Converting W²/NJHz to MKS: Step-by-Step Guide

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The discussion focuses on converting the expression W²/NJHz into MKS (meters, kilograms, seconds) units. Participants clarify the definitions of watts, joules, newtons, and hertz, and how to express them in fundamental units. The correct simplification leads to the result of m/s, confirming that the final answer is indeed meters per second. Several users share their calculations and methods, highlighting the importance of careful exponent management during the conversion process. Ultimately, the consensus is that the correct and simplified answer to the conversion is m/s.
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Given: W = J/s, J = Nm, N = kgm/s2, Hz = 1/s
Convert W²/NJHz to mks and simplify. Make sure to list every step.

can someone please guide me...this is what I have so far...im not sure if I am doing it right though...


J/s² underline meaning over...
kgm/s² * Nm* 1/s



is this right...
 
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joejo said:
Given: W = J/s, J = Nm, N = kgm/s2, Hz = 1/s
Convert W²/NJHz to mks and simplify. Make sure to list every step.
What is an mks?

joejo said:
can someone please guide me...this is what I have so far...im not sure if I am doing it right though...
I can but I need a little help from you.

Also try this thread.

The Bob (2004 ©)
 
thx bob...i don't know but I will email my teacher and ask him...
 
PM me when you have found out.

The Bob (2004 ©)
 
MKS it a measure system. You are already working with MKS.
 
Last edited:
Maybe you need to convert to cgs...
 
No, I think he's just supposed to reduce it to the simplest possible combination of meters, kilograms, and seconds, after substituting for watts, joules, etc.
 
Werg22 said:
MKS it a measure system. You are already working with MKS.
In that case it is fairly simple.

joejo said:
Given: W = J/s, J = Nm, N = kgm/s2, Hz = 1/s
Convert W²/NJHz to mks and simplify. Make sure to list every step.

Write W as Js-1, J as Nm, N as kg ms-2 and Hz as s-1.

\frac{W^2}{N \times J \times Hz} = \frac{(Js^{-1})^2}{kg \ ms^{-2} \ \times \ Nm \ \times \ s^{-1}}

Now it is time to start using your basic knowledge of maths to multiply and divide this out.

\frac{J^2s^{-2}}{\frac{kg \m}{s^2} \ \times \ Nm \ \times \ \frac{1}{s}} = \frac{J^2}{s^2} \div (\frac{kg \m}{s^2} \ \times \ Nm \ \times \ \frac{1}{s}}) = \frac{J^2}{s^2} \div \frac{kg \m \times Nm}{s^2 \times s}

= \frac{J^2}{s^2} \times \frac{s^3}{kg \m \times Nm} = \frac{J ^2 \ s^3}{kg \m \times Nm \times s^2} = \frac{J ^2 \ s}{kg \m \times Nm} = J^2 \ s \ kg^{-1} \ Nm^{-1}

This is what I get however I feel that the (Js-1)2 = J2s-2 might be said to be wrong by someone more in the know than I am on units. However this is what I think until I am told otherwise.

If this does not make too much sense then look at the thread I mentioned in the second post. It will explain things more clearly.

The Bob (2004 ©)
 
First convert all your 'givens' to MKS, starting with the most basic (Hz).

MKS means everything is meters, kilograms, and seconds. If something is on the 'bottom' of a fraction, rewrite it as being multiplied times the top, only make the exponent negative. Example: \frac{4}{x^2} = 4x^{-2}

Hz=(s^{-1})
N=(kg)(m)(s^{-2})
J=(kg)(m^2)(s^{-2})
W=(kg)(m^{2})(s^{-3})

Now just plug them into your big equation:

(W^2)(N^{-1})(J^{-1})(Hz^{-1})

For (W^2), all you have to do is double all the exponents of that part once you substitute, so (W^2)=(kg^2)(m^4)(s^{-2}).

Now, that's one part that I've done for you. I'd like to see you substitute the rest in on your own. Just remember, that when you've determined, for example, J=(kg)(m^2)(s^{-2}), then J^{-1}=(kg^{-1})(m^{-2})(s^{2})...all the exponents switch signs. When multiplying like terms, add their exponents. For example, (m^4)(m^{-3})=(m^{-1}). I put parentheses around everything so I don't get confused by two-letter variables.
 
  • #10
lol...still kinda confused..but this seems righter then me (oops is righter a word)...

so this is right...
 
  • #11
oh thanks kingnothing i just saw ur reply...i will work it out now...and post my answer...thanks to all that replied
 
  • #12
I see what you have done. My method, in my eyes, is fine but I simply need to learn the conversions (as N and J are the same as you have written and I would not have expected).

Cheers.

The Bob (2004 ©)
 
  • #13
so is ur way right too the bob?! u guys r losing me
 
  • #14
hey guys i did all the work on paper...my final answer is

(kg4)(m7)(s-11)

please tell me that's right??
 
