Mechanics question..

A diver of mass 60kg dives from a diving board at a height of 4m. she hits the water travelling at 8ms^-1 and descends to a depth of 2m in the diving pool. Model the diver as a particle.

a) find the work done against air resistance before the diver hits the water and the average magnitude of the air resistance force.

b) find the average magnitude of the force exerted by the water on the diver as she is brought to rest in the diving pool.

for a) i did G.P.E = 60x9.8x4 = 2352j
K.E - when hittin water = 1/2 x 60 x 64 = 1920
so 2352-1920 = 432j
fxd = 432
f=432/4 = 108

can't do b) ?? any help? thanks..

the asnwers 1548

A diver of mass 60kg dives from a diving board at a height of 4m. she hits the water travelling at 8ms^-1 and descends to a depth of 2m in the diving pool. Model the diver as a particle.

a) find the work done against air resistance before the diver hits the water and the average magnitude of the air resistance force.

b) find the average magnitude of the force exerted by the water on the diver as she is brought to rest in the diving pool.

Use energy considerations again. You cant use the constant acceleration equations because there isnt constant acceleration.

Use energy considerations again. You cant use the constant acceleration equations because there isnt constant acceleration.

that is wot it thought....c my answer up a few posts

that is wot it thought....c my answer up a few posts

Yea, i was too slow to post.

Modelling downwards as positive, the diving board as 0 displacement.

Resultant force = W - R
W = mg = 9.8*60
R = ?
F = 588 - R

F = ma
588 - R = 60a

a = ?
u = 0
v = 8
s = 4

[v^2]
64 = 0 + 2*4a
a = 8 m s^-2

588 - R = 60a
588 - R = 60*8
R = 588 - 60*8 = 108N

b) Resultant force = W - P <<(P is the reaction of the water, assumed to be constant)

F = 588 - P
F = ma
60a = 588 - P
P = 588 - 60a

u = 4
v = 0
s = 2
a = ?

[v^2]
0 = 16 + 2*2a
4a = -16
a = -4 m s^-2

P = 588 - 60*-4
P = -828N
|P| = 8.3*10^2 N (2s.f.)

I think...?

Why not, the air resistance is constant, and so is his weight, therefore the resultant force on both occasions is constant.

It says:

"find the average magnitude of the force"

So we assume it is constant i think.

Why not, the air resistance is constant, and so is his weight, therefore the resultant force on both occasions is constant.

It says:

"find the average magnitude of the force"

So we assume it is constant i think.

when the diver hits the water i think he will accelerate by less and less, as when travelling at 8ms^-1 he goes through a certain amount of water in a given time but when at 4ms^1, goes through less in this time and thus there is less resisitive force and as F=ma the magnitude acceleration is less. And F=ma can only be used when acc is constant.

blah I think...?

i think u can do it this way...

F=ma
resistance from water be R#

R-mg=ma
R=m(a+g)
=60(16+9.8)
=1548

cos the force is stopping the diver by accelerating her/him, yet the diver's weight is still trying to accelerate her/him

so u were right mik1a

b)

Thanks, just need to make sure I can read before the next exams.

lol...have u finished then mik1a?? u seem quite good at p2 already so how long have u been doing that