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.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
Back
Top