Work and Energy-Combining equations

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In summary, the equations for this problem involve power (P), work (W), time (t), and force (F), and can be combined to solve for unknown values. To combine the equations, you substitute equal quantities for equal quantities and use basic algebraic principles. It is important to review and practice algebra in order to successfully solve problems involving equations.
  • #1
Petrucciowns
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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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  • #2


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.
 
  • #3


Well, woudn't P=W/t translate to: t=PxW and where does Fs/P come from?
 
  • #4


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?
 
Last edited:
  • #5


I see, yeah it has never been my strong point, but how is FS/P found?
 
  • #6


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?
 
  • #7


W = Fs

EDIT: Nevermind. Somebody got there before me.
 
  • #8


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.
 
  • #9


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
 
  • #10


It was previously in the equation. You see the P in t=W/P? It's that P.
 
  • #11


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.
 
  • #12


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.
 
  • #13


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?
 
  • #14


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.
 
  • #15


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?
 
  • #16


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.
 
  • #17


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
 
  • #18


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.
 

1. What is the relationship between work and energy?

The relationship between work and energy is that work is the transfer of energy from one form to another. In other words, work is the process of using a force to move an object, and this results in the object gaining or losing energy. Energy, on the other hand, is the ability to do work.

2. How do you combine equations for work and energy?

To combine equations for work and energy, you need to use the fundamental equation of work and energy, which is W = ΔK + ΔU, where W is the work done, ΔK is the change in kinetic energy, and ΔU is the change in potential energy. By rearranging this equation and substituting other equations for work and energy, you can combine them to solve for a specific variable.

3. Can you provide an example of combining equations for work and energy?

One example of combining equations for work and energy is calculating the work done on an object by a constant force. The equation for work is W = Fd, where F is the force applied and d is the displacement of the object. The equation for kinetic energy is K = 1/2mv^2, where m is the mass of the object and v is its velocity. By substituting these equations into the fundamental equation, you can solve for the change in kinetic energy (ΔK) or the force applied (F).

4. What are the units for work and energy?

The units for work are Joules (J), which is equivalent to kg*m^2/s^2. The units for energy are also Joules (J) since energy and work are directly related. However, different types of energy may have different units, such as potential energy having units of Joules per kilogram (J/kg) and kinetic energy having units of Joules per second (J/s).

5. How is the principle of conservation of energy related to combining equations for work and energy?

The principle of conservation of energy states that energy cannot be created or destroyed, but it can be transferred from one form to another. When combining equations for work and energy, you are essentially showing how energy is transferred and conserved in a system. By using the fundamental equation and substituting other equations, you can see how the energy changes form but remains constant overall. This principle is crucial in understanding the relationship between work and energy.

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