# Kinetic energy in ocean current

1. Apr 3, 2017

### pinkfishegg

1. The problem statement, all variables and given/known data
Consider an Ocean Current flowing at 2.5 m/s. a) How much energy is contained in a cubical block of water 1m on the side. (The density of the water is about 1030kg/m^3.) b) If the flow is perpendicular to one of the cube faced, what is the rate at which current flow carries kinetic energy across each square meter? Your answer gives an upper limit for the power that could be extracted from the flow-although an unrealistic limit because you'd have to stop the entire flow

2. Relevant equations
KE=(1/2)m*v^2

3. The attempt at a solution
I got part A using m=(pho*volume) and subbing in to get
KE=.5*(1030km/m^3)*1m^3(25m/s)^2=3219J

for part B I'm not sure if I can just divide by 1m^2 to get 3219 J/m^2 or if i have to divide by 6 because there are 6 sides. Can someone help me with this reasoning?

2. Apr 4, 2017

### Staff: Mentor

Consider that all the water is moving in the same direction. So you need to consider only one face of the cube.

The problem asks for a rate, so you have to incorporate time in your thinking.

3. Apr 4, 2017

### haruspex

Further to DrClaude's reply, I see no reason to divide by an area at all. The first part has already limited consideration to a metre cube, so the flow through the face is already "per sq m".