Calculating Work from Force as a Function of Time Graphs

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SUMMARY

This discussion focuses on calculating work from force as a function of time (F(t)) graphs. The key insight is that integrating F(t) over time yields momentum, which can then be used to determine kinetic energy (KE) using the formula KE = p²/2m. However, it is crucial to account for the initial momentum to avoid inaccuracies in calculating work, as the relationship between momentum and work is not linear. The correct approach involves using the equation ΔW = E_f - E_0 = p_f²/2m - p_0²/2m.

PREREQUISITES
  • Understanding of force as a function of time (F(t)) and distance (F(x)) graphs
  • Knowledge of momentum and its relation to force (F = dp/dt)
  • Familiarity with kinetic energy calculations (KE = p²/2m)
  • Basic integration techniques for graphical analysis
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  • Study the principles of graphical analysis for force-time graphs
  • Learn about the relationship between momentum and work in physics
  • Explore advanced integration techniques for calculating work from non-linear functions
  • Review examples of calculating kinetic energy from momentum changes
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Students and professionals in physics, particularly those studying mechanics, as well as educators looking to enhance their understanding of work-energy principles in relation to force-time graphs.

twiztidmxcn
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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 [itex]\Delta W = (\Delta p)^2/2m[/itex] you'll get into trouble because this equation is not linear in p. That is, [itex]\Delta W_{ac} = (\Delta p_{ac})^2/2m = (p_c-p_a)^2/2m[/tex] and [itex]\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: [itex]\Delta W = E_{f} - E_{0} = p_f^2/2m - p_0^2/2m[/itex].[/itex][/itex]
 
you, my friend, i thank you for your help
 

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