# Homework Help: Can someone help me answer this question about work?

1. Jan 25, 2014

### xphysics

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

3. The attempt at a solution

2. Jan 25, 2014

### Staff: Mentor

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

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

3. Jan 25, 2014

### 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 Fx ∫ dsx = Fx times x

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.

4. Jan 25, 2014

### 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)?

5. Jan 25, 2014

### 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.

6. Jan 25, 2014

### Staff: Mentor

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

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.)

7. Jan 25, 2014

### xphysics

ahhh ok thank you guys i understand it now appreciate it