Chapter 6 Soln - Chapter 6 Work and Kinetic Energy 6.1...

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Unformatted text preview: Chapter 6 Work and Kinetic Energy 6.1: a) (2.40 N)(1.5 m) = 3.60 J b) (—0.600 N)(1.50 1n) = —0.900 J c) 3.60 J — 0.720 .1 = 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 = F3 = mgs = (6.75 kg)(9.80 m/sz)(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 = mm. m itka = (0.25)(30.0 kg)(9.80 111/32) = 73.5 N, or 74 N to two figures. b) From Eq. (6.1), F3 = (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) m —331 J. (1) 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 F — w. “my... 9.3.5.290 REW ‘ c0505 — pk sii'IE ‘ cos 30° — (0.25) sin 30° = 99.2 N, keeping extra figures. b) F3 coed) : (99.2 N)(4.50 In) cos 30° = 386.5 J, again keeping an extra figure. c) The normal force is mg + Fsin 42, and so the work done by friction is —(4.50 m)(0.25)((30 kg)(9.80 m/s2) + (99.2 N) sin 30°) = —386.5 J. (1) 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), F's cosgb = (180 N)(300 m) cos 15.0° = 5.22 x 104 J. 155 156 Chapter 6 6.7: 2Fscosqb = 2(1.80 x 106 N)(O.75 ><103 m) cos 14° = 2.62 x 109 J, or 2.6 x 109 J to two places. 6.8: The work you do is: _. F .3 = ((30 N): — (40 mi) - ((—90 m)2 - (3-0 H03“) = (30 N)(—9.0 m) + (—40 N)(—3.0 m) = _270 N.m + 120N111 = —150J 6.9: a) Tension force is always perpendicular to the displacement and does no work. (ii) Work done by gravity is —mg(y2 —— yl). When y1 = y2, me == 0. b) Tension does no work. (ii) Let I be the length of the string. ng = —mg(y2 — 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), 2 $0600 kg) ((50.0 km/h) = 1.54 x105 J. 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. 6-11: F01“ the T'Rex: K = gig-fl? = 4.32 x 103 J. The person's velocity would be '0 = 2(4.32 x 103 J)/70 kg = 11.1 m/s, or about 40 km/h. 6.12: (3.) Estimate: 1) % lm/s (walking) 1) a: 2m/s (running) K: m m 70 kg Walking: KE = ém’u2 = %(70kg)(1m/s)2 = 35J Running: KE = -;—(70 kg)(2 m/s)2 = 140.] (b) Estimate: 1) m 60 mph = 88 ft/s z 30 m/s m z 2000 kg 1 KE 2 5(2000 kg)(30 m/s)2 = 9 x 105J (c) K E = ngdty = mgh Estimate h. a: 2 m KE = (1 kg)(9.8m/sz)(2 m) z 20 J 6.13: Let point 1 be at the bottom of the incline and let point 2 be at the skier. Wtot : K2 — K1 K1 z émvg, K2 = 0 Work is done by gravity and friction, so Wm = ng + Wf. Work and Kinetic Energy 157 Wm = —mg(y2 — in) = -m9h W; = —fs = -(,ukmgcosot)(h/ sin a) = —,ukmgh/ tana Substituting these expressions into the worleenergy theorem and solving for v0 gives v0 = mdi'tmmj 6.14: (a) W = AKE —mgh = émvfz — émvfi ‘00 = x/U? +297i WWWW = ‘/ (25.0 m/s)2 + 2(9.80m/82)(15.0 m) = 30.3 m/s 0”) W = AKE 1 2 1 2 —mgh = émvf w Emvo h _ sired _ QBEEL-ilifl 2g 2(9.80 111/52) = 46.8 m 6.15: a) parallel to incline: force component 2 my sin a, down incline; displacement = h/ sina, down incline W" = (my sin a)(h/ sin a) = mgh perpendicular to incline: no displacement in this direction, so W_L = 0. Wm = W" + W; = mgh, same as falling height h. b) Wm = K2 — K1 gives mgh = aims2 and v = £575, 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 1) = V295 = 17.1s 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 inde- pendent of speed, the distance must also increase by a factor of four. 158 Chapter 6 6.17: Barring a balk, the initial kinetic energy of the ball is zero, and so W = (1/2)m'u2 = (1/2)(0.145 kg)(32.0 111/3)? = 74.2 J. 6.18: As the example explains, the boats have the same kinetic energy K at the finish line, so (1/3771waa = (1/2)mgv§, or, with m3 = 2mA, v3 = 21%. 