What Are the Units of Constant C in a Force Equation?

[skull]

Ok I need this answered by whoever can.
The force on an object is F=C/x^2
A) what are the units of C?
b)How much work is done by this force when the object is moved from x=a to x=3a.
How much work is done by F when the object is moved from x=3a to x=infinite

units of x are in meters

I have tried this countless time, and still have no answer. All help is greatly appreciated.

 

[stunner5000pt]

What are the units of Force? what are the units of x^2? CAn you find the units of C from this?
what is the formula for work? Are you familiar with this definition
W = \int_{i}^{f} F \bullet dx or the non calculus version?

 

[skull]

What are the units of Force? what are the units of x^2? CAn you find the units of C from this?
what is the formula for work? Are you familiar with this definition
W = \int_{i}^{f} F \bullet dx or the non calculus version?

units of force are Newtons
units of x^2 are meters
the formula that I was given is W=Fdcosx
where f is force, d is distance, and x is the angle

 

[skull]

anyone know?
By the way, this is Gade 11 physics, so keep that in mind when answering

 

[G01]

If F is Newtons like you said, and x is m^2 like you said. C divided by m^2 must give you force in Newtons. If this is the case what must the units of C be so that the answer comes out in Newtons?

 

[skull]

the question says that c=a constant, if the helps

 

[G01]

the question says that c=a constant, if the helps

Well you don't need that fact to get the units at least. Maybe later. To get the units for c all you need are the units of the other two variables. Solve for c and what do you get as the units? That gives you the answer for part a.

Tell me what you get when you do that.

 

[skull]

Well you don't need that fact to get the units at least. Maybe later. To get the units for c all you need are the units of the other two variables. Solve for c and what do you get as the units? That gives you the answer for part a.
Tell me what you get when you do that.

solving for c would simply be c=n/m^2
where n is Newtons and m is metres
I don't really understand what you are saying

 

[G01]

solving for c would simply be c=n/m^2
where n is Newtons and m is metres
I don't really understand what you are saying

No. solve for c from this equation:
F = \frac{c}{x^2} gives:

c = Fx^2

Now fill in the units:

c = Nm^2 C then, has the units N times m^2. You screwed up your algebra. Where you should have multiplied both sides by x^2, you divided.

 

[civil_dude]

Don't you have to use calculus to solve for this since F(x) is not a constant?

 

[G01]

Don't you have to use calculus to solve for this since F(x) is not a constant?

Yeah I was just thinking that. But you can find the units of C without calc. All you need is dimensional analysis.

 

[skull]

No. solve for c from this equation:
F = \frac{c}{x^2} gives:
c = Fx^2
Now fill in the units:
c = Nm^2 C then, has the units N times m^2. You screwed up your algebra. Where you should have multiplied both sides by x^2, you divided.

ok, and how can I get units with the equation you gave?

 

[G01]

I just did that. do you agree that c = Fx^2? If so then c must have the same units as Fx^2 since it is equal to it. What are the units of F times x^2

You MUST have done this at some time during you physics studies. Dimensional Analysis is something physics teachers never let you forget. That is what were doing here to find the units.

 

[skull]

Don't you have to use calculus to solve for this since F(x) is not a constant?

I said F is constant

 

[G01]

the question says that c=a constant, if the helps

You said c was constant not F. I don't know how F can be constant if it equals C/x^2. If that's true then F is not constant it changes with respect to x.

 

[skull]

You said c was constant not F. I don't know how F can be constant if it equals C/x^2. If that's true then F is not constant it changes with respect to x.

Oh my bad, i just misread

 

[G01]

Anyway do u understand how to find the units?

 

[skull]

Anyway do u understand how to find the units?

I thought your earlier post was implying that i did not give enough information. Can I still find the units?

 

[G01]

I just did that. do you agree that c = Fx^2? If so then c must have the same units as Fx^2 since it is equal to it. What are the units of F times x^2

You MUST have done this at some time during you physics studies. Dimensional Analysis is something physics teachers never let you forget. That is what were doing here to find the units.

I've shown you how to find the units MULTIPLE TIMES including the post I just quoted. I've told you how to do it and actually given the answer in one post which I'm not supposed to do since you didn't so any work at all.

I will tell you the method ONE more time and that's it. Take the formula for force and solve for c. You won't get a numeric answer just work with the variables. What is c equal to? It must have the same units as whatever it is equal to.

 

[skull]

I've shown you how to find the units MULTIPLE TIMES including the post I just quoted. I've told you how to do it and actually given the answer in one post which I'm not supposed to do since you didn't so any work at all.
I will tell you the method ONE more time and that's it. Take the formula for force and solve for c. You won't get a numeric answer just work with the variables. What is c equal to? It must have the same units as whatever it is equal to.

Ok, I got joule metre as the unit. Is this right? Is this also know as a Newton?

 

[G01]

Yes those are the right units. A joule-meter is not a Newton a joule/m would be a Newton a Joule-meter is just a joule-meter. The units could also be read as Newton-meter^2 or a kilogram-meter^3/s^2

The rest of the problem I wouldn't be able to do without Calc since the Force isn't constant. At least I don't think there's a way to do it without calc.

 

[skull]

Yes those are the right units. A joule-meter is not a Newton a joule/m would be a Newton a Joule-meter is just a joule-meter. The units could also be read as Newton-meter^2 or a kilogram-meter^3/s^2
The rest of the problem I wouldn't be able to do without Calc since the Force isn't constant. At least I don't think there's a way to do it without calc.

are you sure it can't be done w/o calculus? My friend did it and got
-C/(3a) + C/a
Is this right?
If it is, i don't understand why there is a -c in the equation.

 

[G01]

Honestly I'm not sure. Maybe its just because I'm tired, but I can't see a way to solve this correctly without calc. Maybe someone else can help. You can make a new thread focused on this part of the problem. Sry. I'm lost without calc. Probably missing something stupid though. I don't think your friend's answer is right. That answer would equal the force at the start of the trip plus the force at the end. Its has nothing to do with work at all.

 

[stunner5000pt]

your friend is right

if you were to plot your force function then the area under the graph between a and 3a would the be work done.
idont think that it is possible to actually find the area without calculus, as any method of finding the area under a curve without calc is unknown to me.

are you sure this problem was meant for your grade? DO you learn calc in your school?

 

[G01]

The area under the curve can be found easily if its a geometric shape. but this isn't so you can't find the area without calc.

 

[civil_dude]

units of C * 1/m^2 = Units of Force

Units of C = mass * length/time^2 * length^2