How Does Laminar Flow Resistance Affect Barge Movement in a Canal?

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Problem:
Two horses pull a barge from rest in a canal filled with a viscous fluid that provides laminar resistance (FR = –kv). The two horses walk on each side of the canal so that their net force is applied exactly forward. The barge has a mass of 3000 kg and the donkeys pull forward with a constant net forward force of 300 Newtons.

a. If the maximum attainable speed by the barge is 2.5 m/s, what is the resistive coefficient, k (include units)?
b. How long does it take the barge to reach a speed of 1 m/s?

Once the barge reaches 2.5 m/s, at a certain time (call it t = 0) the horses instantaneously stop pulling, and the barge is allowed to drift forward.

c. How long does it take for the barge to slow to 1 m/s?
d. How far does the barge drift before coming to rest?
e. In principle, how long does it take for the barge to come to rest? Briefly explain.
f. Estimate how long it takes for the bar to be, for all practical purposes, at rest. Discuss how you decided on this estimate.

Equations:
F=-kv

Attempt at a solution:
I know how to do the integrations of the laminar flow F=-kv. Other than that I really don't know how to do this.

 

[member 553083]

Please help!a. k = -300 N / (2.5 m/s) = -120 N/m/sb. t = 1/k * ln(v1/v2) = 1/-120 N/m/s * ln(1/2.5) = 2.30 seconds c. t = 1/k * ln(v2/v1) = 1/-120 N/m/s * ln(2.5/1) = 2.30 seconds d. d = v2*t = 2.5 m/s * 2.30 s = 5.75 me. It takes an infinite amount of time for the barge to come to rest, since the force of viscous friction is proportional to the velocity and decreases as the velocity approaches zero. f. This is dependent on the magnitude of the resistive coefficient, k. An estimate of the time it would take for the barge to be practically at rest can be calculated by assuming that the resistance force is equal to the net forward force at a certain velocity. For example, if we assume that the resistance force is equal to the net forward force at 0.10 m/s, then the time it would take for the barge to be practically at rest can be calculated by solving the following equation: Fnet = Fres = -kv = 300 N. Solving this equation for v gives us v = 300 N/-120 N/m/s = 2.50 m/s. The time it would take for the barge to reach this velocity can be found by using the equation t = 1/k * ln(v1/v2) which gives us t = 1/-120 N/m/s * ln(0.10/2.50) = 1.94 seconds. Therefore, it would take approximately 1.94 seconds for the barge to be practically at rest.