3.) Solving for the ratio of the speeds, 12A / v3 = 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 tB/tA = vA/vg = 6.19: a) From Eq. (6.5), K2 2 113/16, and from Eq. (6.6), W = —(15/16)K1. 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, 2 95 2 in»: - v?) : iEEQPng-OO wetnegnefz = 32 0 N F W s s (2.50 m) M = W_Mfifi;fiflflflfifil 2 16 8 cm 0.21: = .— 3 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 _ §f§_ 2(36.0 N)(1.20 m) _ y _ m _ m _ (4.30 kg) — 4.48 m/s. b) The net work is F3 — fks = (F - nkmg)s, so v = MW...WWW. m “3230.0 N (0.30)(4.30 kg) (9.80 m/sz))(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/s2)(20.0 m) = “28.4 J. W 2(—.28..4_J) b = 2 _ = 2 m...“ ) 02 v1 + 2m (25.0 m/s) + (0.145 kg) = 15.26 m/s. Work and Kinetic Energy 159 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 = F3 = mgs = (4.80 kg)(9.80 m/s2)(25.0 m) = 1176 J. b) Since the melon is released from rest, K1 = 0, and Eq. (6.6) gives 6.25: a) Combining Equations (6.5) and (6.6) and solving for v2 algebraically, v2 = \/v;" + 2&5 = \/ (4.00 m/s)2 + 39100 NIQQMEQ = 4.96 m/s. (7.00 kg) Keeping extra figures in the intermediate calculations, the acceleration is a = (10.0 kg-m/sg)/(7.00 kg) = 1.429 m/s2. From Eq. (2.13), with appropriate change in notation, v3 = vf + 20.3 = (4.00 m/s)2 + 2(1429 m/s2)(3.0 In), giving the same result. 6.26: The normal force does no work. The work—energy theorem, along with Eq. (6.5), gives 1) = ([35 = 3g = = m where h = Lsin6 is the vertical distance the block has dropped, and 9 is the angle the plane makes with the horizontal. Using the given numbers, '0 = \/§(9.80 m/s2)(0.75 iii)”§ihw'§éf§”5 = 2.97 m/s. 6.27: a) The friction force is nkmg, which is directed against the car‘s motion, so the net work done is —-,ukmgs. The change in kinetic energy is AK = -K1 = —(1/2)mv§, and so 5 = 413/an9. b) Fromthe result of part (a), the stopping distance is pro- portional to the square of the. initial speed,.and so foran initial speed of 60 km/h, 3 = (91.2 In)(60.0/80.0)2 = 51.3 In. (This method avoids the intermediate calculation of ,uk, which in this case is about 0.279.) 6.28: The intermediate calculation .of thespring'. 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) = 21.3 J. 160 Chapter 6 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 1 1 0 N that to stretch the spring the original 0.050 In requires 5 (0.050 1n)2 = 4 J, so that stretching 0.015 m requires (4 J)(0.015/0.050)2 = 0.360 J and compressing 0.020 111 requires (4 J )(0.020/ 0.050)”! = 0.64 J. Another is to find the spring constant k = (160 N)+ (0.050 1n) = 3.20 X 103 N/m, from which (1/2)(3.20 x 103 N/m)(0.015 m)2 = 0.360 J and (1/2)(3.20 x103 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 x 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)(4O J) = *,.