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1. 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..
2. (Original post by ShOcKzZ)
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..
v^2=u^2+2as
a= 0-8^2/2*2
a=-16ms^-2

F=ma
F=-16*60
=-960N ?? woa
3. the asnwers 1548
4. (Original post by ShOcKzZ)
the asnwers 1548
using cons of energy

energy before water=k.e +g.pe
=0.5mv^2+mgh
=3096J

wd=F.d
so F=wd/d [work done by water against diver]
=3096/2
=1548J
5. (Original post by ShOcKzZ)
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.
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 = u^2 + 2as]
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 = u^2 + 2as]
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...?
6. Use energy considerations again. You cant use the constant acceleration equations because there isnt constant acceleration.
7. (Original post by JamesF)
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
8. (Original post by JamesF)
Use energy considerations again. You cant use the constant acceleration equations because there isnt constant acceleration.
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.
9. (Original post by lgs98jonee)
that is wot it thought....c my answer up a few posts
ok, right i didnt think about that,i get it now, thanks
10. (Original post by lgs98jonee)
that is wot it thought....c my answer up a few posts
Yea, i was too slow to post.
11. (Original post by JamesF)
Yea, i was too slow to post.
yeh dont worry ...comment aimed at personwho started thread
12. (Original post by mik1a)
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 = u^2 + 2as]
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 = u^2 + 2as]
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...?
that's not right. u =/= 4
13. (Original post by mik1a)
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.
Air resistance isnt constant. Just because you have taken an average, doesnt mean that you can model the situation as a constant, with the same value.
14. (Original post by mik1a)
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.
15. (Original post by lgs98jonee)
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.
But you are finding the average force. Therefore, even though they vary in real life, the average force will have the same overall affect on the average acceleration, and cause the same distances to be covered.

If you measured the velocity one-third the way through, you'd get the wrong answer though.
16. (Original post by mik1a)
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
17. 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 = 8
v = 0
s = 2
a = ?

[v^2 = u^2 + 2as]
0 = 64 + 2*2a
4a = -64
a = -16 m s^-2

P = 588 - 60*-16
P = -1548N
|P| = 1.6 kN (2s.f.)

Thanks hihihihi for spotting my silly reading error.

(Original post by lgs98jonee)
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
Thanks, just need to make sure I can read before the next exams.
18. (Original post by 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
19. (Original post by lgs98jonee)
lol...have u finished then mik1a?? u seem quite good at p2 already so how long have u been doing that
I finished my exams last wednesday and have been browsing the P2-P4 books on and off. I get good practice here though. M2 I always got, because it's basically a mixture of M1 and AS physics (with g.p.e. and stuff). I don't get the centre of mass parts, but that's what the book is for.

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