How do I calculate the work done by a force field using the dot product?

[annamal]

Homework Statement

A particle moves along a curved path y = 10*(1 + cos(0.1*x))
from x = 0 to x = 10*pi subject to a tangential force of variable magnitude F = 10*sin(0.1*x)
How much work does the force do?

Relevant Equations

F dot product dr = dW

y = 10*(1 + cos(0.1*x)) --> dy/dx = -sin(0.1x)
dW = F*dx + F*dy = 10*sin(0.1*x)dx + 10*sin(0.1*x)*-sin(0.1x)
integrating we have -100*cos(0.1*x) -10*sin(0.1x)^2 from 0 to 10*pi = W = 43 J. The answer says 257 J. Where am I wrong here?

 

[Orodruin]

You seem to have taken both x- and y-components of the force to be equal to the magnitude of the force.

 

[annamal]

Yes, how would I split F(x) into x and y components?

I also tried
dW = xdx + F(x)*dy and didn't get the right answer either.

 

[Orodruin]

Yes, how would I split F(x) into x and y components?

I also tried
dW = xdx + F(x)*dy and didn't get the right answer either.

Ask yourself if you need to know the components.

 

[annamal]

Ask yourself if you need to know the components.

I don't know. The only way I wouldn't need the know the components is with |F||r|cos(theta) = |F||r| = F(x)*(sqrt(y^2 + x^2))??

 

[Orodruin]

Almost. But it is not $\vec F\cdot \vec x$, it is $\vec F \cdot d\vec x$ ...

 

[SammyS]

I don't know. The only way I wouldn't need the know the components is with |F||r|cos(theta) = |F||r| = F(x)*(sqrt(y^2 + x^2))??

The problem statement does say something about how the force is related to the path, doesn't it ?

 

[annamal]

The problem statement does say something about how the force is related to the path, doesn't it ?

Yes, only with x though F = 10*sin(0.1*x). Still don't know how to figure this problem out.

 

[haruspex]

Yes, only with x though F = 10*sin(0.1*x). Still don't know how to figure this problem out.

If it moves dx along the x axis, how far does it move, approximately, in the y direction? What is that movement as a vector?
What is the corresponding vector form of the force?
How much work is done?

 

[Delta2]

It says the force is tangential to the path. Therefore the infinitesimal work dW is equal to the magnitude of the force multiplied by the infinitesimal length dr (no need to take vectors and dot products).Therefore $$dW=Fdr=F\sqrt{dy^2+dx^2}$$

Now replace $$dy=-\sin (0.1x) dx$$ in the above equation and then calculate $$W=\int dW$$

If you want to know where you are wrong, it is in the equation for the infinitesimal work $dW$. It is $$dW=F_xdx+F_ydy$$ and it is a mini trouble to calculate the components of the force $F_x,F_y$ from the data of the magnitude of the force and from being tangential to the given path.

 

[Delta2]

Almost. But it is not $\vec F\cdot \vec x$, it is $\vec F \cdot d\vec x$ ...

In order to be perfectly consistent with the notation already used, you should have said "not $\vec F\cdot \vec r$ but $\vec F\cdot d\vec r$" where $\vec r$ the position vector $\vec r=x\hat x+y\hat y$ and $d\vec r=dx\hat x+dy\hat y$.

 

[kuruman]

It seems to me that the integral can be set up quite easily by noting that $\vec F$ is in the same direction as the tangential element $d\vec s$ in which case $$dW=F~ds=F\sqrt{(dx)^2+(dy)^2}=Fdx\sqrt{1+\left(\frac{dy}{dx}\right)^2}.$$Doing the ensuing integral is a task in itself.

 

[annamal]

It seems to me that the integral can be set up quite easily by noting that $\vec F$ is in the same direction as the tangential element $d\vec s$ in which case $$dW=F~ds=F\sqrt{(dx)^2+(dy)^2}=Fdx\sqrt{1+\left(\frac{dy}{dx}\right)^2}.$$Doing the ensuing integral is a task in itself.

Yes that leads to the correct answer. Thank you!