(0.5)(58.6kg)(10.1 m/s)^2
  13. KE=2988.893 J
  14. KE=2990
  15. KE= 2.99 E 4 J
  16.  
  17. 2. Matthew just threw a 0.15 kg ball straight up in the air.
  18. A. What is the weight of the ball?
  19. Givens:
  20.         m=0.15kg
  21. g=9.81 m/s^2
  22. w=mg
  23. w=0.15kg*9.81 m/s^2
  24. w=1.4715 N
  25. w=1.5 N
  26. B. If the ball reaches a height of 9.0 m, what is its gravitational potential energy at this height?
  27. Givens:
  28. w=1.4715 N
  29. h=9.0m
  30.  
  31. GPE=?
  32. GPE=wf
  33.  
  34. Work:
  35. GPE=1.4715 N*9.0 M
  36. GPE=13.2435 J
  37. GPE=1.3 E 1 J
  38. C. What will happen to the gravitational potential energy of the ball as it falls?
  39. It will decrease as the height decreases
  40. D. What will happen to the kinetic energy of the ball as it falls?
  41. It will increase as the ball accelerates due to gravity.
  42. 3. Christy just got a new car for her 16th birthday. If she is traveling through town at a constant speed of 10.0 m/s, and the kinetic energy of the car is 5.18 E 4 J, what is the mass of the car and contents?
  43.         Givens:
  44.         v=10.0m/s
  45.         KE=5.18 E 4 J
  46.         m=?
  47.        
  48.         KE=1/2mv^2
  49.          KE*2=1/2mv^2*2
  50.         2KE=mv^2
  51.         2KE=mv^2
  52.         [2KE]/v^2=m
  53.         m=[2KE]/v^2=m
  54.         Work:
  55.         M=[(2* 5.18 E 4) /(10.0m/s)^2
  56.         M=1036 kg
  57.         M=1.04 E 3 Kg
  58.        
  59.         4.08
  60. Here is the URL for the Washington Monument. Go to the web site to retrieve the information for part A.
  61. A. roblems:
  62. 1. Here is the URL for the Washington Monument. Go to the web site to retrieve the information for part A.
  63. A. If 1 meter = 3.28 feet, what is the height of the Washington Monument in meters?
  64. 555' 5(/12) 1/8
  65. 1/12 / 1/8= 0.66
  66. 0.66+5=5.66
  67. 555 5.66/12
  68. 555.4716
  69. Givens
  70. 1m=3.28ft
  71. Total ft= 555.4716 ft
  72. 1m=3.28ft
  73. Height(m)=?
  74.  Height(m)=Height(ft)/3.28m
  75.  
  76. Work:
  77. Height=555.425ft/3.28m
  78. Height=169.337
  79. Height=
  80. 169m(rounded to sig figs)
  81. 2.
  82.  
  83.  
  84. Here is the URL for the Washington Monument. Go to the web site to retrieve the information for part A.
  85. A. roblems:
  86. 1. Here is the URL for the Washington Monument. Go to the web site to retrieve the information for part A.
  87. A. If 1 meter = 3.28 feet, what is the height of the Washington Monument in meters?
  88. 555' 5(/12) 1/8
  89. 1/12 / 1/8= 0.66
  90. 0.66+5=5.66
  91. 555 5.66/12
  92. 555.4716
  93. Givens
  94. 1m=3.28ft
  95. Total ft= 555.4716 ft
  96. 1m=3.28ft
  97. Height(m)=?
  98.  Height(m)=Height(ft)/3.28m
  99.  
  100. Work:
  101. Height=555.425ft/3.28m
  102. Height=169.337
  103. Height=
  104. 169m(rounded to sig figs)
  105. 2.
  106.  
  107.  
  108. A. A worker assigned to the restoration of the Washington Monument is checking the condition of the stone at the very top of the monument. A nickel with a mass of 0.005 kg is in her shirt pocket. What is the gravitational potential energy (GPE) of the nickel at the top of the monument?
  109.                 m=0.005kg
  110.                 g=9.81m/s^2
  111.  
  112. GPE=mgh
  113. GPE=(0.005kg)*(9.81m/s^2)*(169.337m)
  114. GPE=8.306 J
  115.  
