To find work done by a variable force

[Rhdjfgjgj]

Homework Statement

To find work done by a variable force

Relevant Equations

W=integral(F.dr)

Guys, please look at the following question

I am aware of how to take a dot product of 2 vectors whose i,j,k components are given. My issue is with both integrals. I feel it should be '-dx ' rather than dx because x is decreasing , same as in '-dy'.and I'm taught that if x is decreasing then we should take '-dx' , but why is it not so . Please tell me why . I do understand that my questions are a little too fundamental but please answer them

 

[berkeman]

Your image is very hard to read, even when I open it in a new tab and zoom in. Could you please try to post a better image (or better, use a flatbed scanner)? You could also type the math into the forum using LaTeX, which is the preferred way to post math equations here at PF. (see the LaTeX Guide link below the Edit window)

 

[Rhdjfgjgj]

Your image is very hard to read, even when I open it in a new tab and zoom in. Could you please try to post a better image (or better, use a flatbed scanner)? You could also type the math into the forum using LaTeX, which is the preferred way to post math equations here at PF. (see the LaTeX Guide link below the Edit window)

Here's a better one. In addition , my doubt is with both the integrals

 

[erobz]

Something seems strange to me. How does a particle move from A to B by applying $\vec F$, which always points in the opposite directions in which it needs to travel. To me it seems like some other force is moving the particle from A to B and $\vec F$ is resisting it?

 

[Steve4Physics]

My issue is with both integrals. I feel it should be '-dx ' rather than dx because x is decreasing , same as in '-dy'.and I'm taught that if x is decreasing then we should take '-dx' , but why is it not so . Please tell me why . I do understand that my questions are a little too fundamental but please answer them

Note that the diagram is misleading. The usual convention is that the +x direction is to the right, and the +y direction is up. So the displacement vector AB should be drawn pointing down and to the left (225º w.r.t. the +x-axis).

(Alternatively, AB could be as drawn but with the xy axes suitably rotated.)

We use 'dx' rather than '-dx' because the direction of movement is taken into account by the integration limits. Note that for example $\int_a^b f(x)dx = -\int_b^a f(x)dx$.

The sign of the x-contribution to work is determined by the initial and final values of x. Similarly for y.

 

[Steve4Physics]

Something seems strange to me. How does a particle move from A to B by applying $\vec F$, which always points in the opposite directions in which it needs to travel. To me it seems like some other force is moving the particle from A to B and $\vec F$ is resisting it?

For example $\vec F$ could be friction slowing the (initially moving) particle down.

Or there may be other forces involved - but we are only calculating $\vec F$'s contribution.

 

[PeroK]

Here's a better one. In addition , my doubt is with both the integralsView attachment 333892

You have a choice in these cases. You can use $dx$ and have the integral go from 1 to 0. Or, you can use $-dx$ and have the integral go from 0 to 1. You get the same answer either way.

Note that the work done is negative, which highlights that the particle was not accelerated from rest between these two points. It must have had an initial velocity in a direction roughly opposite the force. We can let @erobz figure out what it must have been!

 

[erobz]

For example $\vec F$ could be friction slowing the (initially moving) particle down.

Or there may be other forces involved - but we are only calculating $\vec F$'s contribution.

I get that. Please see the problem statement:

An object is displaced from point A(1,2,) to B(0,1) by applying force $\vec F = x \boldsymbol i + 2y \boldsymbol j$

 

[Steve4Physics]

I get that. Please see the problem statement:

Sorry. Should have read the whole problem-statement.

 

[erobz]

Sorry. Should have read the whole problem-statement.

So you agree that while the intent is what we suspect it to be (given the solution), the problem statement is sloppily composed?

 

[PeroK]

So you agree that while the intent is likely what we think it to be, the problem statement is sloppily composed?

There are quite a few homework problems where the question setter hasn't thought through the problem fully and is simply looking for a plug-and-chug answer. In this case, I agree the question should have been phrased differently.

 

[PeroK]

There's also no acknowledgement that the units of force and displacement are different.

