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  • Subject: I'm so confused with this Physics problem it's not even funny
Subject: I'm so confused with this Physics problem it's not even funny

Now go make me some tea.

I'm working with mechanical energy, and the formula is ME = PEg(m*g*h) + PEe(1/2*k*x^2) + KE(1/2*m*v^2), and the problem says: A 755N diver drops a board at 10m above the water surface. Find the speed when the board is 5m above the surface, then find the speed just before it hits.

So, what I got was 3775J/38.5kg = 9.9m/s for the first part

I don't know how to get the second one it's asking for.

Am I doing it right?

[Edited on 12.10.2012 9:49 AM PST]

  • 12.10.2012 9:48 AM PDT

Holy fack OP

What grade/standard of Physics is this!?

  • 12.10.2012 9:49 AM PDT

http://i.imgur.com/fsISj.png

You're gonna have to define the equation terms for me.

EDIT: the equation you're using is beyond my level, so this might be a really simplistic and stupid answer, but could you use SUVATs? You have the initial velocity (0 ms^-1), the displacement (10 m) and the acceleration (9.81 ms^-2) if you're allowed to ignore air resistance.

[Edited on 12.10.2012 9:53 AM PST]

  • 12.10.2012 9:49 AM PDT

Now go make me some tea.


Posted by: PhyscoRaiderS97
Holy fack OP

What grade/standard of Physics is this!?

High School.

My teacher is a Physicist.


Posted by: annoyinginge
You're gonna have to define the equation terms.

m=mass, g=gravity, h=height, k=N/m and x= something I don't know, m=mass again, and v=speed

[Edited on 12.10.2012 9:51 AM PST]

  • 12.10.2012 9:50 AM PDT

I'm British. Whats High School?

Does it come after Primary?

  • 12.10.2012 9:50 AM PDT


Posted by: PhyscoRaiderS97
I'm British. Whats High School?

Does it come after Primary?
0/10, we call it high school in Britain.

  • 12.10.2012 9:51 AM PDT

Now go make me some tea.


Posted by: PhyscoRaiderS97
I'm British. Whats High School?

Does it come after Primary?

Right before collage.

  • 12.10.2012 9:51 AM PDT

in·dif·fer·ence
the fact or state of being indifferent; lack of care or concern and empathy demonstrated by an absence of emotional reactions.


Posted by: DarkBen64

Posted by: PhyscoRaiderS97
I'm British. Whats High School?

Does it come after Primary?
0/10, we call it high school in Britain.

  • 12.10.2012 9:53 AM PDT

Now go make me some tea.


Posted by: annoyinginge
You're gonna have to define the equation terms for me.

EDIT: the equation you're using is beyond my level, so this might be a really simplistic and stupid answer, but could you use SUVATs? You have the initial velocity (0 ms^-1), the displacement (10 m) and the acceleration (9.81 ms^-2 if you're allowed to ignore air resistance).

I have no initial velocity of 0, since it's asking me for the speed when it's halfway down

  • 12.10.2012 9:53 AM PDT


Posted by: DarkBen64

Posted by: PhyscoRaiderS97
I'm British. Whats High School?

Does it come after Primary?
0/10, we call it high school in Britain.


What part of Britain are you from?

High Schools then, are also know as Secondary mind.

  • 12.10.2012 9:55 AM PDT

This is the way the world ends,
This is the way the world ends,
This is the way the world ends;
Not with a bang but a whimper.

Don't worry OP, this problem isn't that bad. The second equation you have, though, is the potential energy of a spring, so I'd through that out. You shoul know how heavy the board is, so I think maybe you wrote the problem wrong, but that's okay, as mass cancels out anyway. All you do is set:mgh=.5mv^2So if the distance h is 5 meters (when the board has fallen half way), your velocity should be sqrt(2gh). When it falls all the way, h becomes 10, not five. So, ultimately, your two answers should be 9.9m/s (like you said) and 14m/s. It's as easy as that. I knew college taught me something. Let me know if you have any questions.

[Edited on 12.10.2012 9:57 AM PST]

  • 12.10.2012 9:55 AM PDT

Now go make me some tea.