  116. GPE= 8 J
  117.  
  118.  
  119.  
  120. B. What is the kinetic energy (KE) of the nickel in her shirt pocket at the top of the monument?
  121. Givens:
  122. v=0m/s
  123. m=0.005kg
  124. KE=1/2mv^2
  125. Givens:
  126. KE= (0.5)(0.05kg)(0)^2
  127. KE=(0.5)(0.05kg)(0)
  128. KE=(0.5)(0)
  129. KE=0 J
  130.  
  131.  
  132. C. If the nickel accidentally falls out of her pocket, what will happen to the gravitational potential energy (GPE) of the nickel as it falls to the ground?
  133. decrease as the height decreases
  134. D. If the nickel accidentally falls out of her pocket, what will happen to the kinetic energy (KE) of the nickel as it falls to the ground?
  135. Increase as it accelerates due to gravity.
  136.       
  137. 2. Consider the concepts of kinetic energy (KE) and gravitational potential energy (GPE) as you complete these questions. A ball is held 1.4 meters above the floor. Use the terms KE of GPE as your answers.
  138. A. When the ball is held motionless above the floor, the ball possesses only gravitational potential energy.
  139. B. If the ball is dropped, its  gravitational potential energy energy decreases as it falls.
  140. C. If the ball is dropped, its kinetic energy energy increases as it falls.
  141. D. In fact, in the absence of air resistance, the amount of gravitational potential energy when the ball is held motionless above the floor equals the amount of kinetic energy at impact with the floor.
  142.  
  143. 3. We will now use energy considerations to find the speed of a falling object at impact. Artiom is on the roof replacing some shingles when his 0.55 kg hammer slips out of his hands. The hammer falls 3.67 m to the ground. Neglecting air resistance, the total mechanical energy of the system will remain the same. The kinetic energy and the gravitational potential energy possessed by the hammer 3.67 m above the ground is equal to the kinetic energy and the gravitational potential energy of the falling hammer as it falls. Upon impact, all of the energy is in a kinetic form. The following equation can be used to represent the relationship:
  144.  
  145. GPE + KE (top) = GPE + KE (at impact)
  146.  
  147. Because the hammer is dropped from rest, the KE at the top is equal to zero.
  148. Because the hammer is at base level, the height of the hammer is equal to zero; therefore the PE upon impact is zero.
  149.  
  150. We may write our equation like this:
  151. GPE (top) = KE (at impact)
  152. This gives us the equation:
  153. ( mgh) (top) = 1/2 mv2 (at impact)
  154. A. Notice that the mass of the hammer "m" is shown on both sides of the equation. According to the math rules we have learned, what does this mean?
  155. M cancels out.
  156. B. Manipulate the equation (rearrange the variables) to solve for v. (Remember that manipulating an equation does not involve numbers and substitutions. You just rearrange the equation. v = ?)
  157. gh=1/2v^2
  158. 2gh=v^2
  159. Sqrt(2gh)=v
  160. v=Sqrt(2gh)
  161. C. Use your equation from part B to find the speed with which the hammer struck the ground.
  162.         G=9.81m/s^2
  163.         h=3.97m
  164.                 m=0.55kg
  165.         V=?
  166.         ( mgh) (top) = 1/2 mv2 (at impact)
  167. v=Sqrt(2gh)
  168.  
  169.  
  170.         Work:          
  171.        
  172.  
  173.         v=Sqrt[(2)*(9.81m/s^2)*(3.67m)]
  174.         V=Sqrt(72.0054)
  175.         V=8.4856
  176.         V=8.49 m/s
  177.        
  178.        
  179. 4. Eight million (8,000,000) kg of water are at the verge of dropping over Niagara Falls to the rocks 50.0 meters below. What is the change in gravitational potential energy as the water splashes on the rocks below?
  180. Givens:
  181. G=9.81m/s^2
  182. H=50m
  183. M=8,000,000 kg
  184. GPE=mgh
  185.  
  186. GPE=?
  187. ΔGPE=?
