- #1

jisbon

- 476

- 30

- 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

$$ma=pg\left( L^{2}\right) \left( L-y\right) -\dfrac {1}{2}pgL^{3} $$

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