Can someone help me answer this question about work?

[xphysics]

Homework Statement

I've been watching calculus-based classical physics from MITcourseware just to learn physics. and I've been having this question that professor walter lewin did not address neither drphysicsa on youtube
the question is: knowing that W(from A to B)= the Integral of (F dx) and eventually u derive that you'll get change in kinetic energy ok i got that. But what is the different between that and W=Fx which is work=force times displacement?

there's also another equation that prof.WL addressed: dw=F(x.component)dx+F(y.component)dy+F(z.component)dz
and if you take the integral of that you get W by itself? so the rate of change of work is Force times displacement? isn't it just work itself?

Please help!

Homework Equations

The Attempt at a Solution

 

[Doc Al]

the question is: knowing that W(from A to B)= the Integral of (F dx) and eventually u derive that you'll get change in kinetic energy ok i got that. But what is the different between that and W=Fx which is work=force times displacement?

W = Fx applies when the force is constant. For a varying force, you need the integral.

there's also another equation that prof.WL addressed: dw=F(x.component)dx+F(y.component)dy+F(z.component)dz
and if you take the integral of that you get W by itself? so the rate of change of work is Force times displacement? isn't it just work itself?

That's not the 'rate of change' of work, that's just work. Note that the integral is with respect to displacement, not time.

 

[BvU]

Just adding: dW is a small change in work, namely over a small change in path ds = [dx,dy,dz].
Work is the integral of dW over the path.

Now, dW = dF . ds
dW is a number and dF and ds are both vectors.
The only contribution of F to W is the part of F that is along the path ds. That's where the dw=F(x.component)dx+F(y.component)dy+F(z.component)dz expression comes from.

So a force in the y direction does not do work when the path is in the x direction. This is very important to understand.

As the doc says, W = Fx is for a very simple case: F constant and F in the same direction as the x. Then ∫ dF.ds from 0 to x is F_x ∫ ds_x = F_x times x

Kudos for your curiosity and even more for asking good questions!
If you want more, more advanced, or more understandable explanations:
The reason the template has 2 and 3 is that it helps helpers provide assistance at a level that the asking person may understand. In this case there is no homework problem but a question to help understand.
Even more important to get help you can do something with.

 

[xphysics]

Ahhhh ok i kinda understand it now actually i understand it alot! I thought about it for a minute and i saw why:
If you graph F(x) with x as distance on the x-axis and F as force on the y-axis, and you apply a constant force on it, you just eventually get the amount of force times x to find the work so therefore we have the simple equation W=Fx? But if you make a graph with changing forces, you need to find the integral of that graph inable to find the work, and this is what confuse me again(sorry). In able to do that mathematically don't you need F with x as an independent variable instead of F=ma(a as an independent variable)?

 

[lightgrav]

uh-oh. Total applied Force is what causes mass to accelerate ; so in Newton's 2nd, "a" is the DEpendent variable.
Each individual force is caused by some external subject (spring via contact, Earth via gravity, rope via contact, pressure via contact) ... many of these forces ARE functions of location x , in a particular situation.
The ones that are not (viscous friction and drag, especially) are a real pain to deal with, because , yes, you can't calculate their Work until you can re-write their force dependence from F(v) => F(x) ... essentially you need to know v(x) to do this, which you can't until you know the viscous force. So you average and approximate, until you learn how to solve differential equations.

 

[Doc Al]

Ahhhh ok i kinda understand it now actually i understand it alot! I thought about it for a minute and i saw why:
If you graph F(x) with x as distance on the x-axis and F as force on the y-axis, and you apply a constant force on it, you just eventually get the amount of force times x to find the work so therefore we have the simple equation W=Fx?

Right. In that simple case the graph is a rectangle, so the integral is just Fx.

But if you make a graph with changing forces, you need to find the integral of that graph inable to find the work, and this is what confuse me again(sorry). In able to do that mathematically don't you need F with x as an independent variable instead of F=ma(a as an independent variable)?

Of course, to compute the work integral directly you would have to have the force as a function of x. (F = ma is something else.)

 

[xphysics]

ahhh ok thank you guys i understand it now appreciate it