C. Work Homework
5.4 Potential Energy
Problems: 44, 45, 47, 48 51, 52, 54
51: V2 = Vo2 + 2gh. V = 0 m/s at the top of the flight. Vo = 7.5 m/s. Solve for h. Then use U = mgh where h = 1.2 m + what you found in the previous equation)
54: assume first book is on surface so its work = 0 J. To raise 2nd book, it must have its center of gravity lifted 0.020 m + 0.040 m = 0.060 m. So, books 3, 4, 5 and 6 must be lifted 0.1m, 0.14 m, 0.18 m and 0.22m respectively. W = mgh = 30 N * (S heights) )
5.5 Conservation of Energy
Problems: 55, 56, 57, 60, 61, 62, 63, 64, 66, 69, 70, 71
57: Let’s give this one a whirl! Mr. Proodian’s room has a bowling ball suspended from a wire. Bring your helmets!
62: Remember, U at the top = K at the bottom. U = mgh)
63: use K = 1/2mvo2 to solve for vo (you’ll find vo to be 17.9 m/s).
Then, use V2 = Vo2 + 2ghmax to determine the maximum height. Of course, V = 0 m/s at the max height.
64: At the top of her swing, U = mgh. At the bottom of her swing, K = 1/2 mv2. Utop = Kbottom. Therefore, we can say: mgh = 1/2mv2. MASS CANCELS!!!!! LIFE IS GOOD! h = 2m – 0.5 m = 1.5 m. Solve for v.
66: Ktop + Utop = Kbottom + Ubottom since energy is conserved. Substitute 1/2mv2 for K and mgh for U. MASS CANCELS!!!!! LIFE IS GOOD!
69: First, find the height of the “bob”. L – (LCosq) = height (which is 0.38 m). Next, DU = DK so, mgh = 1/2 mv2 Solve for v. PART B: The law of conservation of energy says that it will basically go to the same height. Of course it doesn’t do this exactly because of friction, but in a “perfect” world, it would….
70: First, find K. Then, assume that all K is imparted to the spring as U. U = 1/2kx2. For part b, the new velocity can be found by vo2 – (vo/2)2
71: Work done by a non-conservative force such as friction equals the change in mechanical energy (DE). E – Eo = K + U – (Ko + Uo). Therefore, -2500 J = DE. Ko and Uo are at the top of the hill. K and U are at the bottom. Recall from problem #66 that he had a vo of 5 m/s) see p. 159 for help in understanding this one.
73: Watts are units of power. You are paying for watts * hours. That’s energy you’re paying for, not power.
76: 746 watts = 1 hp
77: There are 24 hours in a day. Convert these into seconds
78: P = work / time. Work = F * d. F = m * a. a = Dv/Dt.
d = x = vot + 1/2at2. This is a problem of multiple equation substitutions.
Realize that vo = 0 m/s and v = 25 m/s – you need to convert km/h into m/s. First, solve for a using a = Dv/Dt. Then, plug that into the kinematic equation x = vot + 1/2at2 and solve for x. Find force by F = ma and then multiply x by the force to find work.
81: Convert 2.0 hp into watts. Power = Work * time. Since it is 45% efficient, you multiply by 0.45.
84: First, convert everything into useable units. Next, find work done by the plane (Wplane = DE = K + U. Next, find energy output of the plane (which is Power * time) The power is 1.12 x 106 J and the time is 750 seconds (I converted these from hp and minutes respectively). Finally, Efficiency = Wplane/Energy output.
48) 0.21 m
52) a) 2.9 x 103J, -1.8 x 103 J. b) -4.7 x 103 J for all
54) 21 J
60) a) 282J b) U = 98.0 J, K = 184 J c) v = 47.5 m/s , E = 282 J
For letter d) a) 0 J b) U = -184 J, K = 184 J c) v = 47.5 m/s, E = 0, K = 282 J
62) 5.10 m
64) 5.42 m/s at the bottom of the swing
66) 15 m/s
70) a) 0.33 m b) 0.28 m
76) 187 W
78) 9.4 x 104 W = 1.3 x 102 hp