C. Work Homework

Part II Sections 5.4-5.6

 

5.4  Potential Energy

Problems:  44, 45, 47, 48 51, 52, 54

Helpful Tips:

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

Helpful Tips:

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.

 

5.6  Power

 Problems:  72, 73, 75, 76, 77, 78, 81, 84

Helpful Tips:

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.

 

Even Answers:

44)  d

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

56)  c

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

72)  b

76)  187 W

78)  9.4 x 104 W = 1.3 x 102 hp

84)  48.7%