Work and Energy-Combining equations

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Work and Energy--Combining equations

Hi, I have the following sample problem that I am having difficulty combining equations:

The power expended in lifting an 825-lb girder to the top of a building 100 ft. high is 10.0 hp. How much time is required to raise the girder.

So the data is:

F= 825
s= 100 ft
P= 10.0
t= ?

Now the following two equations contain all the values we need:

P=W/t and W= Fs

In the text the combined equations are shown as:

t= W/P = Fs/P ----- How was this found and combined?


Thanks.
 
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Well, combining the equations is just minor league algebra. P=W/t is the same as saying t=W/P, right? Then put W=Fs. Is that really what your question is? To really finish the problem you need to look up what a horsepower is in ft*lb/s.
 


Well, woudn't P=W/t translate to: t=PxW and where does Fs/P come from?
 


How would P=W/t turn into t=P*W?? Take P=W/t, and multiply both sides by t, getting P*t=(W/t)*t=W. Now divide both sides by P, getting (P*t)/P=t=W/P. You have a wrong idea about algebra. What is it?
 
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I see, yeah it has never been my strong point, but how is FS/P found?
 


Take t=W/P. You know W=Fs. Just substitute Fs for W. You can always substitute equal quantities for equal quantities. You aren't exaggerating about algebra being a weak point. No offense, but it's REALLY weak. Problems like this are going to take more than you've got. Can you take some time to review and practice?
 


Yeah, I know they are and it hasn't bothered me until now. Ok, so because they are equal quantities it becomes just FS. Where does the P come from if it wasn't previously in the equation.
 


Petrucciowns said:
Yeah, I know they are and it hasn't bothered me until now. Ok, so because they are equal quantities it becomes just FS. Where does the P come from if it wasn't previously in the equation.

What? The P was in the equation previously!

t = W/P [1]

W = Fs [2]

THEREFORE SUBSTITUTING Fs for W in [1],

t = Fs/P
 


Got it. Thanks, both of you very much. I haven't had to use algebra in so long, and I was never really good at it. I will do some practicing.

Thanks again.
 


Actually I have one more question.

With the values

P= known
s= known
t= known
F= unknown

and the equations P= W/t and W= Fs

how does the working equation become F= Pt/sI was doing it this way F= s * W then after this point I was lost.
 


Petrucciowns said:
I was doing it this way F= s * W then after this point I was lost.

Your equation is wrong...
You can look from the previous post that W = F*s , not F = s*W

W = F*s
F = W/s

Do you get it?
 


Ok substitute again that W=Fs into P=W/t.
This gets P=Fs/t. Got it? To get F by itself we multiply both sides by t, Pt=Fs, and then you divide by s on both sides. Pt/s=F or F=Pt/s.
 


See, that's what I was doing before isolating f by dividing both sides by s ,but

how come P=W/t solving for t = t=W/P when using the above approach multiplying both sides by W gets PxW=t What is the difference with these two problems that makes calculating the answer different?
 


Oh sorry! I didn't know the equations were wrong! I just saw two equations and you needed help to get a third one. I don't know the equations or what they mean. Sorry for your trouble.
 


Petrucciowns said:
how come P=W/t solving for t = t=W/P when using the above approach multiplying both sides by W gets PxW=t

P =W/t, multiplying both sides by W, we get :

P x W = W^2 / t , not PxW=t
 


Petrucciowns said:
Actually I have one more question.

With the values

P= known
s= known
t= known
F= unknown

and the equations P= W/t and W= Fs

how does the working equation become F= Pt/s


I was doing it this way F= s * W then after this point I was lost.

W=F*s does NOT turn into F=s*W when you solve for F. It turns into F=W/s. You have to DIVIDE both sides by s. There's really no point in asking these question until you review your algebra.