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AS-level SUVAT / Newtonian physics-based question

  • Thread starter DJsTeLF
  • Start date
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1. Homework Statement
Diver enters the water at 7.92 [ms][/-1] and comes to rest 1.6m below the surface.

Calculate the average total upward force acting on the diver which brings his verticle velocity to zero.

From an earlier part of the questions the divers mass is 72kg.

2. Homework Equations
Thus far I have solved this by equating the verticle kinetic engery of the diver on entering the water with the work done by the water in decelerating him to a stop.

The answer I get is [average decelerating force]\approx[/1411N]

The model answer however states that the total average upward force is 1411 + 706 = 2117N.

My question is therefore from where do they / should I get the other 706N from?


3. The Attempt at a Solution
[tex]\Delta[/\frac{1}{2}m[v][/2]] = decelerating force x 1.6m

Decelerating force = 0.5 x 72 x [7.92][/2] / 1.6 = 1411N
 

Answers and Replies

gneill
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From an earlier part of the questions the divers mass is 72kg.
...

My question is therefore from where do they / should I get the other 706N from?
It's weighty problem.
 
47
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It's weighty problem.
Is this meant to be a joke or a hint at the solution?

It did make me consider whether they expect the student to realise that to come to rest the diver would have to be neutrally bouyant in order to properly come to rest under water.

Downward force due to gravity = mass x gravity = 72 x 9.8 = 705.6N

Therefore equivalent upward force would be necessary for diver to come to a stop.

It appears I've found a route to the answer but this seems like a hell-of-an insight for a 16yr old student to have without any 'nudges' towards considering neutral bouyancy in the question??

Comments welcome
 
gneill
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You don't need the details about buoyancy to solve the problem. All you need to know is that the work done to bring him to a halt has to take into account the change in potential energy due to gravity as well as the kinetic energy he needs to lose. Hence the hint about gravity (okay, it was a joke, too). The result is that there's an extra force acting upward (you don't care about the source) to counteract the downward force due to his weight.
 

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