What Is the Initial Downward Acceleration of an Iron Ball Submerged in Water?

[Bizznaatch]

"Find the initial downward acceleration a of an iron ball of density p =7.60 g/cm3 when submerged in water and released from rest. (Use pwater = 1000 kg/m3)."

My steps so far:

1. I converted p=7.60g/cm3 to p=.76kg/m3.
2. I drew a free-body diagram, taking the buoyant force upwards as positive and the weight of the ball as negative, and set this equal to the mass of the ball times some acceleration:

F(buoyant)-mg=ma

3. I also know that the buoyant force = density(fluid) * Volume(fluid displaced) * gravity, as well as density(ball)=mass(ball) * Volume(ball). I rewrote the above equation to take this into account:

[p(fluid)*V(fluid)*gravity]-[p(ball)*V(ball)*gravity]=p(ball)*V(ball)*a

4. Since we know that the volume of the fluid displaced equals the volume of the ball, then all of the V's in the above equation are equal and will cancel out:

[p(fluid)*gravity]-[p(ball)*gravity]=p(ball)*a

5. This reduces to:

[gravity(p(fluid)-p(ball)]/p(ball)=a

6. Plugging in my numbers, I get a=12884.93684 m/s^2. This is apparently incorrect. Also taking the negative of this number doesn't work (in case they wanted a negative acceleration). Are my steps/thinking correct here? Any help is greatly appreciated!

 

[Danger]

Welcome to PF, Bizz.
This appears to be homework problem, so I've alerted the Mentors to move the thread if they so deem.

 

[alphysicist]

Hi Bizznaatch,

In your step 1 you wrote that the density of the iron ball was 0.76 kg/m^3, but the density of water is 1000 kg/m^3. Iron is definitely denser than water so it looks like a unit conversion problem.

Remeber that since the original density was in terms of g/cm^3, you have one factor of grams to convert to kilograms but three factors of centimeters to convert to meters.