Comparing Work on Two Carts with Constant Force

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Two carts of the same mass are accelerated by the same constant force on frictionless tracks, with cart A receiving the force for twice the time of cart B. The discussion centers on determining the work done on each cart, leading to the conclusion that the work on cart A (WA) is four times that of cart B (WB), making the correct answer D (WA = 4 WB). The reasoning involves using kinematic equations, specifically how distance (x) changes with time when acceleration is constant. As time doubles, the distance traveled by cart A increases by a factor of four, resulting in greater work done. Understanding these relationships is crucial for solving similar physics problems effectively.
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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= vot + 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 = Vot + 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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