our cousin Jeremy has asked you to bankroll his proposed business painting houses in the summer. He plans to operate the business for 5 years to pay his way through college. He needs $5000 to purchase an old pickup, some ladders, a paint sprayer, and some other equipment. He is promising to pay you $1500 at the end of each summer (for 5 years) in return. Calculate your annual rate of return. Contributed by Paul R. McCright, University of South Florida

Prob
Solve the problem using EXCEL functions:
Year	Cash Flow
0	-$5,000
1	1500
2	1500
3	1500
4	1500
5	1500
r	15.24%
Period
0
1
2
3
4
5
6
7
8
9
10
i
r
I eff

Transcribed Image Text:

EK & EK K The following problems relate to the figure on the left. The rod is uniform and is of length I and mass m. It is hinged at the middle and is free to rotate around the hinge in the vertical plane. It is attached with 4 springs of constant K at its ends as seen in the figure. At t = 0 it is set into oscillation from the equilibrium position (0 = 0) with an initial angular speed o. Take the potential energy at equilibrium position as zero (Uequili = 0). K Utot n mx 16) What is conserved (remains constant) during the oscillation of the rod? a) MEtot only b) Ptot and KEtot d) Ltot only a) m 1/3 b) ml/3 18) What is the maximum potential energy Utot mx when the springs are maximally compressed or stretched and the rod is rotated 80 from the equilibrium position? a) 2K02 b) 4K102 c) O d) Kl02 e) 2K103 19) What is the maximum kinetic energy KEmx, of the rod when it is going through the equilibrium position (8 = 0)? a) 1 b) 21.02 c) 23/2 d) 12/2 e) 0 20) What is the relation between the maximum kinetic energy KEmx and the maximum potential energy Utot mx? a) KEx c) KEmx = Utot mx/2 = d) KEmx = Utot mx 17) What is the moment of inertia of the rod around the pivot (its center of mass)? c) m 1/12 c) ml/12 e) 1/12 b)KEmx = 2Utot mx 21) What is the restoring torque when the rod is rotated through angle 8? a) K10/4 b) K128 c) 4K18 22) What is the equation of motion for the oscillation? a) = -(12K/m)0 d) 8 = cos(wt) e) KEmx = Utot mx b) = (K/12m)0 24) The solution that satisfies the initial conditions is; a) 8 = o sin(wt) + no cos(wt) e) = -(K/m)0 e) Ltot and Prot d) Kl0 b) = (o/w) cos(wt) d) = (12K/m)20 23) The general solution of the equation is 8 = A sin(wt) + B cos(wt). Here the angular frequency w is equal to; c) K/m a) 0 b) 12K/m d)12Km e) 12K/m e) 8 = (no/w) sin(wt) c) KEtot and MEtot e) 0 c) 0 = (12K/m 1)8 c) 8 = sin(at) EK & EK K The following problems relate to the figure on the left. The rod is uniform and is of length I and mass m. It is hinged at the middle and is free to rotate around the hinge in the vertical plane. It is attached with 4 springs of constant K at its ends as seen in the figure. At t = 0 it is set into oscillation from the equilibrium position (0 = 0) with an initial angular speed o. Take the potential energy at equilibrium position as zero (Uequili = 0). K Utot n mx 16) What is conserved (remains constant) during the oscillation of the rod? a) MEtot only b) Ptot and KEtot d) Ltot only a) m 1/3 b) ml/3 18) What is the maximum potential energy Utot mx when the springs are maximally compressed or stretched and the rod is rotated 80 from the equilibrium position? a) 2K02 b) 4K102 c) O d) Kl02 e) 2K103 19) What is the maximum kinetic energy KEmx, of the rod when it is going through the equilibrium position (8 = 0)? a) 1 b) 21.02 c) 23/2 d) 12/2 e) 0 20) What is the relation between the maximum kinetic energy KEmx and the maximum potential energy Utot mx? a) KEx c) KEmx = Utot mx/2 = d) KEmx = Utot mx 17) What is the moment of inertia of the rod around the pivot (its center of mass)? c) m 1/12 c) ml/12 e) 1/12 b)KEmx = 2Utot mx 21) What is the restoring torque when the rod is rotated through angle 8? a) K10/4 b) K128 c) 4K18 22) What is the equation of motion for the oscillation? a) = -(12K/m)0 d) 8 = cos(wt) e) KEmx = Utot mx b) = (K/12m)0 24) The solution that satisfies the initial conditions is; a) 8 = o sin(wt) + no cos(wt) e) = -(K/m)0 e) Ltot and Prot d) Kl0 b) = (o/w) cos(wt) d) = (12K/m)20 23) The general solution of the equation is 8 = A sin(wt) + B cos(wt). Here the angular frequency w is equal to; c) K/m a) 0 b) 12K/m d)12Km e) 12K/m e) 8 = (no/w) sin(wt) c) KEtot and MEtot e) 0 c) 0 = (12K/m 1)8 c) 8 = sin(at)

Answer rating: 100% (QA)

There are two ways to solve this problem Method 1 Manually using the Time Value of Money TVM formula The Time Value of Money TVM formula considers the cash flow over time and helps calculate the inter... View the full answer