Posted by: xXIHAYD0IXx
Don't worry OP, this problem isn't that bad. The second equation you have, though, is the potential energy of a spring, so I'd through that out. You shoul know how heavy the board is, so I think maybe you wrote the problem wrong, but that's okay, as mass cancels out anyway. All you do is set:mgh=.5mv^2So if the distance h is 5 meters (when the board has fallen half way), your velocity should be sqrt(2gh). When it falls all the way, h becomes 10, not five. It's as easy as that. I knew college taught me something. Let me know if you have any questions.

How is it 10? It says it's falling so it should be 0 when it hits.

  • 12.10.2012 9:57 AM PDT

http://i.imgur.com/fsISj.png

Posted by: TheDerRitter
Posted by: annoyinginge
You're gonna have to define the equation terms for me.

EDIT: the equation you're using is beyond my level, so this might be a really simplistic and stupid answer, but could you use SUVATs? You have the initial velocity (0 ms^-1), the displacement (10 m) and the acceleration (9.81 ms^-2 if you're allowed to ignore air resistance).

I have no initial velocity of 0, since it's asking me for the speed when it's halfway down

How is that relevant? Regardless of where the final velocity is defined, a dropped object has a velocity of 0 at t=0. Thus a SUVAT taking movement from the point the object is dropped at to the point you have to work out the final velocity will have u=0 regardless of where v is.

  • 12.10.2012 9:57 AM PDT

Now go make me some tea.

-blam!- me, I wrote the problem wrong.


A 755N diver drops FROM a board, so the mass is 755N

  • 12.10.2012 9:59 AM PDT

This is the way the world ends,
This is the way the world ends,
This is the way the world ends;
Not with a bang but a whimper.

Posted by: TheDerRitter
Posted by: xXIHAYD0IXx
Don't worry OP, this problem isn't that bad. The second equation you have, though, is the potential energy of a spring, so I'd through that out. You shoul know how heavy the board is, so I think maybe you wrote the problem wrong, but that's okay, as mass cancels out anyway. All you do is set:mgh=.5mv^2So if the distance h is 5 meters (when the board has fallen half way), your velocity should be sqrt(2gh). When it falls all the way, h becomes 10, not five. So, ultimately, your two answers should be 9.9m/s (like you said) and 14m/s. It's as easy as that. I knew college taught me something. Let me know if you have any questions.
How is it 10? It says it's falling so it should be 0 when it hits.
I edited my last post with the answers. Just an FYI. The "h" in the equation is the initial height above the surface from which the object is dropped, not the final height. Technically, the equation should read mgh_i + .5m(v_i)^2 = mgh_f + .5m(v_f)^2From this, it should be easy to see that v_i=0, as the object is initially at rest and h_f is zero, as the object is just barely above the water. Thus, those two terms go to zero, leaving you with mgh_i = .5m(v_f)^2EDIT- Thank you for correcting the problem, but as I said, mass cancels out. Also, you don't have to trust me if you don't want, but I just took a college physics final over (roughly) the same material two days ago, so it is quite fresh in my mind and I assure you this is right. One more thing, annoyinginge is technically right, though his method is a bit more convoluted.

[Edited on 12.10.2012 10:04 AM PST]

  • 12.10.2012 10:00 AM PDT

http://i.imgur.com/fsISj.png

Using SUVATs:

s = 10 m
u = 0 ms^-1
v =
a = 9.81 ms^-2
t =

v^2 = u^2 + 2as

v^2 = 2 x 9.81 x 10
v = sqrt(196.2) = 14.0 ms^-1 (3sf)

[Edited on 12.10.2012 10:02 AM PST]

  • 12.10.2012 10:01 AM PDT

Now go make me some tea.


