Question

A 70kg divers steps off of a 10m tower and drops from rest straight into the water. If he comes to rest 5m beneath the surface, determine the average resistance force exerted on him by the water.

I get the first part, just need help going past to find the force. I have KE final=6860J

Answer 1

KE = mgh1 = 70*9.8*10


KE + PE = F*d
70*9.8*10 + 10*9.8*5 = F*5

Answer 2

where are you getting this equation from...KE=1/2mv^2 btw. and W=KEf-KEo...derive the equation pls.

Answer 3

look at it this way... how far does he fall, in total?

Answer 4

15m

Answer 5

exactly... and what's his KE at the end of that fall?

Answer 6

6860 J

Answer 7

after 15m his velocity is?

Answer 8

0 m/s

Answer 9

so KE=0

Answer 10

all the energy has gone into the work done by the water on the diver...

Answer 11

70*g*15 - F*5 =0

Answer 12

one focuses on KE +PE
one looks at PE alone
either way, they must be equivalent

Answer 13

whoops... that should be a '70' sorry if that was what was causing the confusion...
"70*9.8*10 + 70*9.8*5 = F*5 "

Answer 14

yea i got it...so technically PEtotal=W(done by water)
right?

Answer 15

yep (delta PE = Work)

Answer 16

usually delta KE=net work...i'm still a bit confused as to how the energy is transferred entirely to the PE

Answer 17

isn't PE zero at the end of the path, before hitting the water?

Answer 18

it's all energy, so it doesn't matter what you call it... if he falls 5 m his PE has decreased by 70*5*g and his KE has increased by 70*5*g... we call one delta PE and the other delta KE, but it's all energy

Answer 19

Oh true. all right then, thanks! :)

Answer 20

it's just convenient to often use one or the other (or both combined in a certain way) in order to solve a specific problem...

Answer 21

quick question though, how would i show that in terms of KEf, KEo, PEo, PEf? The final equation you used.

Answer 22

the first one I wrote was KE +PE ... but you'll notice that the KE, I left in terms of delta PE (70kg*10m*g) ..

Answer 23

so KEo+PEf?

Answer 24

this was to show that it really is the same, whether you're talking about KE +PE or just total PE... because his increase in KE is just the decrease in PE :)

Answer 25

or KEf+PEo?

Answer 26

i'm trying to use conservation of energy to solve it, or else my teacher will mark it wrong haha

Answer 27

that's what this is.

Answer 28

Tell you what, you pick a way to solve it, and we'll "do" it that way.

Answer 29

i m solving it ...wait for few moment...

Answer 30

using KEf+PEf=KEo+PEo. Also, this involves a conservative force, correct? or a nonconservative one?

Answer 31

sure so: KEo = 0 PEo = mgh = 70*g*15

KEf = 0 PEf = 0

one thing you omitted? Work :)

so KEo +PEo = 0 + 70*g*15 = KEf + PEf + W = W
70*g*15 = W

Answer 32

ooooohh all right. thanks!

Answer 33

sure:)

Answer 34

apply the energy conservation equation you will simply get the answer.
when the man is standing at the top of the tower there is only potential energy that he has due to his position height of tower...
now just before hitting the water surface, total potential energy will change into kinetic energy because now height become zero.
now this gain in kinetic energy will loss due to work done against resistive force of water.
so equ will be like this
mg(h+d)=f*d
70*10(10+5)=f*5
f=2100 newton

Answer 35

I already got the answer, but thanks!

Answer 36

ok..