  188. GPE(top)=mgh
  189. H=50.0m
  190. GPE(Impact)= mgh
  191. Impact Height=0m
  192. GPE(Impact)-GPE(Top)=ΔGPE
  193. ΔGPE=GPE(Impact)-GPE(Top)
  194. Work:
  195. GPE(top)=(9.81m/s^2)( 8,000,000 kg)(50.0m)
  196. GPE=3924000000 J
  197.  
  198.  
  199.  
  200.  
  201. GPE(Impact)=(0)(8,000,000 kg)(50.0m)
  202. GPE(Impact)=0 Joules
  203. ΔGPE=0 J - 3924000000 J
  204. ΔGPE=-3924000000
  205. The change in gravitational potentional energy is negative -392400000(converted or "lost" to kinetic energy)
  206.  
  207. 4.10
  208.  
  209. Module 4.10                                                                             Karnbir Sandhu                                                                                          Physics
  210.                                                                                        
  211.  
  212. 1.Calculate the work done when Johnny exerts a 20.0 N force to push a cart 3.5 m.
  213. Givens:
  214. F=20.0 N
  215. d=3.5m
  216. W=?
  217. W=F*d
  218. Work:
  219. W=20.0 N * 3.5m
  220. W=70 J
  221. 2. Robert lifts a 53.5 kg barbell 2.2 m above the ground in 2.0 s.
  222. A. What is the weight of the barbell?
  223. Givens:
  224. m=53.5kg
  225. g=9.81m/s^2
  226. H=2.2m
  227. W=?
  228. W=mg
  229. Work:
  230. W=(9.81m/s^2)(53.5kg)
  231. W=524.835 N
  232. W=525 N
  233.  
  234. B. How much work is done by Robert in lifting the barbell?
  235. Givens:
  236. F=524.835 N
  237. d= 2.2m
  238. W=?
  239.                 W=F*d
  240.                 Work:
  241.                 W=524.835 N*2.2m
  242.                 W=1154.637 J
  243.                 W= 1200 J
  244.                 W= 1.2 E3 J
  245. C. What is the gravitational potential energy of the barbell when it is lifted to this height?
  246. Givens:
  247. h=2.2m
  248. w=524.835 N
  249. GPE=?
  250. GPE=wh
  251. Work:
  252. GPE=524.835 N*2.2m
  253. GPE= 1154.637
  254. GPE= 1200
  255. GPE= 1.2 E 3 J
  256.  
  257. D. Calculate the power expended when the barbell is lifted.
  258. Givens:
  259. W=1154.637 J
  260. T=2.0s
  261.         i       P=?
  262. P=W/t
  263. Work:
  264. P=1154.637/2s
  265. P=577.3185 watts
  266. P= 580 watts
  267.                 P= 5.8 E 2 watts
  268.  
  269.  
  270. 3. Lucas can't find Tipsy the cat. He looks all over the house and is about to "give up" when he sees Tipsy resting 5.0 m above the ground on the limb of a big oak tree. If Lucas has a mass of 25 kg, how much work does Lucas do in climbing up to the limb to retrieve Tipsy?
  271.                 Givens:
  272.                 m=25kg
  273.                 d=h=5.0m
  274.                
  275.                 Work=?
  276.                 Work=GPE=mgh
  277.                 Work:
  278.                 Work=(25kg)(9.81m/s^2)(5.0m)
  279.                 Work=125 J
  280.                 Work=130 J
  281.                 Work: 1.3 E 2 J
  282.  
  283.                
  284.                
  285.                
  286. 4. An astronaut in full space gear climbs a vertical ladder on Earth. Later, the astronaut makes the same climb on the moon. In which location does the astronaut do more work? Explain.
  287. g of Earth>g of moon
  288. Since Work=GPE=mgh
  289. Work=mgh
  290. "the same climb"=same height
  291. h and m are the same on the moon and on earth
  292. And since g(acceleration due to gravity) in mgh is less on the moon(1/6 surface gravity than on Earth)
  293. The work equates to a greater quantity on Earth than on the moon.
  294.  
  295. 5. Oliver moves 8 cartons of books, each weighing 65 N, onto a ramp in 7.62 minutes. Katrina moves the same number of cartons of books in 8.41 minutes. Which student does more work? Explain your answer.
