Is My Calculation of Work Along a Defined Curve Correct?

AI Thread Summary
The discussion revolves around calculating work along a defined curve, specifically addressing a solution presented in LaTeX. The original poster seeks validation for their calculation, particularly Equation 1.11, and has provided a PDF for reference. Feedback indicates that the final result has incorrect units, prompting a clarification on the dimensions involved in the calculation. A key point of confusion arises regarding the interpretation of "m" in the denominator, which affects the dimensional analysis of work. The conversation emphasizes the importance of correctly understanding units in physics calculations.
Morle
Messages
2
Reaction score
0
Homework Statement
I would like to confirm whether my solution is correct.
Relevant Equations
vector line integral
I created a problem for myself, which I have documented in LaTeX. I hope it's acceptable that I included snapshots of my pages.

I would like to calculate the Work W along curve C1.
1727268611164.png


I have solved the problem as shown below. Is my solution correct (see Equation 1.11)?
1727268640213.png


The pictures I included are of low quality, so I've attached the main.pdf for your reference.
 

Attachments

Physics news on Phys.org
I am not opening unsolicited pdfs as a matter of principle.

Please note that the forum has LaTeX compatibility. You’ll have to change a couple of delimiters, but displaying equations written in LaTeX generally works quite well.
 
Your final result seems to have the wrong units.
But what's the point of doing this?
 
Hello Nasu, The goal is to improve my understanding by practicing vector line integral problems in physics, specifically those related to electrostatics. Why is the Unit of W not correct? q in [As], Q in [As], e0 in [As]/[Vm] so k in [As]*[Vm] and finally W in [As]*[Vm]/[m]=[VAs]=[Ws]=[Nm]=[J]. So the unit is correct, right?
Thanks and best regards.
 
What is the meaning of "m" in The denominator of your final result? It looks like "k" divided by some quantity in meters. Which does not have dimensions of work.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top