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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?