—~.«uu--—mmw~— = 2' I a) v \/ 10 kg 83 m/s b) At a; = 12 In, the 40 Joules of kinetic energy will have been increased by 20 J, so - (336331 11 —‘/ 10 kgw m 3.46 m/s. 6.32: The work you do with your changing force is 6.9 N 6.9 6.9 F(:r)d:1: =/ (—20.0N)d:r —] 3.0—wdx o o 0 III = (—20.0N)x|3-9 - (3.0%)62/2) 3-9 = —138N-m— 71.4N-m = ——209.4J or —209J The work is negative because the cow continues to advance as you" vainly attempt to push her backward. 6.33: Wm == K2 — K1 K1 2 ém'ug, K2 = 0 Work is done by the spring force. WM 2 —%k:r2, where a: is the amount the spring is compressed. 1 —§k:r2 =-w~2«mv§ and :L' = 'vM/m/k = 8.5 cm Work and Kinetic Energy 161 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 X 103 N / m, giving the same results. 6.35: a) The static friction force would need to be equal in magnitude to the spring force, psmg : kd or a, = tig£436mhfiéfi§g§§7§j = 1.76, which is quite large. (Keeping extra figures in the intermediate calculation for d gives a different answer.) b) In Example 6.6, the relation 1 1 ,ukmgd + ikd2 = Emu? was obtained, and d was found in terms of the known initial speed '01. In this case, the condition on d is that the static friction force at maximum extension just balances the spring force, or kd = aging. Solving for of and substituting, k of = —d2 + 2gdakd m _ m92 — T (P: + 2flsflk) = (nggygfiglfifi) ((0.60)2 + 2(0.60)(0.47)), from which 01 = 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 = %k:r2 = %(200 N/m)(0.025 m)2 = 0.06 J. b) The work-energy theorem gives _ Jam new m 1)... “.- (4; --0.18 /S. 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 x 1.0 J, andsince the curve is below the axis (Fa, < 0), the work is negative, or -1.0 J. (I) The net work is the sum of the results of parts (a), (b) and (c), 3.0 J. (c) +1.0 J — 2.0 J = —1.0 J. 162 Chapter 6' 6.38: a) K = 4.0 J, so 1) = yams”; = \/2(4.0 J)/(2.0 kg) = 2.00 m/s. b) No work is done between :r = 3.0 m and a: = 4.0 m, so the speed is the same, 2.00 m/s. 0) K m 3.0 J, so 1) = = = 1.73 m/s. 6.39: a) The spring does positive work on the sled and rider; (1 / 2)lcal:2 = (1 / 2)mv2, or v = = (0.375 m)\/(4000 N/m)/(70 kg) = 2.83 m/s. b) The net work done by the spring is (1/2)]c — mg), so the final speed is v = \/ = ¢i§fl§lfi ((0.375 m)2 — (0.200 m)2) = 2.40 m/s. 6.40: a) From Eq. (6.14), with all = qub, P: 90 W = Fcosqbdl = 2wa cosqbdgfi = 2szin 90. 0 P1 In an equivalent geometric treatment, when T" is horizontal, d? = Fdx, and the total work is F 2 210 times the horizontal distance, in this case (see Fig. 6.20(a)) Rsin 60, giving the same result. b) The ratio of the forces is flfiw- = 2 cot 60. 210R sin 60 Sill 60 60 c) __.._.......,,......m.......--........m = M......-....m.-..,. = _ wR(1— c0500) 2 (1 -— 008%) 23% 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)k:c2, must be the Opposite of the negative work done by gravity, —mgL sin 6, or x _ \Ffigtsme _ x/EEHBQOOinkszXLS‘O‘Inlein 400° _ .-.......W.-....... WWW-..” .mmmmmm WWMMMM mm. mm _ 5_7 cm_ It 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 —(0.090() kg)(9.80 m/52)(1.80 m — 0.80 m) sin 400° = —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. 