Why is Work Calculated as Fdcosθ Even with Gravity?

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Work is defined as the dot product of force and displacement, represented by the equation Fdcosθ, where θ is the angle between the force and displacement vectors. This definition applies universally, including when calculating work done by gravity. Work is a scalar quantity, meaning it has magnitude but no direction. The cosine factor accounts for the angle between the force applied and the direction of movement, which is crucial for accurate calculations. Understanding this concept is essential for grasping the principles of mechanical work.
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Homework Statement



This is conceptual question...Is work a scalar or vector quantity? Is it force times distance or displacement?

And it just Fdcostheta by definition of the dot product. Why is it cos theta even when you calculate work done by gravity?

Homework Equations



Fdcos theta

The Attempt at a Solution



Just need conceptual clarification.
 
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hello,

just check http://en.wikipedia.org/wiki/Mechanical_work"

I suppose,it helps.

Dot product of two vectors is scalar :smile:
 
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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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