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Samuelb88

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## Homework Statement

Two objects are connected by a massless rope that passes over a massless, frictionless pulley. One of the objects has a fixed mass (m), while the other is a bucket filled with water (M) with initial mass of 45. kg. The objects are releases from the rest (shown below). There is a hole in the bucket and the water leaks out at the rate given (dM/dt).

(a) Determine when the fixed mass (m) reaches its maximum height, this maximum height, and the mass of the bucket (M) at this time.

(b) Determine when the fixed mass (m) reaches the ground, its final speed, and the final mass of the bucket (M).

## Homework Equations

http://img3.imageshack.us/img3/4852/93084392.jpg

## The Attempt at a Solution

I set up my coordinate system with x directed upward and let M(t) function the changing mass of the bucket.

[tex]M(t) = M_0 + \frac{dM}{dt}\right)[/tex]

Since the objects are bound by a massless rope, the acceleration for both objects is the same and the sum of the forces of the system is

[tex] SUM(F_x) = M_t_o_t_a_la_x = M(t)g - mg[/tex]

Solving for acceleration

[tex] a_x = ( \frac{M(t) - m}{M(t) + m} \right) )g[/tex]

[tex] a_x = ( \frac{(45-t) - 40}{(45-t) + 40} \right) )g[/tex]

[tex] a_x = ( \frac{(5-t}{(85-t} \right) )g[/tex]

Now I've explicity defined an equation for the acceleration of the system.

The fixed object m will be at its max height when the velocity v = 0 m/s. Noticing velocity is at a max. at t = 5s, I integrate from 5 to t, and set v = 0 m/s.

[tex] 0 = g \int_5^t \frac{5-t}{85-t}\right) dt = 5\int_5^t \frac{1}{85-t}\right) dt - \int_5^t\frac{t}{85-t}\right) dt[/tex]

(i divided by g, above)

[tex] 0 = (-5(ln|85-t|) - (-t - 85ln|85-t|))[/tex] (from 5 to t)

[tex] 0 = 80ln|\frac{85-t}{80}\right)|+ t - 5[/tex]

Taking t = 0 s to be the moment when the objects are released from rest, the time when the projection of the velocity wrt time re-obtains this value (is at rest again) is at t = 9.8 s. Therefore, the fixed mass (m) is at its maximum height at t = 9.8 s.

[tex]\int_0^xdv = \int_0^{9.8} (80ln|\frac{85-t}{80}\right)| + t - 5) dt[/tex]

[tex] x(t) = \frac{1}{2}\right) (t - 170)t + 80(t- 85)ln|\frac{85-t}{80}\right) |[/tex] (from 0 to 9.8)

[tex] x(9.8) = ... -.482572[/tex]

Since the bucket M started 10. m above the ground, I added 10 to x(9.8) which equals 9.51743m which I took to be the maximum height. Could anyone check my work? The whole thing seems wrong imo, when I graphed the velocity wrt to t, v(0) = -.15. I determined x_max occurred at t=9.8s because v(9.8) = -.15. But shouldn't at v(0) = 0m/s? (neither object was initially moving...) therefore, v(9.8)=0m/s aswell. I'm quite confused about the whole thing.

Any help would in answering part (a) or (b) be greatly appreciated. Thanks.

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