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)kd2 «— mgh = 0, or d = x/2mgh/ = @030 kg)(9.80 m/s2)(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. Whenthis force goes to zero, the spring is at its uncompressed length. Work and Kinetic Energy 163 6.43: Energy = (p0wer)(time) = (100 W)(3600 s) = 3.6 x 105 J K: ém‘u2 sou: \/2K/m= 1005 form: 70 kg. 6.44: Set time to stop: 2F = ma :11.ka = ma 0. = ,ukg = (0.200)(9.80m/sz) = 1.96m/s2 ’U = ’00 + at 0 = 8.00 m/s -— (1.96 m/sz)t t = 4.088 p = _ $323 1(20.0 kg)(8.00 111/82) —— 3...”...WTW = — 4.08 S 157W 6.45: The total power is (165 N)(9.00 m/s) = 1.485 x 103 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). 19 6.46: a) figflgmflfl = 3.2 x1011 W. 3.2 x 1011 W b WW. 3 . . ) 2.6 x 108 folks I 2 kW/pemon 11 c) WWW-r = 8.0 x 108 m2 = 800 km2. (0.40)1.0 x 103 W/m2 6.47: The power is P = F-v. F is the weight, mg, so P = (700 kg)(9.8 III/S2) (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, ‘ (Montage/13% _ 141 /s (30 kg)(9.80 m/s2)(0.90 m) — ' ’ or 84.6 /min. b) Similarly: MQQQmW = 0.378 /s, (30 kg)(9.80 m/s”)(0.90 m) or 22.7 /min. 164 Chapter 6 6.49: The total mass that can be raised is Sigi‘LEBJSZEEXVm/lfillfififl m 2435 kg (9.80 m/s2)(20.0 In) so the maximum number of passengers is m = 28. 6.50: From any of Equations (6.15), (6.16), (6.18) or (6.19), _ Wh _ (3800 N)(2.80 m) _ 3 _ P— t _ (4.003) "2.66x10 W—3.57hp. 651. F = (We... 2 ntzgisesggenzemp) = 8. x we N_ 'u (65 km/h)((1 km/h)/(3.6 m/Eji 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 ? is gravity, and F = N mg, where N is the number of skiers on the rope. The power is then P = (ng)(v) cosqfi 1 m/s = (50)(70 kg)(9.80 m/s2)( 12.0 km/h) (gmémfimfi) cos(90.0° — 15.0°) = 2.96 x 104 W. Note that Eq. (1.18) uses 05 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 9!) 2 900° — 15.0° is used. 6.53: a) In terms of the acceleration a and the time t since the force was. applied, the speed is 'u = at and the force is me, so the power is P = Fv = (ma)(at) = mazt. b) The power at a given time is proportional to the square of the acceleration, tripling the ac- celeration would mean increasing the power by a factor of nine. 0) 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: (1K (1 1 2 is“ a a (5” ) = mi: @ dt = mm = mav = F?) = P 6.55: Work done in each stroke is W = F3 and PM. = W/t = IOOFS/t t= 1.00 s, F = 2mg and s = 0.010 m. P... = 0.20 W. Work and Kinetic Energy 165 6.56: dm ) M, mm“. ““““ P _‘/" “W “* “firm”; L KE: ~;-(dm)v2 M dmzfdx V213 _ T L1 M 2m: 2 1 M) (Ml/L 2 =— —— Mm— wdm 2(AI/l 4T22 0153 2 _,1 7T __ 2 2 2 ~2(L)(T2)(3)*3WML/T 5 revolutions in 3 seconds —> T = 3/ 5 8 KB = §w2(12.0kg)(2.00 m)2/(3/58)2 = 877.1 6.57: a) (140 N)(3.80 m) = 532 J b) (20.0 kg)(9.80 m/82)(3.80 m)(— sin 25°) = —315 J c) The normal force does no work. d) Wf 2': _fk3 = flukns = —,u.kmgs (3089 = —(0.30)(20.0 kg)(9.80 m/s2)(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 (6) is the kinetic energy at the top of the ramp, so the speed is v = ME = = 1.21 m/s. 6.58: The work per unit mass is = 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) x 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) x 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. 