Why Is Mechanical Work Expressed as W=Fx in Physics?

[Ahmed Abdullah]

Can you explain why mechanical work is defined as W=Fx? Where x is the displacement of the point of exertion of the force toward the direction of the force.
Waiting for yr response.
Thx.

 

[HallsofIvy]

Because it works? (Sorry about the pun.) Fx, "force times distance", has the correct units to be an energy and, experimentally, it satisfies "conservation of energy". If an object moves upward with initial velocity v0 (positive: upward) and constant acceleration -a (negative: downward), after time t, it its velocity will decrease to v0-at. In that same time, it will have moved upward a distance x= v0t-(1/2)at^2. That decrease in velocity means that its kinetic energy will have decreased from (1/2)mv0^2 to (1/2)m(v0-at)^2= (1/2)mv0^2- mav0t+ (1/2)ma^2t^2. In other words, it will have decreased by
m(av0t-(1/2)a^2t^2). In order to have conservation of energy, we must argue that the potential energy has increased by that amount: ma(v0t-(1/2)at^2)= max. Of course, the force causing the accelertaion is F= ma so that is just Fx.

In essense all the various definitions of "energy" and "work" are book keeping devices: we are searching for something that remains constant in all our experiments.

 

[Ahmed Abdullah]

I have learned to derive kinetic energy from the relation W=Fx. Like http://physics.about.com/od/energyworkpower/f/KineticEnergy.htm
When we do not know that work W=Fx , how are we suppose to derive KE=1/2mv^2?

 

[rbj]

Can you explain why mechanical work is defined as W=Fx? Where x is the displacement of the point of exertion of the force toward the direction of the force.
Waiting for yr response.
Thx.

probably a good illustration for why work is the product of force and displacement, W=Fx is the use of a lever or see-saw to lift an object that weighs F Newtons up a distance of x meters. i can lift it directly (F Newtons and x meters) or i can use a lever to lighten my load, F/2 Newtons, but then if i want to lift the load the same x meters, i must lift the lever (with a force of F/2 Newtons) up twice the distance or 2x meters. same if i lighten the load to F/3 Newtons, i must lift the lever 3x meters.

the accomplishment or end result of the job is the same in all 3 cases; i lift an object that weighs F Newtons up a distance of x meters. but i didn't use the same force or go the same distance. however, the product of force and distance was the same in all three cases.

 

[russ_watters]

When we do not know that work W=Fx , how are we suppose to derive KE=1/2mv^2?

That's not a complete sentence, so I'm not certain what you are asking, but the kinetic energy equation can be derived from the work equation by inserting the laws of motion into it.

 

[animalcroc]

W=fx can be manipulated to w=fx=max=m[(Vf-Vi)/t ] *[ (Vf+Vi)/2] * t=(1/2)m(Vf^2-Vi^2)
This is important because it relates velocity to force and distance through which force acted.
As HallsofIvy stated, it's conserved, so this relationship is important.As to why it's called "work" that's arbitrary. You can call it anything you want. I don't believe mechanical work is a more important concept than momentum. The belief that it is is what makes it seem mysterious even though it's not. Momentum is also conserved and is just as important.
In fact, work and impulse (change in momentum) are related through the distance=velocity * time relationship

work=fx
impulse=ft

 

[Link-]

This is how was explained to me...

F=M*a -> F=M*dv/dt -> F=M*V*dv/ds -> F*ds=MVdv -> $$\int F*ds$$=$$\int M*V dv$$

If the force is on the direction of the displacement, F*s=1/2M(V$$_{2}$$ $$^{2}$$-V$$_{1}$$ $$^{2}$$)

Now that you have this relation ship between force and velocity think about this, if you have a toy car and you applied a force for x distance you will see that the car's velocity increases. When you see that your car velocity encrease you say that it gain energy, but the energy doesn't appeared from nothing, you applied a force for a x distance and you will say that energy came from the work you made on the car.

You should note that the real definition of work is W=$$\int \vec{F} \bullet d\vec{s}$$

Note: For some reason the subscripts appeared as superscripts. If somebody now why let me know.

 

[rbj]

Fx, "force times distance", has the correct units to be an energy...

Halls, can you explain to me how that is more than a vacuous truth or tautology? the units or dimension for energy are derived units from defining energy to be force times distance.

the fact that this conceptually derived quantity called "energy" satisfies this conservation principle and has been empirically observed to do so, that is no vacuous truth, but to say that units of force times units of distance has the units of energy only restates the definition which motivation behind it was that the OP wanted explained.

i dunno.

 

[Loren Booda]

You supplied the missing, Link.

 

[Ahmed Abdullah]

That's not a complete sentence, so I'm not certain what you are asking, but the kinetic energy equation can be derived from the work equation by inserting the laws of motion into it.

Suppose that you don't know the relation,
mechanical work W=F.S.

In this circumastances can you derive the equation Ek=1/2mv^2 (without using the relation W=F.S)?

I think any of the two relation (W=F.S and Ek=1/2mv^2) is axiomatic and they are circular. So that you can't get one of them without using the other. When one of them is defined axiomaticaly you can get the other.
Am I correct?

 

[AbedeuS]

Because your talking about fundamental equations?

Also I didnt quite understand Link's explination of how $$\frac{1}{2}MV^{2}$$ comes from, maths seems a bit, dodgey?

 

[Link-]

Because your talking about fundamental equations?

Also I didnt quite understand Link's explination of how $$\frac{1}{2}MV^{2}$$ comes from, maths seems a bit, dodgey?

Is not dodgey, just apply differential equations to F=ma.

Here: $$F=\frac{d\vec{P}}{dt}$$
$$\vec{P}$$ is momentum but we are simplifying this so we consider mass as constant, and you will get $$F=\frac{m*d\vec{v}}{dt}$$
dv/dt=acceleration=v*dv/ds
F=m*v*dv/ds
the next step is differential equation and calculus.

As you see in the work and kinetic energy equation you would see a change in velocity that correspond to a force applied times the distance that is applied.
Change in velocity would mean that system increase or decrease its energy.

-link

PS If you still not understanding the explanation of how 1/2*m*v^2 , I think you should try by yourself to apply differential equations and calculus to equations.

 

[AbedeuS]

dv/dt=acceleration=v*dv/ds

Not seeing what this is explaining, and I do both brands of calculus

$$\frac{dv}{dt} = a$$

Is the bit I do understand

but:

$$Acceleration = v*\frac{dv}{ds}$$

Whats "s" and what is it trying to express?

 

[Link-]

dv/dt=acceleration=v*dv/ds

Not seeing what this is explaining, and I do both brands of calculus

$$\frac{dv}{dt} = a$$

Is the bit I do understand

but:

$$Acceleration = v*\frac{dv}{ds}$$

Whats "s" and what is it trying to express?

Oh, I see what you miss.

dv/ds means the change in velocity divide by the change in length. Usually the differential of distance is expresed as ds.

This is how you get to v*dv/ds=acceleration:

$$a=\frac{dv}{dt}=\frac{dv}{dt}*\frac{ds}{ds}=\frac{dv}{ds}*\frac{ds}{dt}$$

Since $$v=ds/dt$$ then, $$a=v\frac{dv}{ds}$$

$$a=v\frac{dv}{ds}$$ is trying to explaing that acceleration is equal to the velocity times the differetial of velocity divided by differential of length. If you put this equation on the force equation you'll get $$F=m*v\frac{dv}{ds}$$.
;)
-Link