WOrk done by a variable force (in two dimension)

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To find the work done by a variable force in two dimensions, it is essential to consider the path taken between the initial and final points, as different paths yield different results. The force is given by F = ax^2 i + b y^3 j, and the work done can be calculated using the definition of infinitesimal work, dW = vector F dot vector dr. The discussion emphasizes that a straight line path should be used to derive the vector expression for dr based on the endpoints (x1, y1) and (x2, y2). Participants express uncertainty about whether the initial approach adequately accounts for the path's influence on the work done. The importance of correctly applying the path in calculations is highlighted throughout the conversation.
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Homework Statement



How to find the work done by a variable force in (two dimension)

When
F = ax^2 i + b y^3 j

If a subject move from (x1,y1) to (x2, y2)

Homework Equations



F = dW/dr

The Attempt at a Solution



I tried to solve them separately by x-direction and y-direction, and then I added them up
like Wx + Wy = W
Am I right?
 
Last edited:
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In this problem the path taken from (x1,y1) to (x2, y2) is important, different paths taken give different results. In this problem I assume the path is a straight line. If you solve the problem as you suggested I'm not sure you took into account the path taken? Can you scan your work and post it?

Hopefully others will help.
 
Spinnor said:
In this problem the path taken from (x1,y1) to (x2, y2) is important, different paths taken give different results. In this problem I assume the path is a straight line. If you solve the problem as you suggested I'm not sure you took into account the path taken? Can you scan your work and post it?

Hopefully others will help.


https://www.physicsforums.com/attachment.php?attachmentid=51764&stc=1&d=1349922834

I hope I solve it right
 

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I don't think what you did is right. You did not take the path into account only the end points. Start from the diffinition of infitesimal work, dW = vector F dot vector dr and plug and chug. You must come up with a vector expression for dr for the straight line path. You can do this as you know the endpoints.
 
Spinnor said:
I don't think what you did is right. You did not take the path into account only the end points. Start from the diffinition of infitesimal work, dW = vector F dot vector dr and plug and chug. You must come up with a vector expression for dr for the straight line path. You can do this as you know the endpoints.

as I stated before, the particle goes from (x1,y1) to (x2, y2)
so the end point is (x2, y2)

So, am I doing this right?

P/s: however, the path between the two points is straight line
 
I don't think so. Hopefully others will confirm.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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