166 Chapter 6 6.59: a) Moving a distance L along the ramp, 5-“, = L, sout = L sin a, so I M A = min. b) If AMA = IMA, (Font/Fin) = (sin/sent) and so (Font)(sout) — (Em)(sin), or want = in- 0) d) E z lilies}. ,__ 1F sutllmsmeeil = Essifis 2 £942 I/Vin (Finxsin) Sin/Soul; w — /s (7.35 ><103 J) 6. : = -— = = —-»-~w ‘WW« 2 41.7 k . 60 a) m g g (9.80 m/s2)(18.0 m) g Wn 8.25 X 103 J b) n — ‘5‘ m 18.63"” ‘ 458 N- e) The weight is my = % = 408 N, so the acceleration is the net force divided by the ass gay—408 me — 1 2 m/s2 m ’ 41.7 kg m“ ‘ ' 6.61: a) 1 2 1 21rR 2 1 27r(6.66 x 106 In) 2 12 _ = _ W E __ 4 WW”.W..W.-WWWW = 2. 1 _ 27m} 2m ( T ) 2(86’ 00 kg) ((90.1 min)(60 s/min) 59 x 0 J b) (1/2)mv2 = (1/2)(86,400 kg) ((1.00 m)/(3.00 3))2 = 4.80 x 103 J. 6.62: 3) WI = .__fk3 = —p,kmg (:0st : -(0.31)(5.00 kg)(9.80 m/s2) cos 12.0°(1.50 m) = w22.3 J (keeping an extra figure) b) (5.00 kg)(9.80 m/sz)sin12.0°(1.50 m) = 15.3 J. c) The normal force does no work. (:1) 15.3 J — 22.3 J = -—7.0 J. e) K2 = K1 + W = (1/2)(5.00 kg)(2.2 m/s)2 — 7.0 J = 5.1 J, and so 112 = = 1.4 m/s. 6.63: See Problem 6.62: The work done is negative, and is proportional to the distance 3 that the package slides along the ramp, W = mg(sin 6 —— ,uk cos 9)3. Setting this equal to the (negative) change. in kinetic energy and solving for 3 gives (1/2)m'vi vi mg(sin 0 — ,uk cos 6) - M?g(sin6 — pk Work and Kinetic Energy 167 (2.2 111/ s)2 M5130§5W3im53 — W As a check of the result of Problem 6.62, (2.2 m/s)\/1: (1.57m)/(2.6 In) 2 1.4 m/s. 6.64: a) From Eq. (6.7), The force is given to be attractive, so Fm < 0, and It must be positive. If :02 > m1, f; < i, 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 F2, and the work done is negative of that found in part (a), or k i w A , which is positive if 2:; > :31. c) The answers have the same 31 532 magnitude but opposite signs; this is to be expected, in that the net work done is zero. 6.65: F = mg(RE/1r)‘z 2 RE 2 W n —f Fds = —/ dr = —ngE(-—(1/r)|:E) = ngE 1 00 = 2.6 m. Wtot = K2 —K1,K1 = 0 This gives K2 = ngE = 1.25 x 1012 J K2 = §mv§ so '02 = fiKg/“Et = 11,000 m/s 6.66: Let x be the distance past P. Mk = 0.100 + A33 when x = 12.5 m,,uk = 0.600 A : 0.500/12.5 m = 0.0400/m (a) W=AKE2Wf=KEf—KEi — / ,ukmgda: = 0 — émv? g/ (0.100 + Am)da: = 50? 0 9 [(0.100)z:f + flag—g] = (9.80 We) [(0.100):..—f + ...
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    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

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    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

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    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

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    Dana University of Pennsylvania ‘17, Course Hero Intern

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    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

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    Jill Tulane University ‘16, Course Hero Intern