# Help finding potential energy of submerged buoyant object

by bgizzle
Tags: buoyant, energy, object, potential, submerged
 P: 718 The buoyant force acting on a object is completely determined by its size, so, no, you don't need more information. If you want to know whether this buoyant force will be enough to support its weight, you need to know its mass too. Thus, whether an object sinks or floats is determined completely by its density. So, you can't make a sinking object float just by making it larger, unless you're adding a lower density material. As for your question: the relevant physics law is Archimedes' principle, which says that the upward buoyant force on an object is equal the weight of the displaced water. The density of salt water is around $1025 kg/m^3$. The pressure at 1000ft probably isn't enough to significantly increase this. So, the buoyant force is: $F_B = \frac{4}{3}\pi(100m/2)^3 \cdot 1025kg/m^3 = 5.4\times10^8N$ where I've used the water's density and the formula for the volume of a sphere (the volume of water displaced). [edit: this is incorrect, I calculated the mass of the water instead of its weight. I missed a factor of $9.8m/s^2$. The rest of calculation proceeds the same way, but DrZoidberg has already done it, so I won't recalculate everything.] Meanwhile, the force of gravity is: $F_g = -mg = -50 000 \cdot 9.8m/s^2 \approx -5\times10^5N$ (Note the minus sign since the weight and buoyant force are in opposite directions.) This object is very buoyant, the magnitude of the weight is much less than the buoyant force. Your object's density is only about $240 kg/m^3$, much less than salt water. So, the total force on the object is the sum, which is still about $5.4\times10^8 N$. This force is approximately constant all the way down (assuming, in addition to the constancy of the water's density that the change in the acceleration due to gravity on the way down is negligible). This means the potential energy has the simple form: $U = -F\Delta x$ where $\Delta x$ is the change in height as the object descends, which is $-300m$ relative to the surface. Therefore: $U = 5.4\times10^8 N \cdot 300m \approx 1.6\times10^{11} J$ which is positive, as you should expect for a buoyant object. That's 160 gigajoules of potential energy, if you define the zero of potential energy to be at sea level.
 P: 718 DrZoidberg's numbers are correct, I made an error when calculating $F_B$. I calculated the mass of the displaced water, not its weight. So, use his/her results. Also, as DrZoidberg pointed out, the result is less accurate due to the ball's size being comparable to the depth. For a rough answer for the time, though, determine the object's acceleration (net force divided by its mass) and plug it and the depth (and a starting velocity of zero) into the appropriate kinematic equation (http://www.physicsclassroom.com/class/1dkin/u1l6a.cfm) to find the time of ascent.