 

[Rhdjfgjgj]

Guys actually what I mean is as I go from a farther to a nearer point wrt reference frame, x decreases, and elemental decrease is given as '-dx' as I go from 1 to 0 . What you said ' you can change limits to 0 to 1 and have -dx work I case of vectors as there is an additional cos(theta) sign that compensates for that. But here these aren't vectors, the integrand is a component

You have a choice in these cases. You can use $dx$ and have the integral go from 1 to 0. Or, you can use $-dx$ and have the integral go from 0 to 1. You get the same answer either way.

Note that the work done is negative, which highlights that the particle was not accelerated from rest between these two points. It must have had an initial velocity in a direction roughly opposite the force. We can let @erobz figure out what it must have been!

 

[Steve4Physics]

So you agree that while the intent is what we suspect it to be (given the solution), the problem statement is sloppily composed?

Yes. It's a bit of mess!

 

[PeroK]

The work done is negative because the direction of the force is opposite to the displacement, in both the $x$ and $y$ directions. So, however you set up the integral you must get a negative answer. You can do that in two ways.

Alternatively, change your coordinate system so that the displacement is in the positive x and y directions, if you don't like negative displacements.

 

[Rhdjfgjgj]

Im again repeating that I don't want to interchange the limits. I mean mathematically that since x and y is decreasing they have to be -dx and -dy . I don't want to change the limit.

 

[erobz]

Im again repeating that I don't want to interchange the limits. I mean mathematically that since x and y is decreasing they have to be -dx and -dy . I don't want to change the limit.

Start with:

$$ dW = ( x ~\hat{\text{i}} +2y ~\hat{\text{j}} ) \cdot ( -dx ~\hat{\text{i}} -dy ~\hat{\text{j}} ) $$

If it pleases you.

EDIT: Wait, I see this doesn't work out either. You would get positive work.

 

[Rhdjfgjgj]

Start with:

$$ dW = ( x ~\hat{\text{i}} +2y ~\hat{\text{j}} ) \cdot ( -dx ~\hat{\text{i}} -dy ~\hat{\text{j}} ) $$

If it pleases you.

If I use this I get the negative of rhe original answer. You may better understand my doubt if you look at
https://www.physicsforums.com/threa...-conducting-uniformly-charged-sphere.1055763/

 

[PeroK]

Im again repeating that I don't want to interchange the limits. I mean mathematically that since x and y is decreasing they have to be -dx and -dy .

This is wrong. $dx$ and $dy$ relate to your coordinate system. Not to any given displacement.

Your argument is like saying that if $x = -1$, then you need to use $-x$. The differentials $dx$ and $dy$ can be positive or negative.

PS they represent the change in $x$, whether positive or negative. It's the same for the finite $\Delta x$. This is the change in $x$ whether positive or negative.

 

[PeroK]

If you integrate $x$ from 0 to 1, then $dx$ is positive. If you integrate from 1 to 0, then $dx$ is negative. And vice versa for $-dx$.

 

[vela]

If I use this I get the negative of rhe original answer. You may better understand my doubt if you look at
https://www.physicsforums.com/threa...-conducting-uniformly-charged-sphere.1055763/

It has to do with how you compute the dot product. In the other thread, you used $\vec E \cdot d\vec r = \lvert \vec E \rvert \lvert d\vec r \rvert \cos \theta$. Both $\lvert \vec E \rvert$ and $\lvert d\vec r \rvert$ are, by definition, positive quantities. Since you were integrating along a path in the radial direction, you had $d\vec r = dr\,\hat r$. It followed that $\lvert d\vec r \rvert = \lvert dr \rvert$. If $dr<0$ (determined by the limits of the integral), then $\lvert dr \rvert = -dr$. If $dr>0$, then you replace $\lvert dr\rvert$ with $+dr$. You're not throwing in signs willy nilly. You put in the minus sign because of the basic definition of the absolute value.

In this problem, you're calculating the dot product using the components of the two vectors. There are no absolute values present. Since $\vec r$ is defined as $\vec r = x \,\hat i + y\,\hat j$, it follows that $d\vec r = dx\,\hat i + dy\,\hat j$. The signs automatically work themselves out.

 

[Mark44]

The differentials dx and dy can be positive or negative.

Related to this is the mistake that some people make by thinking that x must be positive while -x must be negative. I suspect that the OP falls into this category, based on his previous thread about the mirror image of a particle.