  • #15
=(kg^{4})(m^{7})(s^{-11})

i hope this latex stuff works...
 
  • #16
joejo said:
=(kg^{4})(m^{7})(s^{-11})

i hope this latex stuff works...
Personally I got m kg-1 s-1.

The Bob (2004 ©)
 
  • #17
I got m^1 * s^{-1} in other words, meters per second.

Bob, how did you get an extra kg on the bottom?

Watts squared yields kg^2, then on the bottom Joules and Newtons both have kg^1 so there is no kg on top or bottom.
 
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  • #18
Here is my workings:

N = kg \ ms^{-2}

J = Nm = kg \ ms^{-2} \times m = kg \ m^2 s^{-2}

W = Js^{-1} = kg \ m^2 s^{-2} \times s^{-1} = kg \ m^2 s^{-3}

\frac{W^2}{J \times N \times Hz} = \frac{(kg \ m^2 s^{-3})^2}{kg \ m^2 s^{-2} \times kg \ ms^{-2} \times s^{-1}}

= \frac{kg^2 \ m^4 s^{-6}}{kg \ m^2 s^{-2} \times kg \ ms^{-2} \times s^{-1}}

= \frac{kg^2 \ m^4 s^{-6}}{kg^2 \ m^3 s^{-5}} = m s^{-1}

So I agree with KingNothing and I realized my mistake was a missing squared on my kilograms. :frown:

Stupid me.

The Bob (2004 ©)
 
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  • #19
Alright, it's time to check my answer to prove to you villains that I'm right. :P
meters=5
kg=8
seconds=3

Watt=(8)(5^2)(3^-3)
Watt^2 = 54.87 (about)
Newton = (8)(5)(3^-2)
Newton = 4.44
Joule = (8)(5^2)(3^-2)
Joule = 22.22
Hz = (3^-1)
Hz = .33

Newton * Joule * Hz = 32.56
Watt^2 = 54.87

So, Watt^2 over Newton * Joule * Hz = 1.69


And with my simplification, 5/3 = 1.6666666

Tada!
 
  • #20
KingNothing said:
Alright, it's time to check my answer to prove to you villains that I'm right. :P
All credit to you. :biggrin: I said you were right and I was simply checking my method for joejo as I got the impression it made sense to him/her.

The Bob (2004 ©)
 
  • #21
hey kingnothing...ur totally losing me...can you please show me how you got that answer...im so lost...and i have a quiz 2mr?!

thanks...i have my work if you want to see...so don't think I am asking for the answer...ive been stuck on it since friday...

thanks
 
  • #22
Yes, show your work. Just break your 'givens' down into MKS, then substitute them in and use algebra.

As crazy as this may sound, I looked at your numbers and you have 90% of it right! So don't worry. I think a simpler example will clear things up.

Convert \frac{N}{Hz} to MKS.
N = \frac{Kg * m}{s^2}
Now, this is where it might get tricky. Dividing by s^2 is the same as multiplying by s^{-2}!
Therefore, we can rewrite \frac{Kg * m}{s^2} as Kg * m * s^{-2}

So now we're half done. Rewrite the other part, Hz.
Hz = \frac{1}{s} or Hz = s^{-1}.

So now it's converted to MKS, it's just not simplified.
\frac{N}{Hz} = \frac {Kg * m * s^{-2}}{s^{-1}}
Which is just substitution.

Now just think about that part on the bottom. Remember how moving something to the top made the exponent's sign change? You can do the same thing here.

\frac {Kg * m * s^{-2}}{s^{-1}} = Kg * m * s^{-2} * s^1

It's really close, but we have two like terms, the s. When you multiply two like terms with different exponents, you add the exponents. Therefore, s^{-2} * s^1 = s^{-1}.

According to my numbers, switching the negative signs and exponents is the only part you are doing wrong. But the rest you are doing fine.
 
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  • #23
i didddddd...it didn't work...i have it scanned...but it won't fit...please don't make me do the latex stuff again...
 
  • #24
No, don't bother with Latex. We can probably understand what you mean. The reason I say you are close is this:

for my exponents on the Kg unit, they were +2, -1, and -1, meaning they summed to zero and that's why there is no Kg in the final answer. Yours summed to 4.

for my exponents on the meters unit, they were +4, -1, and -2, meaning they summed to +1. Yours summed to 7.

It seems to me like when you are multiplying like terms like y^2 * y^-1, you are adding the exponents like this: 2+1=3, so it is y^3. No. It is y^1.
 
  • #25
W=(kg)(m^{2})(s^{-3})


(W^2)(N^{-1})(J^{-1})(Hz^{-1})

For (W^2), all you have to do is double all the exponents of that part once you substitute, so (W^2)=(kg^2)(m^4)(s^{-2}).

well i added the like terms...but i think you/I made a mistake earlier ...for W2 u said that is was quote above...should it become -6 because you squared it...thats where are answers differ... because s is orginally -3*2= -6

thanks
 
  • #26
so wen i got my answer I used -3*2= s-6

and u used s-2?? why is that...isn't it W2

thanks
 
  • #27
Yes, it was a typo. It's s^{-6}
 
  • #28
therefore making my answer right...right??...because that affects your answer...
 
