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 height h.

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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...

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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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