An object is pushed into the water with a certain amount of force. Calculate the speed of the object after it is released from equilibrium at a...

First calculate how submerged the block is when in equilibrium. Let the up direction be negative and the down direction be positive.

`F_b=F_g+F`

`rho_w*V_d*g=rho*V*g+F, rho=S_g*rho_w`

`rho_w(1^2*x)g=S_g*rho_w(1^3)g+2000` ,Where x=the distance from the surface to the bottom of the block (here is why I have the downward direction as positive).

Then solve for x:

`x=S_g+2000/(rho_w*g)=0.6+2000/(1000*10)=0.8 meters`

Now to calculate the velocity, at a height h, we must neglect the drag force then use the work kinetic energy...

First calculate how submerged the block is when in equilibrium. Let the up direction be negative and the down direction be positive.

`F_b=F_g+F`

`rho_w*V_d*g=rho*V*g+F, rho=S_g*rho_w`

`rho_w(1^2*x)g=S_g*rho_w(1^3)g+2000` ,Where x=the distance from the surface to the bottom of the block (here is why I have the downward direction as positive).

Then solve for x:

`x=S_g+2000/(rho_w*g)=0.6+2000/(1000*10)=0.8 meters`

Now to calculate the velocity, at a height h, we must neglect the drag force then use the work kinetic energy theorem.

`W=Delta*K`

`int_0.8^h F(x) *dx=1/2m*v^2`

The initial velocity is zero. F(x) is the net force on the block after it is released as a function of the distance submerged. We are only considering when h<0.8m, this is the only region where there is a buoyant force.

 `(2/m)int_0.8^h (F_g-F_b(x)) *dx=v^2`

`(2/(S_g*rho_w*1^3))int_0.8^h (S_g*rho_w*g-rho_w*x*g) *dx=v^2 `

`2g int_0.8^h dx -(2g)/(S_g) int_0.8^h x dx=v^2 `

`2g(h-0.8)-g/(S_g)(h^2-0.8^2)=v^2`

`(2g(h-0.8)-(g)/(S_g)(h^2-0.8^2))^(1/2)=v(h)` 

Here is velocity as a function of height where h<0.8m.

The graph starts at height =0.8 m on the x axis. It then goes to zero m/s at height=0.4 m so the block will stop before it breaks free of the surface at height=0.

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