Diving Board to Diver's Stopping Point: Solve the Distance

[princesspriya]

Homework Statement

a 50 kg diver steps off a diving board and drops straight down into the water. The water provides an upward average net force of 1500 N. If the diver comes to rest 5.0m below the water's surface, what is the total distance between the diving board and the diver's stopping point under water?

Homework Equations

The Attempt at a Solution

the total force would be Fn-Fg which is 1500-50*9.81=1009.5N
W=1009.5*(x+5)

i am completely lost on what to do next. please give me a clue or something.

 

[Doc Al]

the total force would be Fn-Fg which is 1500-50*9.81=1009.5N

That's the total force on the diver when he's in the water. What about before he hits the water?

 

[princesspriya]

oo it wud b fg which is 50*9.81 which is 490.5 . but that doesn't help.

 

[Doc Al]

Sure it helps. There are several ways to solve this kind of problem. One way would be to compare the work done on the diver before he hits the water to the work done on him after he hits the water. After all is said and done, what must the total work (by all forces) be on the diver?

 

[princesspriya]

well you don't know the distance so you cannot find the total work done.

 

[Nabeshin]

Think what total work means in reference to his original state (position and velocity) and final state.

 

[Doc Al]

well you don't know the distance so you cannot find the total work done.

The distance is what you are asked to find. Try calling the distance from board to water surface D (or whatever). Now set it up and see if you can solve for D. (Then use it to get the total distance.)

 

[princesspriya]

you cannot because you would have two unsolved, the work net and the distance.

 

[Doc Al]

Try it and see. The only unknown is the distance.

 

[princesspriya]

W=1009.5*(x+5) and W=490.5x
are those two equations correct?

 

[Doc Al]

W=1009.5*(x+5) and W=490.5x
are those two equations correct?

No.

The diver travels a distance x before hitting the water. What net force acts? What's the net work? Is it positive or negative?

The diver travels 5 m under the water. What net force acts? What's the net work? Is it positive or negative?

What must those two work contributions add to? (Hint: What's the change in KE?)

 

[princesspriya]

well when he is in water the work he does would be 1500N*5m because that's the distance he z traveling and the net force is 1500. so the work done would be 7500J.
The work he does under water would not be the same as above water would it?

 

[Doc Al]

well when he is in water the work he does would be 1500N*5m because that's the distance he z traveling and the net force is 1500. so the work done would be 7500J.

1500 N is the force of the water, not the net force. You found the net force in post #1.

The work he does under water would not be the same as above water would it?

Something like that. What are the signs of the two work contributions?

 

[princesspriya]

the work done above water would be negative while the word done below water would be positive. but how can both of them have the same magnitude? that's the part i don't understand.

 

[Doc Al]

the work done above water would be negative while the word done below water would be positive.

Just the opposite. Above the water, the work done on the diver is positive: the force (gravity) acts in the same direction as the displacement. Below the water, the net force acts up while the displacement is still down; so the work on the diver is negative.

but how can both of them have the same magnitude? that's the part i don't understand.

Above the water, the diver's KE increases as he falls (since work is being done on him); below the water, it decreases (since negative work is being done on him). It all has to balance out, since he starts with 0 KE and ends up with 0 KE.