Posted by: xXIHAYD0IXx
Posted by: TheDerRitter
Posted by: xXIHAYD0IXx
Don't worry OP, this problem isn't that bad. The second equation you have, though, is the potential energy of a spring, so I'd through that out. You shoul know how heavy the board is, so I think maybe you wrote the problem wrong, but that's okay, as mass cancels out anyway. All you do is set:mgh=.5mv^2So if the distance h is 5 meters (when the board has fallen half way), your velocity should be sqrt(2gh). When it falls all the way, h becomes 10, not five. So, ultimately, your two answers should be 9.9m/s (like you said) and 14m/s. It's as easy as that. I knew college taught me something. Let me know if you have any questions.
How is it 10? It says it's falling so it should be 0 when it hits.
I edited my last post with the answers. Just an FYI. The "h" in the equation is the initial height above the surface from which the object is dropped, not the final height. Technically, the equation should read mgh_i + .5m(v_i)^2 = mgh_f + .5m(v_f)^2From this, it should be easy to see that v_i=0, as the object is initially at rest and h_f is zero, as the object is just barely above the water. Thus, those two terms go to zero, leaving you with mgh_i = .5m(v_f)^2EDIT- Thank you for correcting the problem, but as I said, mass cancels out. Also, you don't have to trust me if you don't want, but I just took a college physics final over (roughly) the same material two days ago, so it is quite fresh in my mind and I assure you this is right. One more thing, annoyinginge is technically right, though his method is a bit more convoluted.

Alright, I got it, just a few more tests and I'll report back.


Alright, so I think I get it now: H isn't 0, it's 10 since it's the initial and the final is 0. The final would be thrown out, leaving the initial.

With that: mgh_i= (77kg)(9.8m/s)(10m) which would make it 7550J, and since 1/2(77kg)v^2, M for 1/2mv^2 would be 38.5kg
and then:
7550J/38.5=196.1038961 and the sqrt of that would be 14m/s!

[Edited on 12.10.2012 10:15 AM PST]

  • 12.10.2012 10:10 AM PDT

BrAdLeY

the answer is 42....yup definitely 42

  • 12.10.2012 10:18 AM PDT

Expressing my strong liberal views without shame. Favorite quotes below:

"Two things are infinite: the universe and human stupidity; and I'm not sure about the the universe."
"One starts to live when he can live outside himself."

- Albert Einstein

The diver drops a board, or he drops? You need to be more clear in your wording.

Regardless, total mechanical energy is always the sum of the potential and kinetic energies. Since energy is conserved, you can write

PE + KE = PE' + KE' (where ' is "prime," meaning new).

Expand this formula:

mgh + .5v^2 = (mgh)' + (.5v^2)'

The left hand side is when the object is about to be dropped, and the right hand side is when the object is 5 meters down, so

755(10) + .5(0)^2 = 755(5) + .5v^2

Solve for v using algebra.

  • 12.10.2012 10:31 AM PDT


Posted by: TheDerRitter
Alright, I got it, just a few more tests and I'll report back.


Alright, so I think I get it now: H isn't 0, it's 10 since it's the initial and the final is 0. The final would be thrown out, leaving the initial.

With that: mgh_i= (77kg)(9.8m/s)(10m) which would make it 7550J, and since 1/2(77kg)v^2, M for 1/2mv^2 would be 38.5kg
and then:
7550J/38.5=196.1038961 and the sqrt of that would be 14m/s!

H is arbitrary. The change in height is all you're concerned about. You could define where he starts as 0, 10, or 10000m, so long as he is at -5, 5, or 9995m after he drops 5m.

[Edited on 12.10.2012 10:40 AM PST]

  • 12.10.2012 10:39 AM PDT

Now go make me some tea.

Nevermind, I'm an idiot who can't into pythagorean theorem

[Edited on 12.10.2012 10:45 AM PST]

  • 12.10.2012 10:42 AM PDT

Expressing my strong liberal views without shame. Favorite quotes below:

"Two things are infinite: the universe and human stupidity; and I'm not sure about the the universe."
"One starts to live when he can live outside himself."

- Albert Einstein


Posted by: TheDerRitter
So i've ran into another problem: If the diver leaves the board with an initial upward speed of 2.00m/s, find the diver's speed when striking the water.

(755N)(10m) + .5(77kg)(2m/s)^2 =.5(77kg)(v_f)^2
7550J + 154kg m/s =(38.5kg)(v_f)^2
196.1m/s + 4m/s=(v_f)^2
14m/s + 2m/s=v

But, the back of the book says it's 14.1m/s

Well, for one, look at your algebra. sqrt(a^2 + b^2) is not a +b.

  • 12.10.2012 10:45 AM PDT