Calculating Work from Force as a Function of Time Graphs

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To calculate work from a force as a function of time (F(t)) graph, one must integrate the force over time to determine the change in momentum. This change in momentum can then be used to find kinetic energy, but care must be taken as the relationship between momentum and work is not linear. Simply applying the formula for work based on momentum changes can lead to inaccuracies if the initial momentum is not considered. The correct approach involves calculating the initial and final momentum to accurately determine the work done. Understanding this relationship is crucial for solving problems involving F(t) graphs effectively.
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I'm doing problems where I have f(t) (force as a function of time) graphs and f(x) (force as function of distance) graphs.

I am just having one problem, how do I find work for a force as a function of time graph?

Im not sure how work relates at all to force as a function of time. I know that the area under the graph of F(x) curve is work, but in the case of F(t), I have no clue how I would solve it.

Oh, also, I have no F(t) = whatever, I merely have a graph and am doing graphical analysis.

Any help in how to solve for work from an F(t) graph would be much appreciated.

-TwiztidMxcn
 
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F=dp/dt, where p is momentum. The kinetic energy is equal to p^2/2m. So integrating F(t) over t will give momentum, from which you can get the KE. There is a slight problem though. If you just naively assume that a momentum change corresponds directly to the work done by \Delta W = (\Delta p)^2/2m you'll get into trouble because this equation is not linear in p. That is, \Delta W_{ac} = (\Delta p_{ac})^2/2m = (p_c-p_a)^2/2m[/tex] and \Delta W_{ac} = \Delta W_{ab}+\Delta W_{bc} = (p_b-p_a)^2/2m + (p_c-p_b)^2/2m[/tex] are incompatible. So you need to know the initial momentum and use: \Delta W = E_{f} - E_{0} = p_f^2/2m - p_0^2/2m.
 
you, my friend, i thank you for your help
 
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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