# When a diver enters a pool after a dive and stops

Neohm
When a diver enters a pool after a dive and stops in the water after 5 meters, how is the work/energy theorm set up?

I said Wnet = Change in K, so
Wg + Q = K,

where Wg is the work done by gravity, mgh, and Q is the work done by the resistance of the water. Are Wg and Q negative (-Wg - Q = K), since they are in the opposite direction of motion, and oppose the divers motion when he hits the water? Or can they be added becuase they are both negative work? Oh, and is positive or negative 5 meters used in the calculations?

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russ_watters
Mentor
As long as you keep them consistent it doesn't matter what is positive and what is negative for the forces. For the sake of logic, I'd use a force working against you as being negative. And distance is an absolute value, its always positive. So your Q will be a negative number, and you add it.

Also, since the diver stops, he has no kinetic energy and since the surface of the water is our zero altitude, no potential enegy, so k=0.

Also, what about the work done by the diver's legs? Are you assuming a stiff-legged fall forward.

Anyway:

Wg+Q=K=0
Are Wg and Q negative (-Wg - Q = K)
Remember, putting a negative sign in front of a variable doesn't make the variable negative - in fact it doesn't tell you anything at all about the sign of the variable. It just means in your equation you are reversing its sign.

HallsofIvy