Depth using Work, Weight, Height

[randomjunebug]

Homework Statement

A diver weighing 784 N stands on a diving board 13.6 m above the surface of a pool. The diver steps off the edge and falls into the pool. The pool does -12290 J of work on the diver. How far below the surface of the pool (depth) does the diver reach?

Homework Equations

W=F*d

The Attempt at a Solution

I think maybe it is as simple as this: Work is done by water. Weight is the Force of the diver going straight down. Therefore: Work of water = Force of weight * unknown d. -12290/784 = 15.68 m. Is this correct? Or a better way to do this?

 

[hjelmgart]

Hmm, I am not sure about your way of solving it. Initially I was thinking, it sounds OK, but the result seems unrealistic? At least from my own experience with diving.It is also possible, you could find the speed, the diver hits the water with.

Then find the deacceleration of the water from the energy it does on the diver. Thus calculating the distance.

 

[adjacent]

Homework Statement

A diver weighing 784 N stands on a diving board 13.6 m above the surface of a pool. The diver steps off the edge and falls into the pool. The pool does -12290 J of work on the diver. How far below the surface of the pool (depth) does the diver reach?

Is this a textbook question or a question you made yourself?
The force exerted by the water should not be equal to the weight of the diver(If so,he would move down with the same speed as before,without slowing down)
Also,as he slows down,the force exerted by the water becomes smaller and smaller.

 

[rude man]

What is the force the pool exerts on the diver? (Ignore friction forces).

 

[rude man]

Homework Statement

A diver weighing 784 N stands on a diving board 13.6 m above the surface of a pool. The diver steps off the edge and falls into the pool. The pool does -12290 J of work on the diver. How far below the surface of the pool (depth) does the diver reach?

Homework Equations

W=F*d

The Attempt at a Solution

I think maybe it is as simple as this: Work is done by water. Weight is the Force of the diver going straight down. Therefore: Work of water = Force of weight * unknown d. -12290/784 = 15.68 m. Is this correct? Or a better way to do this?

The force exerted on the diver is not just his/her weight. If you hold a light (assume zero mass) object under water, what is the force pushing it up?

P.S. the answer is a lot less than 15.68m! Like, not too much more than 10% of that.

 

[adjacent]

The force exeeted on the diver is not just his/her weight. If you hold a light (assume zero mass) object under water, what is the force pushing it up?

Rude man!By looking at the information given in the OP,this can't be found.

And also,the force exerted on the diver can't be the diver's weight.The force is proportional to his Speed.

 

[rude man]

Rude man!By looking at the information given in the OP,this can't be found.

And also,the force exerted on the diver can't be the diver's weight.The force is proportional to his Speed.

If you neglect friction it can be found. So the force is not a function of his speed. The problem should admittedly have imposed that condition (no friction).

You seem to completely ignore the main force of the water on the diver: buoyancy. Neglecting friction, the only forces acting on the diver underwater are buoyancy and gravity.

 

[adjacent]

You seem to completely ignore the main force of the water on the diver: buoyancy. Neglecting friction, the only forces acting on the diver underwater are buoyancy and gravity.

I know you are talking about the Archimedes Principle.

But how can you find buoyancy?Diver's volume is not given.

 

[rude man]

I know you are talking about the Archimedes Principle.

But how can you find buoyancy?Diver's volume is not given.

You don't need the volume.
Let W = diver's weight

Two equations:
eq. 1: buoyant force Fb * distance dived d = work done by water
eq. 2: (Fb - W)d = kinetic energy when diver hits the surface

2 equations, 2 unknowns: Fb and d.

 

[haruspex]

Homework Statement

A diver weighing 784 N stands on a diving board 13.6 m above the surface of a pool. The diver steps off the edge and falls into the pool. The pool does -12290 J of work on the diver. How far below the surface of the pool (depth) does the diver reach?

Homework Equations

W=F*d

The Attempt at a Solution

I think maybe it is as simple as this: Work is done by water. Weight is the Force of the diver going straight down. Therefore: Work of water = Force of weight * unknown d. -12290/784 = 15.68 m. Is this correct? Or a better way to do this?

Your method is fine, and yes it is that simple. But how exactly are you defining d here?

 

[rude man]

I know you are talking about the Archimedes Principle.

But how can you find buoyancy?Diver's volume is not given.

@adjacent, it dawned on me later that it doesn't really matter what the nature of the forces is that the water exerts on the diver. It can be buoyancy, friction, whatever, and it doesn't have to be constant with d. The problem gives the total work done on the diver by the water, which is ∫F(d)dd. So in my equations in post 9, F_bd is really the work done by the water as given in the problem and should be generalized to ∫F_w(d)dd = 12,290 J where "w" stands for water and F_w is the instantaneous force on the diver in the water for all reasons, not just buoyancy.

 

[haruspex]

@adjacent, it dawned on me later that it doesn't really matter what the nature of the forces is that the water exerts on the diver. It can be buoyancy, friction, whatever, and it doesn't have to be constant with d. The problem gives the total work done on the diver by the water, which is ∫F(d)dd. So in my equations in post 9, F_bd is really the work done by the water as given in the problem and should be generalized to ∫F_w(d)dd = 12,290 J where "w" stands for water and F_w is the instantaneous force on the diver in the water for all reasons, not just buoyancy.

So you agree the method in the OP is ok?

 

[rude man]

So you agree the method in the OP is ok?

Yes, mutatis mutandis, i.e. when d includes the height above water. That equation is then identical to mine.

 

[randomjunebug]

Hey all thanks for the attempts and guidance. My professor gave us all a solution guide to this because no one got it 100% correct.
height of diver = height of tower + depth of water
therefore, depth of water = height of diver - height of tower
Formula for height of diver can be found using PE = mgh
depth of water = height of diver = PE diver/wt of diver - height of tower
depth of water = -(Work of water)/wt of diver - height of tower
depth of water = -(-12290J)/(784N) - (13.6 m) = 2.076 meters

Honestly I hate his method and if someone can walk me through this 'guide' I would really appreciate it.

 

[rude man]

This is indeed a ridiculously long explanation.
Here's the scoop:

Diver dives from a height of 13.6m to a depth of x meters.
So the total potential energy lost is mg(13.6 + x) = 784(h+x).
But if he had dived thru thin air he would have had a kinetic energy of just that: k.e. = 784(h+x).
So the fact that he had ZERO k.e. when he hit depth x meant that the water took all that amount of k.e. from him.
So 784(h+x) = work done by water = 12,290J.

Solve for x: x = 2.08m.