How Do You Apply Calculus to Determine Force in Physics?

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SUMMARY

The discussion focuses on applying calculus to determine force in physics, specifically through the relationship between potential energy (PE) and force (F). The potential energy function provided is 5x² + 3x + 7, and the derivative of this function, dPE/dx, yields the force as a function of x. The participants confirm that the negative derivative of potential energy corresponds to force, aligning with the equation -PE = Fdcos. Understanding these relationships is crucial for solving physics problems involving energy and force.

PREREQUISITES
  • Understanding of calculus derivatives
  • Familiarity with potential energy concepts
  • Knowledge of force and energy units (Newtons and Joules)
  • Basic principles of physics related to work and energy
NEXT STEPS
  • Study the relationship between potential energy and force in physics
  • Learn how to apply derivatives in physics problems
  • Explore the concept of work done by a force over a distance
  • Investigate the integration of force functions to find energy
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High school physics students, calculus learners, and anyone interested in the application of calculus in physics, particularly in understanding force and energy relationships.

glennpagano44
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Right now I am in AP physics in High School, while also taking calculus and I need some assitance on when and how to apply the calculus (once I know to do calculus I can go from there)

Example

If Potential Energy is 5x^2+3x+7 what function is the force.

- I know you have to take derivative of the PE equation because my teacher told me.

Also because -PE=Fdcos the you get -PE/d=F. From there how do you know to put dPE/dd =F. Also how do you know not to integrate.

Thanks a lot
 
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You can look at the units if nothing else.

Forces are in Newtons N and energy is in Joules which are N-m

If you have Joules then the rate of changes in Joules by x should yield Newtons.

If you have Newtons exerted over a distance x, the integral of F⋅X (F as a function of x) yields you Joules - energy.

In your problem you're wanting to know the F as a function of x and you are given the energy potentials PE as a function of x. Hence dPE/dx should yield you your Force as a function of x.
 
Thanks a lot LowlyPion that helped out. Anyone else have any input?
 

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