Calculating Work Done by a Force on a Particle Moving Along the X-Axis

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The problem involves calculating the work done by a force, represented as kx^3, on a particle moving from position x1 to x2 along the x-axis. The initial approach used the formula W=Fd, but it was recognized that the force varies with position. By taking the anti-derivative of kx^3, the correct expression for work done was derived as (kx2^4)/4 - (kx1^4)/4, simplifying to k/4 (x2^4 - x1^4). This method accurately accounts for the changing force as the particle moves. The solution effectively demonstrates the application of calculus in physics.
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Homework Statement



A particle moves only along the x-axis, and is subject to a force towards the origin of magnitude kx^3. if the particle moves from x1 to x2 how much work does this force do on it? (consider the case x1<x2)

Homework Equations



W=Fd

The Attempt at a Solution



kx^3(x2-x1)
 
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I just noticed that the force would change as the particle moved along the x-axis. Taking the anti-derivative of kx^3 from x1 to x2 I got (kx2^4)/4 - (kx1^4)/4 = k/4 (x2^4-x1^4)
 
Looking good to me!
 
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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