Solving for Maximum Speed of Cube: Kinematics & Upthrust

[jisbon]

Homework Statement

1) Cube is initially attached to string on the bottom of a container filled with water
2) The density of cube is half of the density of water
3) The surface of the water is same level of the top surface of the cube
4) The density of water is p
What is the largest speed of cube when travelling upward after cutting the string?

Relevant Equations

-

So, I recognise that:

$$ma=pg\left( L^{2}\right) \left( L-y\right) -\dfrac {1}{2}pgL^{3} $$
whereby $$pg\left( L^{2}\right) \left( L-y\right)$$ is the upthrust while the other is mg.

So, to find the largest speed of cube, I will assume that acceleration is zero since the acceleration slowly decreases after I cut the string (as mg = upthrust)
So, I should take a=0
I found a to be :
$$\dfrac {pg\left( L^{2}\right) \left( L-y\right) -\dfrac {1}{2}pgL^{3}}{m} $$
and y= 1/2L after making the upthrust and mg equals
So using kinematics, I should be able to find v with :
$$v^2=u^2+2as$$
where a is what I found above and s to be 1/2 L. However, the answer comes out as 0.
Not sure what happened here :/
Thanks

 

[haruspex]

So using kinematics,

The equation you quote is only valid for constant acceleration.
You could use energy.

 

[jisbon]

The equation you quote is only valid for constant acceleration.
You could use energy.

What will be the energy involved in this case though?
Initial energy = gpe? Final energy = gpe+ ke?
Thanks

 

[haruspex]

What will be the energy involved in this case though?
Initial energy = gpe? Final energy = gpe+ ke?
Thanks

Yes. Remember to take into account the water's GPE.
An alternative is to solve the differential equation you had for y.

 

[jisbon]

Yes. Remember to take into account the water's GPE.
An alternative is to solve the differential equation you had for y.

By differential equation for y, do you mean integrating $$\dfrac {pg\left( L^{2}\right) \left( L-y\right) -\dfrac {1}{2}pgL^{3}}{m}$$ in terms of t?

 

[haruspex]

By differential equation for y, do you mean integrating $$\dfrac {pg\left( L^{2}\right) \left( L-y\right) -\dfrac {1}{2}pgL^{3}}{m}$$ in terms of t?

No, you clearly cannot do that. I mean solving the differential equation of which that forms part:
$$\ddot y=\dfrac {pg\left( L^{2}\right) \left( L-y\right) -\dfrac {1}{2}pgL^{3}}{m}$$

You may recognise that as a well known form of equation in mechanics.

 

[jisbon]

No, you clearly cannot do that. I mean solving the differential equation of which that forms part:
$$\ddot y=\dfrac {pg\left( L^{2}\right) \left( L-y\right) -\dfrac {1}{2}pgL^{3}}{m}$$

You may recognise that as a well known form of equation in mechanics.

I'm sorry, but I can't seem to recognise any equations here. Do you mean solving
$$ma = {pg\left( L^{2}\right) \left( L-y\right) -\dfrac {1}{2}pgL^{3}}$$

 

[haruspex]

I'm sorry, but I can't seem to recognise any equations here. Do you mean solving
$$ma = {pg\left( L^{2}\right) \left( L-y\right) -\dfrac {1}{2}pgL^{3}}$$

Yes, but a is the same as $\ddot y$, and the nature of the equation is more obvious in that form. To make it more obvious still, we can collapse various groups of constants into single items:
$\ddot y +Ay+B=0$.
And to make it blindingly obvious, substitute z=y+B/A:
$\ddot z +Az=0$.