  • #29
I get m/s

W=kg*m^2/s^3
N=kg*m/s^2
J= kg*m^2/s^2
Hz=1/s

(kg*m^2/s^3)^2/ (Kg*m/s^2* Kg*m^2/s^2*1/s)

So

(kgm^2/s^3)^2/(kg^2*m^3/s^5)

(kg^2*m^4/s^6)(s^5/ kg^2*m^3)

kg^2*m^4*s^5/s^6*kg^2*m^3

m/s
 
  • #30
joejo said:
therefore making my answer right...right??...because that affects your answer...

The answer is ms^{-1}

(W^2)(N^{-1})(J^{-1})(Hz^{-1})

leads to... kg^2m^4s^{-6}*kg^{-1}m^{-1}s^{2}*kg^{-1}m^{-2}s^{2}*s^{1} = ms^{-1}

Watch out for the sign of your exponents.
 
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  • #31
\frac{\mbox{W}^2}{\mbox{NJHz}} = \frac{\left(\frac{\mbox{Nm}}{\mbox{s}}\right)^2}{\left(\frac{\mbox{N}^2\mbox{m}}{\mbox{s}}\right)} = \frac{\mbox{m}}{\mbox{s}}
 
  • #32
Aw Data, you beat me to it. A nice concise simplification. I'm going to post mine anyway since its still in the buffer

joejo said:
Given: W = J/s, J = Nm, N = kgm/s2, Hz = 1/s
Convert W²/NJHz to mks and simplify. Make sure to list every step.

can someone please guide me...this is what I have so far...im not sure if I am doing it right though...


J/s² underline meaning over...
kgm/s² * Nm* 1/s



is this right...

So many posts for such a small problem :rolleyes:

\frac{W^2}{NJHz}

W=\frac{J}{s}

J=Nm

N = \frac{kgm}{s^2}

Hz = \frac{1}{s}

\frac{W^2}{NJHz} = \frac{(\frac{J}{s})^2}{NJ\frac{1}{s}} = \frac{\frac{J}{s}}{N} = \frac{Nm}{Ns} = \frac{m}{s}
 
  • #33
OlderDan said:
So many posts for such a small problem :rolleyes:

Precisely the reason why. Because some people feel they need to re-post the right answer for seemingly no reason ;).

Joejo, answer these problems. Simplify these so there is no fractions, only exponents:

x^2 *x^{-5} = ??


\frac{x^7}{y^4} = ??

Answer those and we will be able to see your problem.
 
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  • #34
KingNothing said:
Precisely the reason why. Because some people feel they need to re-post the right answer for seemingly no reason ;).

Well no doubt mine was more than needed, but at the time I started working on all the laborious tex stuff, somebody needed to post a solution that was concise and to the point eliminating the derived units as much as possible before sustituting in all the m, kg, and s all over the place. Most of the incorrect solutions posted were needlessly complex, and that's probably why they were wrong.
 
  • #35
The Bob said:
Here is my workings:

\frac{W^2}{J \times N \times Hz}

= \frac{kg^2 \ m^4 s^{-6}}{kg^2 \ m^3 s^{-5}} = m s^{-1}

So I agree with KingNothing and I realized my mistake was a missing squared on my kilograms. :frown:

The Bob (2004 ©)
I have already shown how to do the simplification. It is a full example.

The Bob (2004 ©)
 
  • #36
so many answers??!? which one is right guys?!

older dan is that right...because that is easy to understand...

someone please help?!
 
  • #37
joejo said:
so many answers??!? which one is right guys?!

older dan is that right...because that is easy to understand...

someone please help?!

m/s is the correct answer. You can count on it.

Some of the earlier solutions also got the same result. I posted my solution because it is simpler than the approaches posted earlier. Data's is also simple, but it combines two pieces of the given information in one step. Bob's final answer also agrees, but his latest post combines many things in the first substitution. I avoided that in my approach.

There are often many ways to approach a problem, and I'm not claiming mine is the "best", but as a general rule when I have to simplify units I prefer to eliminate as many "derived" units as possible before replacing them with the "fundamental" units of m,kg,s. It usually cuts down on the number of steps and reduces the probability of losing track of something.
 
  • #38
I posted the answer because there are lots of incorrect answers in this thread, which could easily confuse the poster (and apparently have!). m/s is correct.
 
  • #39
Data said:
m/s is correct.
Is 'm/s' standard practise to most people? It was good at GCSE but causes confusion when doing this sort of problem. Making it ms-1 makes more sense as you can mathematically solve it with letters.

This is only my peronsal opinion and if others find m/s easier then that is fine. Just remember that higher levels of education use ms-1. :smile:

The Bob (2004 ©)
 

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