  296. They both do an equal amount of work since they move the same amount of cartons of books that equate to the same mass and weight.(W=F*d)
  297. 6. Kathleen is an experienced archer who aspires to make the Olympic Team. If Kathleen does 17.6 J of work when she holds her bow and carefully pulls her arrow back 0.41 m, what is the average force she exerts?
  298. Givens:
  299.  
  300. W=17.6 J
  301.         d=0.41m
  302.         F=?
  303.         W=F*d
  304.         W/d=F
  305.         F=W/d
  306.         Work:
  307.         F= 17.6 J/ 0.41 M
  308.         F= 42.9268
  309.         F=43 M
  310.         F= 4.3 E 1 N
  311.  
  312.  
  313.        
  314. 4.12
  315. Etotal = KE + PE +Q (In a closed system)
  316.  
  317.  
  318. Examples of conservative forces:
  319. Examples of non-conservative forces:
  320. Gravitational force
  321. Electric force
  322. Frictional forces
  323. Air resistance
  324. As the roller coaster decreases in vertical height, its potential energy decreases, but this is accompanied by an increase in the kinetic energy of the roller coaster as its speed increases.
  325. We can set the total energy at the crest equal to the total energy at the bottom of the drop:
  326. Etot = (PE + KE)top = (PE + KE)bottom
  327.  
  328. Using the bottom of the drop as the base level to determine the potential energy at the bottom equals zero. At the top where the velocity equals zero, the kinetic energy is zero. There is no KE at the crest and there is no PE at the bottom of the drop.
  329. PEtop = KEbottom
  330. Therefore:
  331. (mgh)top = (½ mv2)bottom
  332. The mass cancels out of the equation because it is on both sides of the equation.
  333. (gh)top = (½ v2)bottom
  334. 1 calorie = 4.186 Joules
  335. 1 Kilocalorie = 4186 Joules
  336. Practice Problem 1. A ball is rolling along a frictionless track as shown. How fast must it be going at A in order to just make it to the top of B?
  337.  
  338.  
  339.  
  340. Givens:Etot = (PE + KE)top = (PE + KE)bottom
  341. Etot=(PE+KE)top=(PE+KE)bottom
  342. ΔGEP=+5.0
  343.  
  344. Bh=10.0m
  345. Etot=PEtop=KEbottom
  346. PEtop==mgh=1/2mv^2=KE bottom
  347. gh=1/2v^2
  348. 14.007141035914502420949018641836= m/s
  349.  
  350.  
  351. Sqrt(2gh)=v
  352. Sqrt(2gh)=98.1 m/s
  353. KEBottom=PETOP
  354.  
  355. 1/2mv^2=mgh
  356. 1/2v^2=(9.81m/s^2(10)
  357.  
  358.  
  359.  
  360.  
  361. Practice Problem 2. A 2.0 kg rock is thrown from the top of a hill 60.0 m above sea level using a catapult. The rock leaves the catapult with a speed of 20.0 m/s. As shown in the diagram below, the rock lands 40.0 meters below the foot of the hill. Calculate the speed of the rock just before it makes contact with the ground.
  362. Etot = (PE + KE)top =(PE+KE)bottom
  363.  
  364.         PE+KE
  365.  
  366. Etot=PEtop=KEbottom
  367. Et=PE
  368. Mgh=1/2
  369. Etot=mgh=1/2mv^2
  370. Etot=1177 J=1/2mv^2
  371.         v^2
  372. 34.310348293189913346837442623344
  373.  
  374.  
  375. Practice Problem 3. A 0.145 kg copper projectile, with a Specific Heat Capacity of 387 J/(kg * Co), hits a titanium plate moving at 15.6 m/s. If all of the KE of the projectile is converted into thermal energy of the projectile, what is the increase in temperature of the projectile?
  376. Images: © 2003flvs   
  377.                         KEprojectile=Qbrakes
  378.                         1/2mv^2=0.145kg
  379.  
  380. Loli Lancer
  381. Physics Caryn Flowers
  382. KB>100 Module 4.6-4.12 Unfinished