Comparing Work on Two Carts with Constant Force

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Homework Help Overview

The problem involves two carts of the same mass being accelerated by the same constant force on horizontal frictionless tracks. The force is applied to one cart for twice the duration compared to the other, leading to a discussion about the work done on each cart.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between time, distance, and work, referencing kinematic equations. Some express confusion about the reasoning behind the proposed answer of WA = 4 WB and seek clarification on how to derive this from the equations provided.

Discussion Status

There is an ongoing exploration of the kinematic equations and their relevance to the problem. Some participants are sharing insights and affirmations about the learning process, while others are questioning the interpretations and assumptions made regarding the work done on each cart.

Contextual Notes

Participants note the potential confusion arising from using kinematic equations in a context that may not have been clearly defined as relevant to the problem at hand. There is also mention of external resources that lack solutions, prompting the need for guidance in understanding the concepts involved.

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Homework Statement


The same constant force is used to accelerate two carts of the same mass, initially at rest, on horizontal frictionless tracks. The force is applied to cart A for twice as long a time as it is applied to cart B. The work the force does on A is WA; that on B is WB. Which statement is correct?

a.WA = WB.

b.WA = https://www.physicsforums.com/x-apple-ql-id://75E1D3CA-C6B4-441B-A5A4-36B84CE66FF0/x-apple-ql-magic/C3C64FB0-9E65-46A8-9150-27D9512B39A1.pdf WB.

c.WA = 2 WB.

d.WA = 4 WB.

e.WB = 2WA.

Homework Equations


W = F * d * cos(theta)
x= volt + 0.5 *a * t^2

The Attempt at a Solution


The answer is D, but I don't know where to start to get the answer d.
I thought since it has same mass and acceleration, twice of time will just make it twice of work.
Can anyone help me explaining this with equation??
 
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Oh oh I think I got it. Is it because
x = volt + 0.5at^2 and when I times 2 on t, the x will become 4x and it will eventually make 4 times more work right?
 
How do you figure that the answer is d? Where did you get that answer from?
 
Also those kinematic equations should have been posted in relevant equations if you are still using them in class. If you are doing kinematics, then use that equation.
 
Meesh said:
How do you figure that the answer is d? Where did you get that answer from?
Oh I was studying Energy part by myself and I downloaded a random one in the internet! They have questions with answers, but no solution so if I don't know how to solve it, I just have to ask here to get an advice.
 
Meesh said:
Also those kinematic equations should have been posted in relevant equations if you are still using them in class. If you are doing kinematics, then use that equation.
Haha yeah I'm not in kinematic right now and didn't know I was going to use that equation here. After thinking I kinda figured out I might able to use kinematic equation, but wasn't 100 percent sure if I could, so I wrote in reply. Sorry haha
 
Oh I understand. I think it is great that you are teaching yourself physics! I think the kinematic is your best bet in this case because it is the only way you can take velocity and acceleration into account. Good job. :)
 
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Meesh said:
Oh I understand. I think it is great that you are teaching yourself physics! I think the kinematic is your best bet in this case because it is the only way you can take velocity and acceleration into account. Good job. :)
Thank you! :)
 
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