Potential Energy as a function of x

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Homework Help Overview

The discussion revolves around calculating potential energy as a function of position \( x \) based on a given force expression \( 8e^{-2x} \). The original poster attempts to derive the potential energy function starting from an initial condition of \( U = 5 \) at \( x = 0 \).

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration of the force function and the implications of definite versus indefinite integrals. There are questions about the integration limits and the constant of integration, \( C \), in the context of potential energy calculations.

Discussion Status

Some participants provide guidance on the integration process and the need to consider the constant of integration. There is an acknowledgment of the original poster's confusion regarding the correct form of the potential energy function, and attempts are made to clarify the steps involved.

Contextual Notes

There is mention of an initial condition that must be satisfied, and the discussion includes considerations about the limits of integration that were not initially applied. The participants are exploring different interpretations of the integration process and its impact on the final expression for potential energy.

KvnBushi
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[SOLVED] Potential Energy as a function of x

Homework Statement


Take U = 5 at x = 0 and calculate potential energy as a function of x, corresponding to the force:
[tex]8e^{-2x}[/tex]


Homework Equations


[tex]W_{net} = U_i - U_f[/tex]
[tex]W = \int_a^b F_x dx[/tex]



The Attempt at a Solution



[tex]\int 8e^{-2x} dx = -4e^{-2x} = W(x)[/tex]

[tex]W(x) = -4e^{-2x} = U_i(x) - U_f(x)[/tex]
[tex]U_f(x) = U_i(x) + 4e^{-2x}[/tex]
[tex]U_f(x) = 5 + 4e^{-2x}[/tex]

correct answer: U(x) = [tex]1 + 4e^{-2x}[/tex]

Any ideas how i went wrong?
 
Last edited:
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You were working with definite integral, I guess. when you integrated 8exp(-2x), you should have put the limits (x = 0 and x = x).
 
SOLVED

[tex]W(x) = \int 8e^{-2x} = -4e^{-2x} + C[/tex] ( I forgot the C earlier)
[tex]W(x) = U_i(x) - U_f(x)[/tex]
[tex]-4e^{-2x} + C = 5 - U_f(x)[/tex]
[tex]U_f(x) = 5 - 4e^{-2x} - C[/tex]

SOLVE FOR C

U(0) = 5 = 5 - 4(1) - C
C = 4

SOLVE FOR U(x)

[tex]U_f(x) = 5 - 4e^{-2x} - 4[/tex]

[tex]U_f(x) = 1 - 4e^{-2x}[/tex]
 
Sourabh N said:
You were working with definite integral, I guess. when you integrated 8exp(-2x), you should have put the limits (x = 0 and x = x).

I am going to try this way as well when I get back from eating. Cheers!
 

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