Bouyant force and Water resistance

[theuniverse]

Homework Statement

A buoyant measuring device of m = 0.5 kg is projected straight down into a lake at initial velocity v0 = 20 m/s. Two forces act on the device: (1) buoyancy pulls up with a force equal to the weight of the displaced water, in this case 10 Newtons; (2) water resistance resists the motion of the device in the water with force 1/2 |v|, where v is the velocity. (For this question, we'll ignore the force of gravity on the object.)
(a) Find (analytically) the time at which the device reaches its maximum depth.
(b) Find the maximum depth that is reached.
(c) Find expressions for the velocity and position of the device as it returns to the lake's surface.
(d) Estimate the total time for which the device will be under water2. Attempt to solve:
a) m(dv/dt) = Fb+1/2v
dv/dt = 10+0.5v/m
(1/20+v)dv=dt
ln |20+v| = t +C
v = Ce^t-20 [where C=40 after subbing in initial condition]

so then I sub in v=0 to find the time of max depth and get t=0.7s

b) integral (v) = integral (40e^t-20)
x(t) = 40e^t-20t
and sub in the time from (a) to get max depth of 66.5m

c) m(dv/dt) = Fb-1/2v
(1/20-v)dv = dt
v = -Ce^t+20 [sub in v=0 since that would be the velocity when it's back on surface?]
to get: v=-20e^t+20
Now take the integral of it to find distance: x(t) = -20e^t+20t

d) This is where I got stuck. I think something in my previous solutions is incorrect or I just don't think about it right... I just can't get an answer...

Thanks!

 

[alphysicist]

Hi theuniverse,

Homework Statement

A buoyant measuring device of m = 0.5 kg is projected straight down into a lake at initial velocity v0 = 20 m/s. Two forces act on the device: (1) buoyancy pulls up with a force equal to the weight of the displaced water, in this case 10 Newtons; (2) water resistance resists the motion of the device in the water with force 1/2 |v|, where v is the velocity. (For this question, we'll ignore the force of gravity on the object.)
(a) Find (analytically) the time at which the device reaches its maximum depth.
(b) Find the maximum depth that is reached.
(c) Find expressions for the velocity and position of the device as it returns to the lake's surface.
(d) Estimate the total time for which the device will be under water


2. Attempt to solve:
a) m(dv/dt) = Fb+1/2v
dv/dt = 10+0.5v/m

You have some sign problems here that are giving you the wrong function for v(t). You have Fb=+10, and so you have picked upwards to be positive.

Now for the second term, you have the resistive force to be +1/2v; the problem is that v is downwards, and so is a negative number; so in your equation, the force of +1/2v is downwards, which is incorrect.

(1/20+v)dv=dt
ln |20+v| = t +C
v = Ce^t-20 [where C=40 after subbing in initial condition]

so then I sub in v=0 to find the time of max depth and get t=0.7s

I think this time might actually be the correct answer, but the function you have for v is incorrect and leads to the wrong answer for part b.

b) integral (v) = integral (40e^t-20)
x(t) = 40e^t-20t
and sub in the time from (a) to get max depth of 66.5m

c) m(dv/dt) = Fb-1/2v
(1/20-v)dv = dt
v = -Ce^t+20 [sub in v=0 since that would be the velocity when it's back on surface?]
to get: v=-20e^t+20
Now take the integral of it to find distance: x(t) = -20e^t+20t

d) This is where I got stuck. I think something in my previous solutions is incorrect or I just don't think about it right... I just can't get an answer...

Thanks!